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University of Missouri- Columbia
Physics
Physics 2750
Wexler
Solutions_to_Univ._Physics...pdf
Solutions_to_Univ._Physics...pdf
Physics 2750
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University of Missouri- Columbia
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Gabriel B.
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2012-02-09
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1-1 UNITS, PHYSICAL QUANTITIES AND VECTORS 1.1. IDENTIFY: Convert units from mi to km and from km to ft. SET UP: 1 in. 2.54 cm= , 1 km = 1000 m , 12 in. 1 ft= , 1 mi = 5280 ft . EXECUTE: (a) 2 3 5280 ft 12 in. 2.54 cm 1 m 1 km 1.00 mi (1.00 mi) 1.61 km 1 mi 1 ft 1 in. 10 cm 10 m ? ?? ?? ?? ?? ?= =? ?? ?? ?? ?? ?? ?? ?? ?? ?? ? (b) 3 2 310 m 10 cm 1 in. 1 ft1.00 km (1.00 km) 3.28 10 ft 1 km 1 m 2.54 cm 12 in. ? ?? ?? ?? ?= = ×? ?? ?? ?? ?? ?? ?? ?? ? EVALUATE: A mile is a greater distance than a kilometer. There are 5280 ft in a mile but only 3280 ft in a km. 1.2. IDENTIFY: Convert volume units from L to 3in. . SET UP: 31 L 1000 cm= . 1 in. 2.54 cm= EXECUTE: 33 31000 cm 1 in.0.473 L 28.9 in. . 1 L 2.54 cm ? ? ? ?× × =? ? ? ?? ?? ? EVALUATE: 31 in. is greater than 31 cm , so the volume in 3in. is a smaller number than the volume in 3cm , which is 3473 cm . 1.3. IDENTIFY: We know the speed of light in m/s. /t d v= . Convert 1.00 ft to m and t from s to ns. SET UP: The speed of light is 83.00 10 m/sv = × . 1 ft 0.3048 m= . 91 s 10 ns= . EXECUTE: 98 0.3048 m 1.02 10 s 1.02 ns 3.00 10 m/s t ?= = × =× EVALUATE: In 1.00 s light travels 8 5 53.00 10 m 3.00 10 km 1.86 10 mi× = × = × . 1.4. IDENTIFY: Convert the units from g to kg and from 3cm to 3m . SET UP: 1 kg 1000 g= . 1 m 1000 cm= . EXECUTE: 3 4 3 3 g 1 kg 100 cm kg 11.3 1.13 10 cm 1000 g 1 m m ? ? ? ?× × = ×? ? ? ?? ? ? ? EVALUATE: The ratio that converts cm to m is cubed, because we need to convert 3cm to 3m . 1.5. IDENTIFY: Convert volume units from 3in. to L. SET UP: 31 L 1000 cm= . 1 in. 2.54 cm= . EXECUTE: ( ) ( ) ( ) 33 3327 in. 2.54 cm in. 1 L 1000 cm 5.36 L× × = EVALUATE: The volume is 35360 cm . 31 cm is less than 31 in. , so the volume in 3cm is a larger number than the volume in 3in. . 1.6. IDENTIFY: Convert 2ft to 2m and then to hectares. SET UP: 4 21.00 hectare 1.00 10 m= × . 1 ft 0.3048 m= . EXECUTE: The area is 22 4 2 43,600 ft 0.3048 m 1.00 hectare (12.0 acres) 4.86 hectares 1 acre 1.00 ft 1.00 10 m ? ?? ? ? ? =? ?? ? ? ?×? ? ? ?? ? . EVALUATE: Since 1 ft 0.3048 m= , 2 2 21 ft (0.3048) m= . 1.7. IDENTIFY: Convert seconds to years. SET UP: 91 billion seconds 1 10 s= × . 1 day 24 h= . 1 h 3600 s= . EXECUTE: ( )9 1 h 1 day 1 y1.00 billion seconds 1.00 10 s 31.7 y 3600 s 24 h 365 days ? ?? ?? ?= × =? ?? ?? ?? ?? ?? ? . 1 1-2 Chapter 1 EVALUATE: The conversion 71 y 3.156 10 s= × assumes 1 y 365.24 d= , which is the average for one extra day every four years, in leap years. The problem says instead to assume a 365-day year. 1.8. IDENTIFY: Apply the given conversion factors. SET UP: 1 furlong 0.1250 mi and 1 fortnight 14 days.= = 1 day 24 h.= EXECUTE: ( ) 0.125 mi 1 fortnight 1 day180,000 furlongs fortnight 67 mi/h 1 furlong 14 days 24 h ? ?? ?? ? =? ?? ?? ?? ?? ?? ? EVALUATE: A furlong is less than a mile and a fortnight is many hours, so the speed limit in mph is a much smaller number. 1.9. IDENTIFY: Convert miles/gallon to km/L. SET UP: 1 mi 1.609 km= . 1 gallon 3.788 L.= EXECUTE: (a) 1.609 km 1 gallon55.0 miles/gallon (55.0 miles/gallon) 23.4 km/L 1 mi 3.788 L ? ?? ?= =? ?? ?? ?? ? . (b) The volume of gas required is 1500 km 64.1 L 23.4 km/L = . 64.1 L 1.4 tanks 45 L/tank = . EVALUATE: 1 mi/gal 0.425 km/L = . A km is very roughly half a mile and there are roughly 4 liters in a gallon, so 241 mi/gal km/L? , which is roughly our result. 1.10. IDENTIFY: Convert units. SET UP: Use the unit conversions given in the problem. Also, 100 cm 1 m= and 1000 g 1 kg= . EXECUTE: (a) mi 1h 5280 ft ft60 88 h 3600s 1mi s ? ? ? ?? ? =? ? ? ?? ?? ? ? ? ? ? (b) 2 2 ft 30.48cm 1 m m 32 9.8 s 1ft 100 cm s ? ?? ? ? ? =? ?? ? ? ?? ? ? ?? ? (c) 3 3 3 3 g 100 cm 1 kg kg 1.0 10 cm 1 m 1000 g m ? ?? ? ? ? =? ?? ? ? ?? ? ? ? ? ? EVALUATE: The relations 60 mi/h 88 ft/s= and 3 3 31 g/cm 10 kg/m= are exact. The relation 2 232 ft/s 9.8 m/s= is accurate to only two significant figures. 1.11. IDENTIFY: We know the density and mass; thus we can find the volume using the relation density mass/volume /m V= = . The radius is then found from the volume equation for a sphere and the result for the volume. SET UP: 3Density 19.5 g/cm= and critical 60.0 kg.m = For a sphere 343V r?= . EXECUTE: 3critical 360.0 kg 1000 g/ density 3080 cm19.5 g/cm 1.0 kgV m ? ?? ?= = =? ?? ?? ?? ? . ( )33 33 3 3080 cm 9.0 cm 4 4 V r ? ?= = = . EVALUATE: The density is very large, so the 130 pound sphere is small in size. 1.12. IDENTIFY: Use your calculator to display 710? × . Compare that number to the number of seconds in a year. SET UP: 1 yr 365.24 days,= 1 day 24 h,= and 1 h 3600 s.= EXECUTE: 724 h 3600 s(365.24 days/1 yr) 3.15567... 10 s 1 day 1 h ? ?? ? = ×? ?? ?? ?? ? ; 7 710 s 3.14159... 10 s? × = × The approximate expression is accurate to two significant figures. EVALUATE: The close agreement is a numerical accident. 1.13. IDENTIFY: The percent error is the error divided by the quantity. SET UP: The distance from Berlin to Paris is given to the nearest 10 km. EXECUTE: (a) 3310 m 1.1 10 %.890 10 m ?= ×× (b) Since the distance was given as 890 km, the total distance should be 890,000 meters. We know the total distance to only three significant figures. EVALUATE: In this case a very small percentage error has disastrous consequences. 1.14. IDENTIFY: When numbers are multiplied or divided, the number of significant figures in the result can be no greater than in the factor with the fewest significant figures. When we add or subtract numbers it is the location of the decimal that matters. Units, Physical Quantities and Vectors 1-3 SET UP: 12 mm has two significant figures and 5.98 mm has three significant figures. EXECUTE: (a) ( ) ( ) 212 mm 5.98 mm 72 mm× = (two significant figures) (b) 5.98 mm 0.50 12 mm = (also two significant figures) (c) 36 mm (to the nearest millimeter) (d) 6 mm (e) 2.0 (two significant figures) EVALUATE: The length of the rectangle is known only to the nearest mm, so the answers in parts (c) and (d) are known only to the nearest mm. 1.15. IDENTIFY and SET UP: In each case, estimate the precision of the measurement. EXECUTE: (a) If a meter stick can measure to the nearest millimeter, the error will be about 0.13%. (b) If the chemical balance can measure to the nearest milligram, the error will be about 38.3 10 %.?× (c) If a handheld stopwatch (as opposed to electric timing devices) can measure to the nearest tenth of a second, the error will be about 22.8 10 %.?× EVALUATE: The percent errors are those due only to the limit of precision of the measurement. 1.16. IDENTIFY: Use the extreme values in the piece?s length and width to find the uncertainty in the area. SET UP: The length could be as large as 5.11 cm and the width could be as large as 1.91 cm. EXECUTE: The area is 9.69 ± 0.07 cm2. The fractional uncertainty in the area is 2 2 0.07 cm 0.72%, 9.69 cm = and the fractional uncertainties in the length and width are 0.01 cm 0.20% 5.10 cm = and 0.01 cm 0.53%. 1.9 cm = The sum of these fractional uncertainties is 0.20% 0.53% 0.73%+ = , in agreement with the fractional uncertainty in the area. EVALUATE: The fractional uncertainty in a product of numbers is greater than the fractional uncertainty in any of the individual numbers. 1.17. IDENTIFY: Calculate the average volume and diameter and the uncertainty in these quantities. SET UP: Using the extreme values of the input data gives us the largest and smallest values of the target variables and from these we get the uncertainty. EXECUTE: (a) The volume of a disk of diameter d and thickness t is 2( / 2) .V d t?= The average volume is 2 3(8.50 cm/2) (0.50 cm) 2.837 cm .V ?= = But t is given to only two significant figures so the answer should be expressed to two significant figures: 32.8 cm .V = We can find the uncertainty in the volume as follows. The volume could be as large as 2 3(8.52 cm/2) (0.055 cm) 3.1 cm ,V ?= = which is 30.3 cm larger than the average value. The volume could be as small as 2 3(8.52 cm/2) (0.045 cm) 2.5 cm ,V ?= = which is 30.3 cm smaller than the average value. The uncertainty is 30.3 cm ,± and we express the volume as 32.8 0.3 cm .V = ± (b) The ratio of the average diameter to the average thickness is 8.50 cm/0.050 cm 170.= By taking the largest possible value of the diameter and the smallest possible thickness we get the largest possible value for this ratio: 8.52 cm/0.045 cm 190.= The smallest possible value of the ratio is 8.48/ 0.055 150.= Thus the uncertainty is 20± and we write the ratio as 170 20.± EVALUATE: The thickness is uncertain by 10% and the percentage uncertainty in the diameter is much less, so the percentage uncertainty in the volume and in the ratio should be about 10%. 1.18. IDENTIFY: Estimate the number of people and then use the estimates given in the problem to calculate the number of gallons. SET UP: Estimate 83 10× people, so 82 10× cars. EXECUTE: ( ) ( )Number of cars miles/car day / mi/gal gallons/day× = ( ) ( )8 82 10 cars 10000 mi/yr/car 1 yr/365 days / 20 mi/gal 3 10 gal/day× × × = × EVALUATE: The number of gallons of gas used each day approximately equals the population of the U.S. 1.19. IDENTIFY: Express 200 kg in pounds. Express each of 200 m, 200 cm and 200 mm in inches. Express 200 months in years. SET UP: A mass of 1 kg is equivalent to a weight of about 2.2 lbs. 1 in. 2.54 cm= . 1 y 12 months= . EXECUTE: (a) 200 kg is a weight of 440 lb. This is much larger than the typical weight of a man. (b) 4 3 1 in. 200 m (2.00 10 cm) 7.9 10 inches 2.54 cm ? ?= × = ×? ?? ? . This is much greater than the height of a person. (c) 200 cm 2.00 m 79 inches 6.6 ft= = = . Some people are this tall, but not an ordinary man. 1-4 Chapter 1 (d) 200 mm 0.200 m 7.9 inches= = . This is much too short. (e) 1 y 200 months (200 mon) 17 y 12 mon ? ?= =? ?? ? . This is the age of a teenager; a middle-aged man is much older than this. EVALUATE: None are plausible. When specifying the value of a measured quantity it is essential to give the units in which it is being expressed. 1.20. IDENTIFY: The number of kernels can be calculated as bottle kernel/ .N V V= SET UP: Based on an Internet search, Iowan corn farmers use a sieve having a hole size of 0.3125 in. ? 8 mm to remove kernel fragments. Therefore estimate the average kernel length as 10 mm, the width as 6 mm and the depth as 3 mm. We must also apply the conversion factors 31 L 1000 cm and 1 cm 10 mm.= = EXECUTE: The volume of the kernel is: ( )( )( ) 3kernel 10 mm 6 mm 3 mm 180 mmV = = . The bottle?s volume is: ( ) ( ) ( ) ( ) ( )3 33 6 3 bottle 2.0 L 1000 cm 1.0 L 10 mm 1.0 cm 2.0 10 mmV ? ?? ?= = ×? ? ? ? . The number of kernels is then ( ) ( )6 3 3 kernels bottle kernels/ 2.0 10 mm 180 mm 11,000 kernelsN V V= ? × = . EVALUATE: This estimate is highly dependent upon your estimate of the kernel dimensions. And since these dimensions vary amongst the different available types of corn, acceptable answers could range from 6,500 to 20,000. 1.21. IDENTIFY: Estimate the number of pages and the number of words per page. SET UP: Assuming the two-volume edition, there are approximately a thousand pages, and each page has between 500 and a thousand words (counting captions and the smaller print, such as the end-of-chapter exercises and problems). EXECUTE: An estimate for the number of words is about 610 . EVALUATE: We can expect that this estimate is accurate to within a factor of 10. 1.22. IDENTIFY: Approximate the number of breaths per minute. Convert minutes to years and 3cm to 3m to find the volume in 3m breathed in a year. SET UP: Assume 10 breaths/min . 524 h 60 min1 y (365 d) 5.3 10 min 1 d 1 h ? ?? ?= = ×? ?? ?? ?? ? . 210 cm 1 m= so 6 3 310 cm 1 m= . The volume of a sphere is 3 34 13 6V r d? ?= = , where r is the radius and d is the diameter. Don?t forget to account for four astronauts. EXECUTE: (a) The volume is 5 6 3 4 35.3 10 min(4)(10 breaths/min)(500 10 m ) 1 10 m / yr 1 y ? ? ?×× = ×? ?? ? . (b) 1/ 31/ 3 4 36 6[1 10 m ] 27 m V d ? ? ? ?×? ?= = =? ?? ?? ? ? ? EVALUATE: Our estimate assumes that each 3cm of air is breathed in only once, where in reality not all the oxygen is absorbed from the air in each breath. Therefore, a somewhat smaller volume would actually be required. 1.23. IDENTIFY: Estimate the number of blinks per minute. Convert minutes to years. Estimate the typical lifetime in years. SET UP: Estimate that we blink 10 times per minute. 1 y 365 days= . 1 day 24 h= , 1 h 60 min= . Use 80 years for the lifetime. EXECUTE: The number of blinks is 860 min 24 h 365 days(10 per min) (80 y/lifetime) 4 10 1 h 1 day 1 y ? ?? ?? ? = ×? ?? ?? ?? ?? ?? ? EVALUATE: Our estimate of the number of blinks per minute can be off by a factor of two but our calculation is surely accurate to a power of 10. 1.24. IDENTIFY: Estimate the number of beats per minute and the duration of a lifetime. The volume of blood pumped during this interval is then the volume per beat multiplied by the total beats. SET UP: An average middle-aged (40 year-old) adult at rest has a heart rate of roughly 75 beats per minute. To calculate the number of beats in a lifetime, use the current average lifespan of 80 years. EXECUTE: ( ) 9beats 60 min 24 h 365 days 80 yr75 beats/min 3 10 beats/lifespan1 h 1 day yr lifespanN ? ?? ?? ?? ?= = ×? ?? ?? ?? ?? ?? ?? ?? ? ( ) 93 7 blood 3 1 L 1 gal 3 10 beats 50 cm /beat 4 10 gal/lifespan 1000 cm 3.788 L lifespan V ? ?×? ?? ?= = ×? ?? ?? ?? ?? ?? ? EVALUATE: This is a very large volume. Units, Physical Quantities and Vectors 1-5 1.25. IDENTIFY: Estimation problem SET UP: Estimate that the pile is 18 in. 18 in. 5 ft 8 in..× × Use the density of gold to calculate the mass of gold in the pile and from this calculate the dollar value. EXECUTE: The volume of gold in the pile is 318 in. 18 in. 68 in. 22,000 in. .V = × × = Convert to 3cm : 3 3 3 5 322,000 in. (1000 cm / 61.02 in. ) 3.6 10 cm .V = = × The density of gold is 319.3 g/cm , so the mass of this volume of gold is 3 5 3 6(19.3 g/cm )(3.6 10 cm ) 7 10 g.m = × = × The monetary value of one gram is $10, so the gold has a value of 6 7($10/ gram)(7 10 grams) $7 10 ,× = × or about 6$100 10× (one hundred million dollars). EVALUATE: This is quite a large pile of gold, so such a large monetary value is reasonable. 1.26. IDENTIFY: Estimate the diameter of a drop and from that calculate the volume of a drop, in 3m . Convert 3m to L. SET UP: Estimate the diameter of a drop to be 2 mmd = . The volume of a spherical drop is 3 34 13 6V r d? ?= = . 3 310 cm 1 L= . EXECUTE: 3 3 316 (0.2 cm) 4 10 cmV ? ?= = × . The number of drops in 1.0 L is 3 5 3 3 1000 cm 2 10 4 10 cm? = ×× EVALUATE: Since 3V d? , if our estimate of the diameter of a drop is off by a factor of 2 then our estimate of the number of drops is off by a factor of 8. 1.27. IDENTIFY: Estimate the number of students and the average number of pizzas eaten by each student in a school year. SET UP: Assume a school of thousand students, each of whom averages ten pizzas a year (perhaps an underestimate) EXECUTE: They eat a total of 104 pizzas. EVALUATE: The same answer applies to a school of 250 students averaging 40 pizzas a year each. 1.28. IDENTIFY: The number of bills is the distance to the moon divided by the thickness of one bill. SET UP: Estimate the thickness of a dollar bills by measuring a short stack, say ten, and dividing the measurement by the total number of bills. I obtain a thickness of roughly 1 mm. From Appendix F, the distance from the earth to the moon is 83.8 10 m.× EXECUTE: 8 3 12 12 bills 3.8 10 m 10 mm 3.8 10 bills 4 10 bills 0.1 mm/bill 1 m N ? ?? ?×= = × ? ×? ?? ?? ?? ? EVALUATE: This answer represents 4 trillion dollars! The cost of a single space shuttle mission in 2005 is significantly less ? roughly 1 billion dollars. 1.29. IDENTIFY: The cost would equal the number of dollar bills required; the surface area of the U.S. divided by the surface area of a single dollar bill. SET UP: By drawing a rectangle on a map of the U.S., the approximate area is 2600 mi by 1300 mi or 3,380,000 2mi . This estimate is within 10 percent of the actual area, 3,794,083 2mi . The population is roughly 83.0 10× while the area of a dollar bill, as measured with a ruler, is approximately 186 in. by 582 in. EXECUTE: ( ) ( ) ( )[ ] ( ) ( ) 222 16 2U.S. 3,380,000 mi 5280 ft / 1 mi 12 in. 1 ft 1.4 10 in.A ? ?= = ×? ? ( )( ) 2 bill 6.125 in. 2.625 in. 16.1 in.A = = ( ) ( )16 2 2 14 bills U.S. billTotal cost 1.4 10 in. 16.1 in. / bill 9 10 billsN A A= = = × = × 14 8 6Cost per person (9 10 dollars) /(3.0 10 persons) 3 10 dollars/person= × × = × EVALUATE: The actual cost would be somewhat larger, because the land isn?t flat. 1.30. IDENTIFY: The displacements must be added as vectors and the magnitude of the sum depends on the relative orientation of the two displacements. SET UP: The sum with the largest magnitude is when the two displacements are parallel and the sum with the smallest magnitude is when the two displacements are antiparallel. EXECUTE: The orientations of the displacements that give the desired sum are shown in Figure 1.30. EVALUATE: The orientations of the two displacements can be chosen such that the sum has any value between 0.6 m and 4.2 m. Figure 1.30 1-6 Chapter 1 1.31. IDENTIFY: Draw each subsequent displacement tail to head with the previous displacement. The resultant displacement is the single vector that points from the starting point to the stopping point. SET UP: Call the three displacements A? , B? , and C? . The resultant displacement R? is given by R = A + B + C? ?? . EXECUTE: The vector addition diagram is given in Figure 1.31. Careful measurement gives that R? is 7.8 km, 38 north of east? . EVALUATE: The magnitude of the resultant displacement, 7.8 km, is less than the sum of the magnitudes of the individual displacements, 2.6 km 4.0 km 3.1 km+ + . Figure 1.31 1.32. IDENTIFY: Draw the vector addition diagram, so scale. SET UP: The two vectors A? and B? are specified in the figure that accompanies the problem. EXECUTE: (a) The diagram for C = A + B? ? is given in Figure 1.32a. Measuring the length and angle of C? gives 9.0 mC = and an angle of 34? = ° . (b) The diagram for ?D = A B?? is given in Figure 1.32b. Measuring the length and angle of D? gives 22 mD = and an angle of 250? = ° . (c) ( )? ? ?A B = A + B? ? , so ? ?A B? ? has a magnitude of 9.0 m (the same as A + B? ? ) and an angle with the x+ axis of 214° (opposite to the direction of A + B? ? ). (d) ( )? ? ?B A = A B?? , so ?B A?? has a magnitude of 22 m and an angle with the x+ axis of 70° (opposite to the direction of ?A B? ? ). EVALUATE: The vector ?A? is equal in magnitude and opposite in direction to the vector A? . Figure 1.32 1.33. IDENTIFY: Since she returns to the starting point, the vectors sum of the four displacements must be zero. SET UP: Call the three given displacements A? , B? , and C? , and call the fourth displacement D? . 0A + B + C + D = ? ? . EXECUTE: The vector addition diagram is sketched in Figure 1.33. Careful measurement gives that D? is144 m, 41 south of west.? Units, Physical Quantities and Vectors 1-7 EVALUATE: D? is equal in magnitude and opposite in direction to the sum A + B + C? ?? . Figure 1.33 1.34. IDENTIFY and SET UP: Use a ruler and protractor to draw the vectors described. Then draw the corresponding horizontal and vertical components. EXECUTE: (a) Figure 1.34 gives components 4.7 m, 8.1 m. (b) Figure 1.34 gives components 15.6 km,15.6 km? . (c) Figure 1.34 gives components 3.82 cm, 5.07 cm? . EVALUATE: The signs of the components depend on the quadrant in which the vector lies. Figure 1.34 1.35. IDENTIFY: For each vector V? , use that cosxV V ?= and sinyV V ?= , when ? is the angle V ? makes with the x+ axis, measured counterclockwise from the axis. SET UP: For A? , 270.0? = ° . For B? , 60.0? = ° . For C? , 205.0? = ° . For D? , 143.0? = ° . EXECUTE: 0xA = , 8.00 myA = ? . 7.50 mxB = , 13.0 myB = . 10.9 mxC = ? , 5.07 myC = ? . 7.99 mxD = ? , 6.02 myD = . EVALUATE: The signs of the components correspond to the quadrant in which the vector lies. 1.36. IDENTIFY: tan y x A A ? = , for ? measured counterclockwise from the x+ -axis. SET UP: A sketch of xA , yA and A ? tells us the quadrant in which A ? lies. EXECUTE: (a) 1.00 m tan 0.500 2.00 m y X A ? A ?= = = ? . ( )1tan 0.500 360 26.6 333? ?= ? = ° ? ° = ° . (b) 1.00 m tan 0.500 2.00 m y x A ? A = = = . ( )1tan 0.500 26.6? ?= = ° . (c) 1.00 m tan 0.500 2.00 m y x A ? A = = = ?? . ( )1tan 0.500 180 26.6 153? ?= ? = ° ? ° = ° . (d) 1.00 m tan 0.500 2.00 m y x A ? A ?= = =? . ( )1tan 0.500 180 26.6 207? ?= = ° + ° = ° EVALUATE: The angles 26.6° and 207° have the same tangent. Our sketch tells us which is the correct value of ? . 1.37. IDENTIFY: Find the vector sum of the two forces. SET UP: Use components to add the two forces. Take the -directionx+ to be forward and the -directiony+ to be upward. 1-8 Chapter 1 EXECUTE: The second force has components 2 2 cos32.4 433 NxF F= ° = and 2 2 sin32.4 275 N.yF F= ° = The first force has components 1 725 NxF = and 1 0.yF = 1 2 1158 Nx x xF F F= + = and 1 2 275 Ny y yF F F= + = The resultant force is 1190 N in the direction 13.4° above the forward direction. EVALUATE: Since the two forces are not in the same direction the magnitude of their vector sum is less than the sum of their magnitudes. 1.38. IDENTIFY: Find the vector sum of the three given displacements. SET UP: Use coordinates for which x+ is east and y+ is north. The driver?s vector displacements are: 2.6 km, 0 of north; 4.0 km, 0 of east; 3.1 km, 45 north of east= ° = ° = °A B C? ? . EXECUTE: ( ) ( )0 4.0 km 3.1 km cos 45 6.2 kmx x x xR A B C= + + = + + =? ; y y y yR A B C= + + = ( )2.6 km 0 (3.1 km) sin45 4.8 km+ + =? ; 2 2 7.8 kmx yR R R= + = ; ( ) ( )1tan 4.8 km 6.2 km? ? ? ?= ? ? 38= ? ; 7.8 km, 38 north of east.=R ?? This result is confirmed by the sketch in Figure 1.38. EVALUATE: Both xR and yR are positive and R ? is in the first quadrant. Figure 1.38 1.39. IDENTIFY: If C = A + B? ? , then x x xC A B= + and y y yC A B= + . Use xC and yC to find the magnitude and direction of C ? . SET UP: From Figure 1.34 in the textbook, 0xA = , 8.00 myA = ? and sin30.0 7.50 mxB B= + =° , cos30.0 13.0 myB B= + =° . EXECUTE: (a) C = A + B? ? so 7.50 mx x xC A B= + = and 5.00 my y yC A B= + = + . 9.01 mC = . 5.00 m tan 7.50 m y x C C ? = = and 33.7? = ° . (b) B + A = A + B ?? , so B + A ?? has magnitude 9.01 m and direction specified by 33.7° . (c) ?D = A B?? so 7.50 mx x xD A B= ? = ? and 21.0 my y yD A B= ? = ? . 22.3 mD = . 21.0 mtan 7.50 m y x D D ? ?= = ? and 70.3? = ° . D? is in the 3rd quadrant and the angle ? counterclockwise from the x+ axis is 180 70.3 250.3+ =° ° ° . (d) ( )? = ? ?B A A B?? , so ?B A?? has magnitude 22.3 m and direction specified by 70.3? = ° . EVALUATE: These results agree with those calculated from a scale drawing in Problem 1.32. 1.40. IDENTIFY: Use Equations (1.7) and (1.8) to calculate the magnitude and direction of each of the given vectors. SET UP: A sketch of xA , yA and A ? tells us the quadrant in which A ? lies. EXECUTE: (a) 2 2( 8.60 cm) (5.20 cm) 10.0? + = cm, 5.20arctan 148.8 8.60 ? ? = °? ??? ? (which is 180 31.2° ? ° ). (b) 2 2( 9.7 m) ( 2.45 m) 10.0 m,? + ? = 2.45arctan 14 180 194 . 9.7 ?? ? = ° + ° = °? ??? ? (c) 2 2(7.75 km) ( 2.70 km) 8.21 km,+ ? = 2.7arctan 340.8 7.75 ?? ? = °? ?? ? (which is 360 19.2° ? ° ). EVALUATE: In each case the angle is measured counterclockwise from the x+ axis. Our results for ? agree with our sketches. Units, Physical Quantities and Vectors 1-9 1.41. IDENTIFY: Vector addition problem. We are given the magnitude and direction of three vectors and are asked to find their sum. SET UP: 3.25 kmA = 4.75 kmB = 1.50 kmC = Figure 1.41a Select a coordinate system where x+ is east and y+ is north. Let ,A? B? and C? be the three displacements of the professor. Then the resultant displacement R ? is given by .= + +R A B C ? ?? ? By the method of components, x x x xR A B C= + + and .y y y yR A B C= + + Find the x and y components of each vector; add them to find the components of the resultant. Then the magnitude and direction of the resultant can be found from its x and y components that we have calculated. As always it is essential to draw a sketch. EXECUTE: 0,xA = 3.25 kmyA = + 4.75 km,xB = ? 0yB = 0,xC = 1.50 kmyC = ? x x x xR A B C= + + 0 4.75 km 0 4.75 kmxR = ? + = ? y y y yR A B C= + + 3.25 km 0 1.50 km 1.75 kmyR = + ? = Figure 1.41b The angle ? measured counterclockwise from the -axis.x+ In terms of compass directions, the resultant displacement is 20.2 N° of W. EVALUATE: 0xR < and 0,yR > so R ? is in 2nd quadrant. This agrees with the vector addition diagram. 1.42. IDENTIFY: Add the vectors using components. ( )? ?B A = B + A? ?? ? . SET UP: If C = A + B? ? then x x xC A B= + and y y yC A B= + . If ?D = B A ?? ? then x x xD B A= ? and y y yD B A= ? . EXECUTE: (a) The x- and y-components of the sum are 1.30 cm 4.10 cm 5.40 cm,+ = 2.25 cm ( 3.75 cm) 1.50 cm.+ ? = ? (b) Using Equations (1.7) and (1.8), 2 2(5.40cm) ( 1.50 cm) 5.60 cm,? = 1.50arctan 344.5 5.40 ?? ? = °? ?+? ? ccw. 2 2 2 2( 4.75 km) (1.75 km)x yR R R= + = ? + 5.06 kmR = 1.75 km tan 0.3684 4.75 km y x R R ? = = = ?? 159.8? = ° Figure 1.41c 1-10 Chapter 1 (c) Similarly, ( )4 10 cm 1 30 cm 2 80 cm,. ? . = . ( )3 75 cm 2 25 cm 6 00 cm. ? . = . .2 2 (d) 2 2(2.80cm) ( 6.00cm) 6.62+ ? = cm, 6.00arctan 295 2.80 ?? ? = °? ?? ? (which is 360 65° ? ° ). EVALUATE: We can draw the vector addition diagram in each case and verify that our results are qualitatively correct. 1.43. IDENTIFY: Vector addition problem. ( ).? ?A B = A + B? ?? ? SET UP: Find the x- and y-components of A? and .B? Then the x- and y-components of the vector sum are calculated from the x- and y-components of A ? and .B ? EXECUTE: cos(60.0 )xA A= ° (2.80 cm)cos(60.0 ) 1.40 cmxA = ° = + sin(60.0 )yA A= ° (2.80 cm)sin(60.0 ) 2.425 cmyA = ° = + cos( 60.0 )xB B= ? ° (1.90 cm)cos( 60.0 ) 0.95 cmxB = ? ° = + sin( 60.0 )yB B= ? ° (1.90 cm)sin( 60.0 ) 1.645 cmyB = ? ° = ? Note that the signs of the components correspond to the directions of the component vectors. Figure 1.43a (a) Now let .= +R A B ?? 1.40 cm 0.95 cm 2.35 cm.x x xR A B= + = + + = + 2.425 cm 1.645 cm 0.78 cm.y y yR A B= + = + ? = + 2 2 2 2(2.35 cm) (0.78 cm)x yR R R= + = + 2.48 cmR = 0.78 cm tan 0.3319 2.35 cm y x R R ? += = = ++ 18.4? = ° Figure 1.43b EVALUATE: The vector addition diagram for = +R A B?? ? is R ? is in the 1st quadrant, with ,y xR R< in agreement with our calculation. Figure 1.43c Units, Physical Quantities and Vectors 1-11 (b) EXECUTE: Now let .?=R A B?? 1.40 cm 0.95 cm 0.45 cm.x x xR A B= ? = + ? = + 2.425 cm 1.645 cm 4.070 cm.y y yR A B= ? = + + = + 2 2 2 2(0.45 cm) (4.070 cm)x yR R R= + = + 4.09 cmR = 4.070 cm tan 9.044 0.45 cm y x R R ? = = = + 83.7? = ° Figure 1.43d EVALUATE: The vector addition diagram for ( )?= +R A B?? ? is R ? is in the 1st quadrant, with ,x yR R< in agreement with our calculation. Figure 1.43e (c) EXECUTE: ( )? ? ?B A = A B? ?? ? ?B A?? and ?A B? ? are equal in magnitude and opposite in direction. 4.09 cmR = and 83.7 180 264? = ° + ° = ° Figure 1.43f 1-12 Chapter 1 EVALUATE: The vector addition diagram for ( )?= +R B A?? ? is R ? is in the 3rd quadrant, with ,x yR R< in agreement with our calculation. Figure 1.43g 1.44. IDENTIFY: The velocity of the boat relative to the earth, B/Ev? , the velocity of the water relative to the earth, W/Ev? , and the velocity of the boat relative to the water, B/Wv ? , are related by B/E B/W W/Ev = v + v ? ? . SET UP: W/E 5.0 km/h=v? , north and B/W 7.0 km/h=v? , west. The vector addition diagram is sketched in Figure 1.44. EXECUTE: 2 2 2B/E W/E B/Wv v v= + and 2 2B/E (5.0 km/h) (7.0 km/h) 8.6 km/hv = + = . W/E B/W 5.0 km/h tan 7.0 km/h v v ? = = and 36? = ° , north of west. EVALUATE: Since the two vectors we are adding are perpendicular we can use the Pythagorean theorem directly to find the magnitude of their vector sum. Figure 1.44 1.45. IDENTIFY: Let 625 NA = and 875 NB = . We are asked to find the vector C? such that 0A + B = C =? ? . SET UP: 0xA = , 625 NyA = ? . (875 N)cos30 758 NxB = =° , (875 N)sin30 438 NyB = =° . EXECUTE: ( ) (0 758 N) 758 Nx x xC A B= ? + = ? + = ? . ( ) ( 625 N 438 N) 187 Ny y yC A B= ? + = ? ? + = + . Vector C ? and its components are sketched in Figure 1.45. 2 2 781 Nx yC C C= + = . 187 Ntan 758 N y x C C ? = = and 13.9? = ° . C ? is at an angle of 13.9° above the x? -axis and therefore at an angle 180 13.9 166.1? =° ° ° counterclockwise from the -axisx+ . EVALUATE: A vector addition diagram for A + B + C? ?? verifies that their sum is zero. Figure 1.45 Units, Physical Quantities and Vectors 1-13 1.46. IDENTIFY: We know the vector sum and want to find the magnitude of the vectors. Use the method of components. SET UP: The two vectors A? and B? and their resultant C? are shown in Figure 1.46. Let y+ be in the direction of the resultant. A B= . EXECUTE: y y yC A B= + . 372 N 2 cos43.0A= ° and 254 NA = . EVALUATE: The sum of the magnitudes of the two forces exceeds the magnitude of the resultant force because only a component of each force is upward. Figure 1.46 1.47. IDENTIFY: Find the components of each vector and then use Eq.(1.14). SET UP: 0xA = , 8.00 myA = ? . 7.50 mxB = , 13.0 myB = . 10.9 mxC = ? , 5.07 myC = ? . 7.99 mxD = ? , 6.02 myD = . EXECUTE: ?( 8.00 m)?A = j? ; ? ?(7.50 m) (13.0 m)B = i + j? ; ? ?( 10.9 m) ( 5.07 m)? ?C = i + j? ; ? ?( 7.99 m) (6.02 m)?D = i + j? . EVALUATE: All these vectors lie in the xy-plane and have no z-component. 1.48. IDENTIFY: The general expression for a vector written in terms of components and unit vectors is ? ?x yA AA = i + j ? SET UP: ? ?5.0 5.0(4 6 ) 20 30? = ?B = i j i j?? EXECUTE: (a) 5.0xA = , 6.3yA = ? (b) 11.2xA = , 9.91yA = ? (c) 15.0xA = ? , 22.4yA = (d) 20xA = , 30yA = ? EVALUATE: The components are signed scalars. 1.49. IDENTIFY: Use trig to find the components of each vector. Use Eq.(1.11) to find the components of the vector sum. Eq.(1.14) expresses a vector in terms of its components. SET UP: Use the coordinates in the figure that accompanies the problem. EXECUTE: (a) ( ) ( ) ( ) ( )? ? ? ?3.60 m cos70.0 3.60 m sin 70.0 1.23 m 3.38 m° °A = i + j = i + j? ( ) ( ) ( ) ( )? ? ? ?2.40 m cos 30.0 2.40 m sin 30.0 2.08 m 1.20 m? ° ? ° ? ?B = i j = i + j? (b) ( ) ( )3.00 4.00?C = A B? ? ( )( ) ( )( ) ( )( ) ( )( )? ? ? ?3.00 1.23 m 3.00 3.38 m 4.00 2.08 m 4.00 1.20 m? ? ? ?= i + j i j ? ?(12.01 m) (14.94)= +i j (c) From Equations (1.7) and (1.8), ( ) ( )2 2 14.94 m12.01 m 14.94 m 19.17 m, arctan 51.2 12.01 m C ? ?= + = = °? ?? ? EVALUATE: xC and yC are both positive, so ? is in the first quadrant. 1.50. IDENTIFY: Find A and B. Find the vector difference using components. SET UP: Deduce the x- and y-components and use Eq.(1.8). EXECUTE: (a) ? ?4.00 3.00 ;= +A i j? 4.00;xA = + 3.00yA = + 2 2 2 2(4.00) (3.00) 5.00x yA A A= + = + = 1-14 Chapter 1 ? ?5.00 2.00 ;= ?B i j? 5.00;xB = + 2.00yB = ? 2 2 2 2(5.00) ( 2.00) 5.39x yB B B= + = + ? = EVALUATE: Note that the magnitudes of A? and B? are each larger than either of their components. EXECUTE: (b) ( )? ? ? ? ? ?4.00 3.00 5.00 2.00 (4.00 5.00) (3.00 2.00)? = + ? ? = ? + +A B i j i j i j? ? ? ?1.00 5.00? = ? +A B i j? ? (c) Let ? ?1.00 5.00 .? = ? +=R A B i j?? Then 1.00,xR = ? 5.00.yR = 2 2 x yR R R= + 2 2( 1.00) (5.00) 5.10.R = ? + = 5.00 tan 5.00 1.00 y x R R ? = = = ?? 78.7 180 101.3 .? = ? ° + ° = ° Figure 1.50 EVALUATE: 0xR < and 0,yR > so R ? is in the 2nd quadrant. 1.51. IDENTIFY: A unit vector has magnitude equal to 1. SET UP: The magnitude of a vector is given in terms of its components by Eq.(1.12). EXECUTE: (a) 2 2 2? ? ? 1 1 1 3 1= + + = ?i + j + k so it is not a unit vector. (b) 2 2 2x y zA A A= + +A ? . If any component is greater than 1+ or less than 1,? 1>A? , so it cannot be a unit vector. A ? can have negative components since the minus sign goes away when the component is squared. (c) 1=A? gives ( ) ( )2 22 23.0 4.0 1a a+ = and 2 25 1a = . 1 0.20 5.0 a = ± = ± . EVALUATE: The magnitude of a vector is greater than the magnitude of any of its components. 1.52. IDENTIFY: If vectors A? and B? commute for addition, A + B = B + A? ?? ? . If they commute for the scalar product, ? = ?A B B A? ? . SET UP: Express the sum and scalar product in terms of the components of A? and B? . EXECUTE: (a) Let ? ?x yA AA = i + j ? and ? ?x yB BB = i + j ? . ( ) ( )? ?x x y yA B A B+ +A + B = i + j? ? . ( ) ( )? ?x x y yB A B A+ +B + A = i + j?? . Scalar addition is commutative, so A + B = B + A? ?? ? . x x y yA B A B? = +A B ? ? and x x y yB A B A? = +B A ?? . Scalar multiplication is commutative, so ? = ?A B B A? ? . (b) ( ) ( ) ( )? ? ? y z z y z x x z x y y xA B A B A B A B A B A B? ? ?A ? B = i + j + k? ? . ( ) ( ) ( )? ? ? y z z y z x x z x y y xB A B A B A B A B A B A? ? ?B ? A = i + j + k?? . Comparison of each component in each vector product shows that one is the negative of the other. EVALUATE: The result in part (b) means that A ? B? ? and B ? A?? have the same magnitude and opposite direction. 1.53. IDENTIFY: cosAB ?? =A B? ? SET UP: For A? and B? , 150.0? = ° . For B? and C? , 145.0? = ° . For A? and C? , 65.0? = ° . EXECUTE: (a) 2(8.00 m)(15.0 m)cos150.0 104 m? = = ?A B? ? ° (b) 2(15.0 m)(12.0 m)cos145.0 148 m? = = ?B C?? ° (c) 2(8.00 m)(12.0 m)cos65.0 40.6 m? = =A C? ° EVALUATE: When 90? < ° the scalar product is positive and when 90? > ° the scalar product is negative. 1.54. IDENTIFY: Target variables are ?A B? ? and the angle ? between the two vectors. SET UP: We are given A? and B? in unit vector form and can take the scalar product using Eq.(1.19). The angle ? can then be found from Eq.(1.18). Units, Physical Quantities and Vectors 1-15 EXECUTE: (a) ? ?4.00 3.00 ,= +A i j? ? ?5.00 2.00 ;= ?B i j? 5.00,A = 5.39B = ? ? ? ?(4.00 3.00 ) (5.00 2.00 ) (4.00)(5.00) (3.00)( 2.00)? = + ? ? = + ? =A B i j i j? ? 20.0 6.0 14.0.? = + (b) 14.0 cos 0.519; (5.00)(5.39)AB ? ?= = =A B ? ? 58.7 .? = ° EVALUATE: The component of B? along A? is in the same direction as ,A? so the scalar product is positive and the angle ? is less than 90 .° 1.55. IDENTIFY: For all of these pairs of vectors, the angle is found from combining Equations (1.18) and (1.21), to give the angle? as arccos arccos x x y yA B A B AB AB ? +? ? ? ??= =? ? ? ?? ?? ? A B ? ? . SET UP: Eq.(1.14) shows how to obtain the components for a vector written in terms of unit vectors. EXECUTE: (a) 22, 40, 13,A B? = ? = =A B? ? and so 22arccos 165 40 13 ? ? ??= = °? ?? ? . (b) 60, 34, 136,A B? = = =A B? ? 60arccos 28 34 136 ? ? ?= = °? ?? ? . (c) 0? =A B? ? and 90? = ° . EVALUATE: If 0? >A B? ? , 0 90?? < ° . If 0?
, the direction of D ? is 10.5° west of north. EVALUATE: The four displacements add to zero. 1.77. IDENTIFY and SET UP: The vector A ? that connects points 1 1( , )x y and 2 2( , )x y has components 2 1xA x x= ? and 2 1yA y y= ? . EXECUTE: (a) Angle of first line is 1 200 20tan 42 . 210 10 ? ? ?? ?= = °? ??? ? Angle of second line is 42 30 72 .° + ° = ° Therefore 10 250 cos 72 87X = + ° = , 20 250 sin 72 258Y = + ° = for a final point of (87,258). (b) The computer screen now looks something like Figure 1.77. The length of the bottom line is ( ) ( )2 2210 87 200 258 136? + ? = and its direction is 1 258 200tan 25 210 87 ? ?? ? = °? ??? ? below straight left. EVALUATE: Figure 1.77 is a vector addition diagram. The vector first line plus the vector arrow gives the vector for the second line. Figure 1.77 Units, Physical Quantities and Vectors 1-23 1.78. IDENTIFY: Let the three given displacements be A? , B? and C? , where 40 stepsA = , 80 stepsB = and 50 stepsC = . R = A + B + C?? . The displacement C? that will return him to his hut is ?R? . SET UP: Let the east direction be the -directionx+ and the north direction be the -direction.y+ EXECUTE: (a) The three displacements and their resultant are sketched in Figure 1.78. (b) ( ) ( )40 cos45 80 cos 60 11.7xR = ° ? ° = ? and ( ) ( )40 sin 45 80 sin 60 50 47.6.yR = ° + ° ? = The magnitude and direction of the resultant are 2 2( 11.7) (47.6) 49,? + = 47.6arctan 76 11.7 ? ? = °? ?? ? , north of west. We know that R ? is in the second quadrant because 0xR < , 0yR > . To return to the hut, the explorer must take 49 steps in a direction 76° south of east, which is 14° east of south. EVALUATE: It is useful to show xR , yR and R ? on a sketch, so we can specify what angle we are computing. Figure 1.78 1.79. IDENTIFY: Vector addition. One vector and the sum are given; find the second vector (magnitude and direction). SET UP: Let x+ be east and y+ be north. Let A? be the displacement 285 km at 40.0° north of west and let B? be the unknown displacement. + =A B R? ? where 115 km,=R? east = ?B R A?? ,x x xB R A= ? y y yB R A= ? EXECUTE: cos40.0 218.3 km,xA A= ? ° = ? sin 40.0 183.2 kmyA A= + ° = + 115 km,xR = 0yR = Then 333.3 km,xB = 183.2 km.yB = ? 2 2 380 km;x yB B B= + = tan / (183.2 km)/(333.3 km)y xB B? = = 28.8 ,? = ° south of east Figure 1.79 EVALUATE: The southward component of B? cancels the northward component of .A? The eastward component of B ? must be 115 km larger than the magnitude of the westward component of .A ? 1.80. IDENTIFY: Find the components of the weight force, using the specified coordinate directions. SET UP: For parts (a) and (b), take x+ direction along the hillside and the y+ direction in the downward direction and perpendicular to the hillside. For part (c), 35.0? = ° and 550 Nw = . EXECUTE: (a) sinxw w ?= (b) cosyw w ?= (c) The maximum allowable weight is ( )sinxw w ?= ( ) ( )550 N sin35.0 959 N= ° = . EVALUATE: The component parallel to the hill increases as ? increases and the component perpendicular to the hill increases as ? decreases. 1-24 Chapter 1 1.81. IDENTIFY: Vector addition. One force and the vector sum are given; find the second force. SET UP: Use components. Let y+ be upward. B ? is the force the biceps exerts. Figure 1.81a E ? is the force the elbow exerts. ,+ =E B R? ? where 132.5 NR = and is upward. ,x x xE R B= ? y y yE R B= ? EXECUTE: sin 43 158.2 N,xB B= ? ° = ? cos43 169.7 N,yB B= + ° = + 0,xR = 132.5 NyR = + Then 158.2 N,xE = + 37.2 NyE = ? 2 2 160 N;x yE E E= + = tan / 37.2 /158.2y xE E? = = 13 ,? = ° below horizontal Figure 1.81b EVALUATE: The x-component of E? cancels the x-component of .B? The resultant upward force is less than the upward component of ,B ? so yE must be downward. 1.82. IDENTIFY: Find the vector sum of the four displacements. SET UP: Take the beginning of the journey as the origin, with north being the y-direction, east the x-direction, and the z-axis vertical. The first displacement is then ?( 30 m) ,? k the second is ?( 15 m) ,? j the third is ?(200 m) ,i and the fourth is ?(100 m) .j EXECUTE: (a) Adding the four displacements gives ? ? ? ? ? ? ?( 30 m) ( 15 m) (200 m) (100 m) (200 m) (85 m) (30 m) .? ? ?k + j + i + j = i + j k (b) The total distance traveled is the sum of the distances of the individual segments: 30 m 15 m 200 m 100 m 345 m+ + + = . The magnitude of the total displacement is: ( )22 2 2 2 2(200 m) (85 m) 30 m 219 m.x y zD D D D= + + = + + ? = EVALUATE: The magnitude of the displacement is much less than the distance traveled along the path. 1.83. IDENTIFY: The sum of the force displacements must be zero. Use components. SET UP: Call the displacements A? , B? , C? and D? , where D? is the final unknown displacement for the return from the treasure to the oak tree. Vectors A ? , B ? , and C ? are sketched in Figure 1.83a. 0A + B + C + D = ? ? says 0x x x xA B C D+ + + = and 0y y y yA B C D+ + + = . 825 mA = , 1250 mB = , and 1000 mC = . Let x+ be eastward and y+ be north. EXECUTE: (a) 0x x x xA B C D+ + + = gives ( ) (0 [1250 m]sin30.0 [1000 m]cos40.0 141 mx x x xD A B C=? + + =? ? + =?° °) . 0y y y yA B C D+ + + = gives ( ) ( 825 m [1250 m]cos30.0 [1000 m]sin 40.0 900 my y y yD A B C= ? + + = ? ? + + = ?° °) . The fourth displacement D ? and its components are sketched in Figure 1.83b. 2 2 911 mx yD D D= + = . 141 m tan 900 m x y D D ? = = and 8.9? = ° . You should head 8.9° west of south and must walk 911 m. Units, Physical Quantities and Vectors 1-25 (b) The vector diagram is sketched in Figure 1.83c. The final displacement D ? from this diagram agrees with the vector D ? calculated in part (a) using components. EVALUATE: Note that D? is the negative of the sum of A? , B? , and C? . Figure 1.83 1.84. IDENTIFY: If the vector from your tent to Joe?s is A? and from your tent to Karl?s is B? , then the vector from Joe?s tent to Karl?s is ?B A?? . SET UP: Take your tent's position as the origin. Let x+ be east and y+ be north. EXECUTE: The position vector for Joe?s tent is ( ) ( )? ? ? ?[21.0 m]cos 23 [21.0 m]sin 23 (19.33 m) (8.205 m) .° ? ° ?i j = i j The position vector for Karl's tent is ( ) ( )? ? ? ?[32.0 m]cos 37 [32.0 m]sin 37 (25.56 m) (19.26 m) .° °i + j = i + j The difference between the two positions is ( ) ( )? ? ? ?19.33 m 25.56 m 8.205 m 19.25 m (6.23 m) (27.46 m) .? ? ? ? ?i + j = i j The magnitude of this vector is the distance between the two tents: ( ) ( )2 26.23 m 27.46 m 28.2 mD = ? + ? = EVALUATE: If both tents were due east of yours, the distance between them would be 32.0 m 21.0 m 17.0 m? = . If Joe?s was due north of yours and Karl?s was due south of yours, then the distance between them would be 32.0 m 21.0 m 53.0 m+ = . The actual distance between them lies between these limiting values. 1.85. IDENTIFY: In Eqs.(1.21) and (1.27) write the components of A? and B? in terms of A, B, A? and B? . SET UP: From Appendix B, cos( ) cos cos sin sina b a b a b? = + and sin( ) sin cos cos sina b a b a b? = ? . EXECUTE: (a) With 0z zA B= = , Eq.(1.21) becomes ( )( ) ( )( ) cos cos sin sinx x y y A B A BA B A B A ? B ? A ? B ?+ = + ( ) ( )cos cos sin sin cos cos x x y y A B A B A BA B A B AB ? ? ? ? AB ? ? AB ?+ = + = ? = , where the expression for the cosine of the difference between two angles has been used. (b) With 0z zA B= = , ?zCC = k ? and zC C= . From Eq.(1.27), ( )( ) ( )( ) cos sin sin cos x y y x A B A AC A B A B A ? B ? A ? B ?= ? = ? ( )cos sin sin cos sin sinA B A B B AC AB ? ? ? ? AB ? ? AB ?= ? = ? = , where the expression for the sine of the difference between two angles has been used. EVALUATE: Since they are equivalent, we may use either Eq.(1.18) or (1.21) for the scalar product and either (1.22) or (1.27) for the vector product, depending on which is the more convenient in a given application. 1.86. IDENTIFY: Apply Eqs.(1.18) and (1.22). SET UP: The angle between the vectors is 20 90 0 140 .° + ° = °°+3 EXECUTE: (a) Eq. (1.18) gives ( )( ) 23.60 m 2.40 m cos 140 6.62 m .? = ° = ?A B? ? (b) From Eq.(1.22), the magnitude of the cross product is ( )( ) 23.60 m 2.40 m sin 140 5.55 m° = and the direction, from the right-hand rule, is out of the page (the -directionz+ ). EVALUATE: We could also use Eqs.(1.21) and (1.27), with the components of A? and B? . 1-26 Chapter 1 1.87. IDENTIFY: Compare the magnitude of the cross product, sinAB ? , to the area of the parallelogram. SET UP: The two sides of the parallelogram have lengths A and B. ? is the angle between A? and B? . EXECUTE: (a) The length of the base is B and the height of the parallelogram is sinA ? , so the area is sinAB ? . This equals the magnitude of the cross product. (b) The cross product A? B ? ? is perpendicular to the plane formed by A ? and B ? , so the angle is 90° . EVALUATE: It is useful to consider the special cases 0? = ° , where the area is zero, and 90? = ° , where the parallelogram becomes a rectangle and the area is AB. 1.88. IDENTIFY: Use Eq.(1.27) for the components of the vector product. SET UP: Use coordinates with the -axisx+ to the right, -axisy+ toward the top of the page, and -axisz+ out of the page. 0xA = , 0yA = and 3.50 cmzA = ? . The page is 20 cm by 35 cm, so 20 cmxB = and 35 cmyB = . EXECUTE: ( ) ( ) ( )2 2122 cm , 70 cm , 0. x y z = = ? =A ? B A ? B A ? B? ?? ? EVALUATE: From the components we calculated the magnitude of the vector product is 2141 cm . 40.3 cmB = and 90? = ° , so 2sin 141 cmAB ? = , which agrees. 1.89. IDENTIFY: A? and B? are given in unit vector form. Find A, B and the vector difference .?A B? ? SET UP: 2.00 3.00 4.00 ,= ? + +A i j k? ? ? 3.00 1.00 3.00= + ?B i j k? ?? Use Eq.(1.8) to find the magnitudes of the vectors. EXECUTE: (a) 2 2 2 2 2 2( 2.00) (3.00) (4.00) 5.38x y zA A A A= + + = ? + + = 2 2 2 2 2 2(3.00) (1.00) ( 3.00) 4.36x y zB B B B= + + = + + ? = (b) ? ? ? ? ? ?( 2.00 3.00 4.00 ) (3.00 1.00 3.00 )? = ? + + ? + ?A B i j k i j k? ? ? ? ? ? ? ?( 2.00 3.00) (3.00 1.00) (4.00 ( 3.00)) 5.00 2.00 7.00 .? = ? ? + ? + ? ? = ? + +A B i j k i j k? ? (c) Let ,= ?C A B? ? so 5.00,xC = ? 2.00,yC = + 7.00zC = + 2 2 2 2 2 2( 5.00) (2.00) (7.00) 8.83x y zC C C C= + + = ? + + = ( ),? = ? ?B A A B?? so ?A B? ? and ?B A?? have the same magnitude but opposite directions. EVALUATE: A, B and C are each larger than any of their components. 1.90. IDENTIFY: Calculate the scalar product and use Eq.(1.18) to determine ? . SET UP: The unit vectors are perpendicular to each other. EXECUTE: The direction vectors each have magnitude 3 , and their scalar product is ( )( ) ( )( ) ( )( )1 1 1 1 1 1 1,+ ? + ? =2 so from Eq. (1.18) the angle between the bonds is 1 1 arccos arccos 109 . 33 3 ?? ? ? ?= ? = °? ?? ? ? ?? ? EVALUATE: The angle between the two vectors in the bond directions is greater than 90° . 1.91. IDENTIFY: Use the relation derived in part (a) of Problem 1.92: 2 2 2 2 cos ,C A B AB ?= + + where ? is the angle between A ? and B ? . SET UP: cos 0? = for 90? = ° . cos 0? < for 90 180?< <° ° and cos 0? > for 0 90?< <° ° . EXECUTE: (a) If 2 2 2 , cos 0,C A B ?= + = and the angle between A? and B? is 90° (the vectors are perpendicular). (b) If 2 2 2, cos 0,C A B ?< + < and the angle between A? and B? is greater than 90° . (c) If 2 2 2 , cos 0,C A B ?> + > and the angle between A? and B? is less than 90 .° EVALUATE: It is easy to verify the expression from Problem 1.92 for the special cases 0? = , where C A B= + , and for 180? = ° , where C A B= ? . 1.92. IDENTIFY: Let C = A + B? ? and calculate the scalar product ?C C? ? . SET UP: For any vector V? , 2V? =V V? . cosAB ?? =A B? ? . EXECUTE: (a) Use the linearity of the dot product to show that the square of the magnitude of the sum A + B? ? is ( ) ( ) 2 2 2 2 2 2 2 cos A B A B AB ? ? = ? + ? + ? + ? = ? + ? + ? = + + ? = + + A+ B A+ B A A A B B A B B A A B B A B A B ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? Units, Physical Quantities and Vectors 1-27 (b) Using the result of part (a), with ,A B= the condition is that 2 2 2 22 cos A A A A ?= + + , which solves for 1 2 2cos ,?= + 12cos ,? = ? and 120 .? = ° EVALUATE: The expression 2 2 2 2 cosC A B AB ?= + + is called the law of cosines. 1.93. IDENTIFY: Find the angle between specified pairs of vectors. SET UP: Use cos AB ? ?= A B ? ? EXECUTE: (a) ?A = k? (along line ab) ? ? ?B = i + j + k ? (along line ad ) 1,A = 2 2 21 1 1 3B = + + = ( )? ? ? ? 1? ? =A B = k i + j + k? ? So cos 1/ 3; AB ? ?= =A B ? ? 54.7? = ° (b) ? ? ?A = i + j + k ? (along line ad ) ? ?B = j + k ? (along line ac) 2 2 21 1 1 3;A = + + = 2 21 1 2B = + = ( ) ( )? ? ? ? ? 1 1 2? ? = + =A B = i + j + k i + j? ? So 2 2 cos ; 3 2 6AB ? ?= = =A B ? ? 35.3? = ° EVALUATE: Each angle is computed to be less than 90 ,° in agreement with what is deduced from Fig. 1.43 in the textbook. 1.94. IDENTIFY: The cross product A? B? ? is perpendicular to both A? and B? . SET UP: Use Eq.(1.27) to calculate the components of A? B? ? . EXECUTE: The cross product is 6.00 11.00? ? ? ? ? ?( 13.00) (6.00) ( 11.00) 13 (1.00) 13.00 13.00 ? ?? ?? ? ? ?? ?? ?? ?? ?i + j+ k = i + j k . The magnitude of the vector in square brackets is 1.93, and so a unit vector in this direction is ? ? ?(1.00) (6.00 /13.00) (11.00/13.00) 1.93 ? ?? ?? ?? ?? ? i + j k . The negative of this vector, ? ? ?(1.00) (6.00/13.00) (11.00 /13.00) 1.93 ? ??? ?? ?? ? i j+ k , is also a unit vector perpendicular to A ? and B ? . EVALUATE: Any two vectors that are not parallel or antiparallel form a plane and a vector perpendicular to both vectors is perpendicular to this plane. 1.95. IDENTIFY and SET UP: The target variables are the components of .C? We are given A? and .B? We also know ?A C? and ,?B C?? and this gives us two equations in the two unknowns xC and .yC EXECUTE: A? and C? are perpendicular, so 0.? =A C? ? 0,x x y yA C A C+ = which gives 5.0 6.5 0.x yC C? = 15.0,? =B C?? so 3.5 7.0 15.0x yC C? + = We have two equations in two unknowns xC and .yC Solving gives 8.0xC = and 6.1yC = EVALUATE: We can check that our result does give us a vector C? that satisfies the two equations 0? =A C? and 15.0.? =B C?? 1.96. IDENTIFY: Calculate the magnitude of the vector product and then use Eq.(1.22). SET UP: The magnitude of a vector is related to its components by Eq.(1.12). 1-28 Chapter 1 EXECUTE: sinAB ?=A? B? ? . ( ) ( )( )( ) 2 25.00 2.00 sin 0.5984 3.00 3.00AB ? ? += = =A? B ? ? and ( )1sin 0.5984 36.8 .? ?= = ° EVALUATE: We haven't found A? and B? , just the angle between them. 1.97. (a) IDENTIFY: Prove that ( ) ( ) .? = ?A B ? C A? B C? ?? SET UP: Express the scalar and vector products in terms of components. EXECUTE: ( ) ( ) ( ) ( )x yx y zA A? = +? ? ? ?? ? ? zA B ? C B ? C B ? C + A B ? C ( ) ( ) ( ) ( )x y z z y y z x x z z x y y xA B C B C A B C B C A B C B C? = ? + ? + ?A B ? C? ? ( ) ( ) ( ) ( )x y zx y zC C C? = + +A? B C A? B A? B A? B? ? ? ?? ? ? ( ) ( ) ( ) ( )y z z y x z x x z y x y y x zA B A B C A B A B C A B A B C? = ? + ? + ?A? B C? ? Comparison of the expressions for ( )?A B ? C? ? and ( ) ?A? B C? ?? shows they contain the same terms, so ( ) ( ) .? = ?A B ? C A? B C? ?? (b) IDENTIFY: Calculate ( ) ,?A? B C? ? given the magnitude and direction of ,A? ,B? and .C? SET UP: Use Eq.(1.22) to find the magnitude and direction of .A? B? ? Then we know the components of A? B? ? and of C ? and can use an expression like Eq.(1.21) to find the scalar product in terms of components. EXECUTE: 5.00;A = 26.0 ;A? = ° 4.00,B = 63.0B? = ° sin .AB ?=A? B? ? The angle ? between A? and B? is equal to 63.0 26.0 37.0 .B A? ? ?= ? = ° ? ° = ° So (5.00)(4.00)sin37.0 12.04,= ° =A? B? ? and by the right hand-rule A? B? ? is in the -direction.z+ Thus ( ) (12.04)(6.00) 72.2? = =A? B C? ? EVALUATE: A? B? ? is a vector, so taking its scalar product with C? is a legitimate vector operation. ( ) ?A? B C? ?? is a scalar product between two vectors so the result is a scalar. 1.98. IDENTIFY: Use the maximum and minimum values of the dimensions to find the maximum and minimum areas and volumes. SET UP: For a rectangle of width W and length L the area is LW. For a rectangular solid with dimensions L, W and H the volume is LWH. EXECUTE: (a) The maximum and minimum areas are ( )( )L l W w LW lW Lw,+ + = + + ( )( )L l W w LW lW Lw,? ? = ? ? where the common terms wl have been omitted. The area and its uncertainty are then ( ),WL lW Lw± + so the uncertainty in the area is .a lW Lw= + (b) The fractional uncertainty in the area is a lW Wl l w A WL L W += = + , the sum of the fractional uncertainties in the length and width. (c) The similar calculation to find the uncertainty v in the volume will involve neglecting the terms lwH, lWh and Lwh as well as lwh; the uncertainty in the volume is ,v lWH LwH LWh= + + and the fractional uncertainty in the volume is v lWH LwH LWh l w h V LWH L W H + += = + + , the sum of the fractional uncertainties in the length, width and height. EVALUATE: The calculation assumes the uncertainties are small, so that terms involving products of two or more uncertainties can be neglected. 1.99. IDENTIFY: Add the vector displacements of the receiver and then find the vector from the quarterback to the receiver. SET UP: Add the x-components and the y-components. Units, Physical Quantities and Vectors 1-29 EXECUTE: The receiver's position is ( ) ( ) ( ) ( )? ? ? ?[ 1.0 9.0 6.0 12.0 yd] [ 5.0 11.0 4.0 18.0 yd] 16.0 yd 28.0 yd+ + ? + ? + + +i + j = i + j . The vector from the quarterback to the receiver is the receiver's position minus the quarterback's position, or ( ) ( )? ?16.0 yd 35.0 ydi + j , a vector with magnitude ( ) ( )2 216.0 yd 35.0 yd 38.5 yd+ = . The angle is 16.0 arctan 24.6 35.0 ? ? = °? ?? ? to the right of downfield. EVALUATE: The vector from the quarterback to receiver has positive x-component and positive y-component. 1.100. IDENTIFY: Use the x and y coordinates for each object to find the vector from one object to the other; the distance between two objects is the magnitude of this vector. Use the scalar product to find the angle between two vectors. SET UP: If object A has coordinates ( , )A Ax y and object B has coordinates ( , )B Bx y , the vector ABr? from A to B has x-component B Ax x? and y-component B Ay y? . EXECUTE: (a) The diagram is sketched in Figure 1.100. (b) (i) In AU, 2 2(0.3182) (0.9329) 0.9857.+ = (ii) In AU, 2 2 2(1.3087) ( 0.4423) ( 0.0414) 1.3820.+ ? + ? = (iii) In AU 2 2 2(0.3182 1.3087) (0.9329 ( 0.4423)) (0.0414) 1.695.? + ? ? + = (c) The angle between the directions from the Earth to the Sun and to Mars is obtained from the dot product. Combining Equations (1.18) and (1.21), ( 0.3182)(1.3087 0.3182) ( 0.9329)( 0.4423 0.9329) (0) arccos 54.6 . (0.9857)(1.695) ? ? ?? ? + ? ? ? += = °? ?? ? (d) Mars could not have been visible at midnight, because the Sun-Mars angle is less than 90o. EVALUATE: Our calculations correctly give that Mars is farther from the Sun than the earth is. Note that on this date Mars was farther from the earth than it is from the Sun. Figure 1.100 1.101. IDENTIFY: Draw the vector addition diagram for the position vectors. SET UP: Use coordinates in which the Sun to Merak line lies along the x-axis. Let A? be the position vector of Alkaid relative to the Sun, M ? is the position vector of Merak relative to the Sun, and R ? is the position vector for Alkaid relative to Merak. 138 lyA = and 77 lyM = . EXECUTE: The relative positions are shown in Figure 1.101. M + R = A?? ? . x x xA M R= + so (138 ly)cos25.6 77 ly 47.5 lyx x xR A M= ? = ? =° . (138 ly)sin 25.6 0 59.6 lyy y yR A M= ? = ? =° . 76.2 lyR = is the distance between Alkaid and Merak. (b) The angle is angle ? in Figure 1.101. 47.5 lycos 76.2 ly xR R ? = = and 51.4? = ° . Then 180 129? ?= ? =° ° . EVALUATE: The concepts of vector addition and components make these calculations very simple. Figure 1.101 1-30 Chapter 1 1.102. IDENTIFY: Define ? ? ?A B CS = i + j+ k? . Show that 0?r S =?? if 0Ax By Cz+ + = . SET UP: Use Eq.(1.21) to calculate the scalar product. EXECUTE: ? ? ? ? ? ?( ) ( )x y z A B C Ax By Cz? = + + ? + + = + +r S i j k i j k?? If the points satisfy 0,Ax By Cz+ + = then 0? =r S?? and all points r? are perpendicular to S? . The vector and plane are sketched in Figure 1.102. EVALUATE: If two vectors are perpendicular their scalar product is zero. Figure 1.102 2-1 M OTION ALONG A STRAIGHT LINE 2.1. IDENTIFY: The average velocity is av-x xv t ?= ? . SET UP: Let x+ be upward. EXECUTE: (a) av- 1000 m 63 m 197 m/s4.75 sxv ?= = (b) av- 1000 m 0 169 m/s 5.90 sx v ?= = EVALUATE: For the first 1.15 s of the flight, av- 63 m 0 54.8 m/s1.15 sxv ?= = . When the velocity isn?t constant the average velocity depends on the time interval chosen. In this motion the velocity is increasing. 2.2. IDENTIFY: av-x xv t ?= ? SET UP: 513.5 days 1.166 10 s= × . At the release point, 65.150 10 mx = + × . EXECUTE: (a) 6 2 1 av- 6 5.150 10 m 4.42 m/s 1.166 10 sx x x v t ? ×= = = ?? × (b) For the round trip, 2 1x x= and 0x? = . The average velocity is zero. EVALUATE: The average velocity for the trip from the nest to the release point is positive. 2.3. IDENTIFY: Target variable is the time t? it takes to make the trip in heavy traffic. Use Eq.(2.2) that relates the average velocity to the displacement and average time. SET UP: av-x xv t ?= ? so av-xx v t? = ? and av- . x x t v ?? = EXECUTE: Use the information given for normal driving conditions to calculate the distance between the two cities: av- (105 km/h)(1 h/60 min)(140 min) 245 km.xx v t? = ? = = Now use av-xv for heavy traffic to calculate ;t? x? is the same as before: av- 245 km 3.50 h 3 h 70 km/hx x t v ?? = = = = and 30 min. The trip takes an additional 1 hour and 10 minutes. EVALUATE: The time is inversely proportional to the average speed, so the time in traffic is (105/ 70)(140 m) 210 min.= 2.4. IDENTIFY: The average velocity is av-x xv t ?= ? . Use the average speed for each segment to find the time traveled in that segment. The average speed is the distance traveled by the time. SET UP: The post is 80 m west of the pillar. The total distance traveled is 200 m 280 m 480 m+ = . EXECUTE: (a) The eastward run takes time 200 m 40.0 s 5.0 m/s = and the westward run takes 280 m 70.0 s 4.0 m/s = . The average speed for the entire trip is 480 m 4.4 m/s 110.0 s = . (b) av- 80 m 0.73 m/s 110.0 sx x v t ? ?= = = ?? . The average velocity is directed westward. 2 2-2 Chapter 2 EVALUATE: The displacement is much less than the distance traveled and the magnitude of the average velocity is much less than the average speed. The average speed for the entire trip has a value that lies between the average speed for the two segments. 2.5. IDENTIFY: When they first meet the sum of the distances they have run is 200 m. SET UP: Each runs with constant speed and continues around the track in the same direction, so the distance each runs is given by d vt= . Let the two runners be objects A and B. EXECUTE: (a) 200 mA Bd d+ = , so (6.20 m/s) (5.50 m/s) 200 mt t+ = and 200 m 17.1 s11.70 m/st = = . (b) (6.20 m/s)(17.1 s) 106 mA Ad v t= = = . (5.50 m/s)(17.1 s) 94 mB Bd v t= = = . The faster runner will be 106 m from the starting point and the slower runner will be 94 m from the starting point. These distances are measured around the circular track and are not straight-line distances. EVALUATE: The faster runner runs farther. 2.6. IDENTIFY: To overtake the slower runner the first time the fast runner must run 200 m farther. To overtake the slower runner the second time the faster runner must run 400 m farther. SET UP: t and 0x are the same for the two runners. EXECUTE: (a) Apply 0 0xx x v t? = to each runner: 0 f( ) (6.20 m/s)x x t? = and 0 s( ) (5.50 m/s)x x t? = . 0 f 0 s( ) ( ) 200 mx x x x? = ? + gives (6.20 m/s) (5.50 m/s) 200 mt t= + and 200 m 286 s6.20 m/s 5.50 m/st = =? . 0 f( ) 1770 mx x? = and 0 s( ) 1570 mx x? = . (b) Repeat the calculation but now 0 f 0 s( ) ( ) 400 mx x x x? = ? + . 572 st = . The fast runner has traveled 3540 m. He has made 17 full laps for 3400 m and 140 m past the starting line in this 18th lap. EVALUATE: In part (a) the fast runner will have run 8 laps for 1600 m and will be 170 m past the starting line in his 9th lap. 2.7. IDENTIFY: In time St the S-waves travel a distance S Sd v t= and in time Pt the P-waves travel a distance P Pd v t= . SET UP: S P 33 st t= + EXECUTE: S P 33 s d d v v = + . 1 1 33 s 3.5 km/s 6.5 km/s d ? ?? =? ?? ? and 250 kmd = . EVALUATE: The times of travel for each wave are S 71 st = and P 38 st = . 2.8. IDENTIFY: The average velocity is av-x xv t ?= ? . Use ( )x t to find x for each t. SET UP: (0) 0x = , (2.00 s) 5.60 mx = , and (4.00 s) 20.8 mx = EXECUTE: (a) av- 5.60 m 0 2.80 m/s2.00 sxv ?= = + (b) av- 20.8 m 0 5.20 m/s 4.00 sx v ?= = + (c) av- 20.8 m 5.60 m 7.60 m/s 2.00 sx v ?= = + EVALUATE: The average velocity depends on the time interval being considered. 2.9. (a) IDENTIFY: Calculate the average velocity using Eq.(2.2). SET UP: av-x xv t ?= ? so use ( )x t to find the displacement x? for this time interval. EXECUTE: 0 :t = 0x = 10.0 s:t = 2 2 3 3(2.40 m/s )(10.0 s) (0.120 m/s )(10.0 s) 240 m 120 m 120 m.x = ? = ? = Then av- 120 m 12.0 m/s. 10.0 sx x v t ?= = =? (b) IDENTIFY: Use Eq.(2.3) to calculate ( )xv t and evaluate this expression at each specified t. SET UP: 22 3 .x dxv bt ctdt= = ? EXECUTE: (i) 0 :t = 0xv = (ii) 5.0 s:t = 2 3 22(2.40 m/s )(5.0 s) 3(0.120 m/s )(5.0 s) 24.0 m/s 9.0 m/s 15.0 m/s.xv = ? = ? = (iii) 10.0 s:t = 2 3 22(2.40 m/s )(10.0 s) 3(0.120 m/s )(10.0 s) 48.0 m/s 36.0 m/s 12.0 m/s.xv = ? = ? = Motion Along a Straight Line 2-3 (c) IDENTIFY: Find the value of t when ( )xv t from part (b) is zero. SET UP: 2 3xv bt ct 2= ? 0xv = at 0.t = 0xv = next when 22 3 0bt ct? = EXECUTE: 2 3b ct= so 2 3 2 2(2.40 m/s ) 13.3 s 3 30(.120 m/s ) b t c = = = EVALUATE: ( )xv t for this motion says the car starts from rest, speeds up, and then slows down again. 2.10. IDENTIFY and SET UP: The instantaneous velocity is the slope of the tangent to the x versus t graph. EXECUTE: (a) The velocity is zero where the graph is horizontal; point IV. (b) The velocity is constant and positive where the graph is a straight line with positive slope; point I. (c) The velocity is constant and negative where the graph is a straight line with negative slope; point V. (d) The slope is positive and increasing at point II. (e) The slope is positive and decreasing at point III. EVALUATE: The sign of the velocity indicates its direction. 2.11. IDENTIFY: The average velocity is given by av-x xv t ?= ? . We can find the displacement t? for each constant velocity time interval. The average speed is the distance traveled divided by the time. SET UP: For 0t = to 2.0 st = , 2.0 m/sxv = . For 2.0 st = to 3.0 st = , 3.0 m/sxv = . In part (b), 3.0 m/sxv = ? for 2.0 st = to 3.0 st = . When the velocity is constant, xx v t? = ? . EXECUTE: (a) For 0t = to 2.0 st = , (2.0 m/s)(2.0 s) 4.0 mx? = = . For 2.0 st = to 3.0 st = , (3.0 m/s)(1.0 s) 3.0 mx? = = . For the first 3.0 s, 4.0 m 3.0 m 7.0 mx? = + = . The distance traveled is also 7.0 m. The average velocity is av- 7.0 m 2.33 m/s 3.0 sx x v t ?= = =? . The average speed is also 2.33 m/s. (b) For 2.0 st = to 3.0 s, ( 3.0 m/s)(1.0 s) 3.0 mx? = ? = ? . For the first 3.0 s, 4.0 m ( 3.0 m) 1.0 mx? = + ? = + . The dog runs 4.0 m in the x+ -direction and then 3.0 m in the x? -direction, so the distance traveled is still 7.0 m. av- 1.0 m 0.33 m/s 3.0 sx x v t ?= = =? . The average speed is 7.00 m 2.33 m/s 3.00 s = . EVALUATE: When the motion is always in the same direction, the displacement and the distance traveled are equal and the average velocity has the same magnitude as the average speed. When the motion changes direction during the time interval, those quantities are different. 2.12. IDENTIFY and SET UP: av, xx va t ?= ? . The instantaneous acceleration is the slope of the tangent to the xv versus t graph. EXECUTE: (a) 0 s to 2 s: av, 0xa = ; 2 s to 4 s: 2av, 1.0 m/sxa = ; 4 s to 6 s: 2av, 1.5 m/sxa = ; 6 s to 8 s: 2 av, 2.5 m/sxa = ; 8 s to 10 s: 2av, 2.5 m/sxa = ; 10 s to 12 s: 2av, 2.5 m/sxa = ; 12 s to 14 s: 2av, 1.0 m/sxa = ; 14 s to 16 s: av, 0xa = . The acceleration is not constant over the entire 16 s time interval. The acceleration is constant between 6 s and 12 s. (b) The graph of xv versus t is given in Fig. 2.12. 9 st = : 22.5 m/sxa = ; 13 st = : 21.0 m/sxa = ; 15 st = : 0xa = . 2-4 Chapter 2 EVALUATE: The acceleration is constant when the velocity changes at a constant rate. When the velocity is constant, the acceleration is zero. Figure 2.12 2.13. IDENTIFY: The average acceleration for a time interval t? is given by av- xx va t ?= ? . SET UP: Assume the car is moving in the x+ direction. 1 mi/h 0.447 m/s= , so 60 mi/h 26.82 m/s= , 200 mi/h = 89.40 m/s and 253 mi/h 113.1 m/s= . EXECUTE: (a) The graph of xv versus t is sketched in Figure 2.13. The graph is not a straight line, so the acceleration is not constant. (b) (i) 2av- 26.82 m/s 0 12.8 m/s 2.1 sx a ?= = (ii) 2av- 89.40 m/s 26.82 m/s 3.50 m/s20.0 s 2.1 sxa ?= =? (iii) 2 av- 113.1 m/s 89.40 m/s 0.718 m/s 53 s 20.0 sx a ?= =? . The slope of the graph of xv versus t decreases as t increases. This is consistent with an average acceleration that decreases in magnitude during each successive time interval. EVALUATE: The average acceleration depends on the chosen time interval. For the interval between 0 and 53 s, 2 av- 113.1 m/s 0 2.13 m/s 53 sx a ?= = . Figure 2.13 Motion Along a Straight Line 2-5 2.14. IDENTIFY: av- xx va t ?= ? . ( )xa t is the slope of the xv versus t graph. SET UP: 60 km/h 16.7 m/s= EXECUTE: (a) (i) 2av- 16.7 m/s 0 1.7 m/s10 sxa ?= = . (ii) 2av- 0 16.7 m/s 1.7 m/s10 sxa ?= = ? . (iii) 0xv? = and av- 0xa = . (iv) 0xv? = and av- 0xa = . (b) At 20 st = , xv is constant and 0xa = . At 35 st = , the graph of xv versus t is a straight line and 2 av- 1.7 m/sx xa a= = ? . EVALUATE: When av-xa and xv have the same sign the speed is increasing. When they have opposite sign the speed is decreasing. 2.15. IDENTIFY and SET UP: Use x dx v dt = and xx dva dt= to calculate ( )xv t and ( ).xa t EXECUTE: 22.00 cm/s (0.125 cm/s )x dxv tdt= = ? 20.125 cm/sxx dv a dt = = ? (a) At 0,t = 50.0 cm,x = 2.00 cm/s,xv = 20.125 cm/s .xa = ? (b) Set 0xv = and solve for t: 16.0 s.t = (c) Set 50.0 cmx = and solve for t. This gives 0t = and 32.0 s.t = The turtle returns to the starting point after 32.0 s. (d) Turtle is 10.0 cm from starting point when 60.0 cmx = or 40.0 cm.x = Set 60.0 cmx = and solve for t: 6.20 st = and 25.8 s.t = At 6.20 s,t = 1.23 cm/s.xv = + At 25.8 s,t = 1.23 cm/s.xv = ? Set 40.0 cmx = and solve for t: 36.4 st = (other root to the quadratic equation is negative and hence nonphysical). At 36.4 s,t = 2.55 cm/s.xv = ? (e) The graphs are sketched in Figure 2.15. Figure 2.15 EVALUATE: The acceleration is constant and negative. xv is linear in time. It is initially positive, decreases to zero, and then becomes negative with increasing magnitude. The turtle initially moves farther away from the origin but then stops and moves in the -direction.x? 2.16. IDENTIFY: Use Eq.(2.4), with 10 st? = in all cases. SET UP: xv is negative if the motion is to the right. EXECUTE: (a) ( ) ( )( ) ( ) 25.0 m/s 15.0 m/s / 10 s 1.0 m/s? = ? (b) ( ) ( )( ) ( ) 215.0 m/s 5.0 m/s / 10 s 1.0 m/s? ? ? = ? (c) ( ) ( )( ) ( ) 215.0 m/s 15.0 m/s / 10 s 3.0 m/s? ? + = ? EVALUATE: In all cases, the negative acceleration indicates an acceleration to the left. 2.17. IDENTIFY: The average acceleration is av- xx va t ?= ? SET UP: Assume the car goes from rest to 65 mi/h (29 m/s) in 10 s. In braking, assume the car goes from 65 mi/h to zero in 4.0 s. Let x+ be in the direction the car is traveling. EXECUTE: (a) 2av- 29 m/s 0 2.9 m/s10 sxa ?= = (b) 2av- 0 29 m/s 7.2 m/s 4.0 sx a ?= = ? 2-6 Chapter 2 (c) In part (a) the speed increases so the acceleration is in the same direction as the velocity. If the velocity direction is positive, then the acceleration is positive. In part (b) the speed decreases so the acceleration is in the direction opposite to the direction of the velocity. If the velocity direction is positive then the acceleration is negative, and if the velocity direction is negative then the acceleration direction is positive. EVALUATE: The sign of the velocity and of the acceleration indicate their direction. 2.18. IDENTIFY: The average acceleration is av- xx va t ?= ? . Use ( )xv t to find xv at each t. The instantaneous acceleration is xx dv a dt = . SET UP: (0) 3.00 m/sxv = and (5.00 s) 5.50 m/sxv = . EXECUTE: (a) 2av- 5.50 m/s 3.00 m/s 0.500 m/s5.00 s x x v a t ? ?= = =? (b) 3 3(0.100 m/s )(2 ) (0.200 m/s )xx dv a t t dt = = = . At 0t = , 0xa = . At 5.00 st = , 21.00 m/sxa = . (c) Graphs of ( )xv t and ( )xa t are given in Figure 2.18. EVALUATE: ( )xa t is the slope of ( )xv t and increases at t increases. The average acceleration for 0t = to 5.00 st = equals the instantaneous acceleration at the midpoint of the time interval, 2.50 st = , since ( )xa t is a linear function of t. Figure 2.18 2.19. (a) IDENTIFY and SET UP: xv is the slope of the x versus t curve and xa is the slope of the xv versus t curve. EXECUTE: 0t = to 5 st = : x versus t is a parabola so xa is a constant. The curvature is positive so xa is positive. xv versus t is a straight line with positive slope. 0 0.xv = 5 st = to 15 st = : x versus t is a straight line so xv is constant and 0.xa = The slope of x versus t is positive so xv is positive. 15 st = to 25 s:t = x versus t is a parabola with negative curvature, so xa is constant and negative. xv versus t is a straight line with negative slope. The velocity is zero at 20 s, positive for 15 s to 20 s, and negative for 20 s to 25 s. 25 st = to 35 s:t = x versus t is a straight line so xv is constant and 0.xa = The slope of x versus t is negative so xv is negative. 35 st = to 40 s:t = x versus t is a parabola with positive curvature, so xa is constant and positive. xv versus t is a straight line with positive slope. The velocity reaches zero at 40 s.t = Motion Along a Straight Line 2-7 The graphs of ( )xv t and ( )xa t are sketched in Figure 2.19a. Figure 2.19a (b) The motions diagrams are sketched in Figure 2.19b. Figure 2.19b EVALUATE: The spider speeds up for the first 5 s, since xv and xa are both positive. Starting at 15 st = the spider starts to slow down, stops momentarily at 20 s,t = and then moves in the opposite direction. At 35 st = the spider starts to slow down again and stops at 40 s.t = 2.20. IDENTIFY: ( )x dxv t dt= and ( ) x x dv a t dt = SET UP: 1( )n nd t nt dt ?= for 1n ? . EXECUTE: (a) 2 6 5( ) (9.60 m/s ) (0.600 m/s )xv t t t= ? and 2 6 4( ) 9.60 m/s (3.00 m/s )xa t t= ? . Setting 0xv = gives 0t = and 2.00 st = . At 0t = , 2.17 mx = and 29.60 m/sxa = . At 2.00 st = , 15.0 mx = and 238.4 m/sxa = ? . (b) The graphs are given in Figure 2.20. 2-8 Chapter 2 EVALUATE: For the entire time interval from 0t = to 2.00 st = , the velocity xv is positive and x increases. While xa is also positive the speed increases and while xa is negative the speed decreases. Figure 2.20 2.21. IDENTIFY: Use the constant acceleration equations to find 0xv and .xa (a) SET UP: The situation is sketched in Figure 2.21. Figure 2.21 EXECUTE: Use 00 ,2 x xv vx x t +? ?? = ? ?? ? so 0 0 2( ) 2(70.0 m) 15.0 m/s 5.0 m/s. 7.00 sx x x x v v t ?= ? = ? = (b) Use 0 ,x x xv v a t= + so 20 15.0 m/s 5.0 m/s 1.43 m/s .7.00 s x x x v v a t ? ?= = = EVALUATE: The average velocity is (70.0 m)/(7.00 s) 10.0 m/s.= The final velocity is larger than this, so the antelope must be speeding up during the time interval; 0x xv v< and 0.xa > 2.22. IDENTIFY: Apply the constant acceleration kinematic equations. SET UP: Let x+ be in the direction of the motion of the plane. 173 mi/h 77.33 m/s= . 307 ft 93.57 m= . EXECUTE: (a) 0 0xv = , 77.33 m/sxv = and 0 93.57 mx x? = . 2 20 02 ( )x x xv v a x x= + ? gives 2 2 2 20 0 (77.33 m/s) 0 32.0 m/s 2( ) 2(93.57 m) x x x v v a x x ? ?= = =? . (b) 00 2 x xv vx x t +? ?? = ? ?? ? gives 0 0 2( ) 2(93.57 m) 2.42 s 0 77.33 m/sx x x x t v v ?= = =+ + EVALUATE: Either 0x x xv v a t= + or 210 0 2x xx x v t a t? = + could also be used to find t and would give the same result as in part (b). 2.23. IDENTIFY: For constant acceleration, Eqs. (2.8), (2.12), (2.13) and (2.14) apply. SET UP: Assume the ball starts from rest and moves in the -direction.x+ EXECUTE: (a) 0 1.50 mx x? = , 45.0 m/sxv = and 0 0xv = . 2 20 02 ( )x x xv v a x x= + ? gives 2 2 2 20 0 (45.0 m/s) 675 m/s 2( ) 2(1.50 m) x x x v v a x x ?= = =? . (b) 00 2 x xv vx x t +? ?? = ? ?? ? gives 0 0 2( ) 2(1.50 m) 0.0667 s 45.0 m/sx x x x t v v ?= = =+ EVALUATE: We could also use 0x x xv v a t= + to find 245.0 m/s 0.0667 s675 m/s x x v t a = = = which agrees with our previous result. The acceleration of the ball is very large. 0 70.0 mx x? = 7.00 st = 15.0 m/sxv = 0 ?xv = Motion Along a Straight Line 2-9 2.24. IDENTIFY: For constant acceleration, Eqs. (2.8), (2.12), (2.13) and (2.14) apply. SET UP: Assume the ball moves in the x+ direction. EXECUTE: (a) 73.14 m/sxv = , 0 0xv = and 30.0 mst = . 0x x xv v a t= + gives 20 3 73.14 m/s 0 2440 m/s 30.0 10 s x x x v v a t ? ? ?= = =× . (b) 300 0 73.14 m/s (30.0 10 s) 1.10 m 2 2 x xv vx x t ? + +? ? ? ?? = = × =? ? ? ?? ? ? ? EVALUATE: We could also use 210 0 2x xx x v t a t? = + to calculate 0x x? : 2 3 21 0 2 (2440 m/s )(30.0 10 s) 1.10 mx x ?? = × = , which agrees with our previous result. The acceleration of the ball is very large. 2.25. IDENTIFY: Assume that the acceleration is constant and apply the constant acceleration kinematic equations. Set xa equal to its maximum allowed value. SET UP: Let x+ be the direction of the initial velocity of the car. 2250 m/sxa = ? . 105 km/h 29.17 m/s= . EXECUTE: 0 29.17 m/sxv = + . 0xv = . 2 20 02 ( )x x xv v a x x= + ? gives 2 2 2 0 0 2 0 (29.17 m/s) 1.70 m 2 2( 250 m/s ) x x x v v x x a ? ?? = = =? . EVALUATE: The car frame stops over a shorter distance and has a larger magnitude of acceleration. Part of your 1.70 m stopping distance is the stopping distance of the car and part is how far you move relative to the car while stopping. 2.26. IDENTIFY: Apply constant acceleration equations to the motion of the car. SET UP: Let x+ be the direction the car is moving. EXECUTE: (a) From Eq. (2.13), with 0 0,xv = 2 2 2 0 (20 m s) 1.67 m s . 2( ) 2(120 m) x x v a x x = = =? (b) Using Eq. (2.14), 02( ) 2(120 m) (20 m s) 12 s.xt x x v= ? = = (c) (12 s)(20 m s) 240 m.= EVALUATE: The average velocity of the car is half the constant speed of the traffic, so the traffic travels twice as far. 2.27. IDENTIFY: The average acceleration is av- xx va t ?= ? . For constant acceleration, Eqs. (2.8), (2.12), (2.13) and (2.14) apply. SET UP: Assume the shuttle travels in the x+ direction. 161 km/h 44.72 m/s= and 1610 km/h 447.2 m/s= . 1.00 min 60.0 s= EXECUTE: (a) (i) 2av- 44.72 m/s 0 5.59 m/s8.00 s x x v a t ? ?= = =? (ii) 2av- 447.2 m/s 44.72 m/s 7.74 m/s 60.0 s 8.00 sx a ?= =? (b) (i) 8.00 st = , 0 0xv = , and 44.72 m/sxv = . 00 0 44.72 m/s (8.00 s) 179 m2 2 x xv vx x t + +? ? ? ?? = = =? ? ? ?? ? ? ? . (ii) 60.0 s 8.00 s 52.0 st? = ? = , 0 44.72 m/sxv = , and 447.2 m/sxv = . 40 0 44.72 m/s 447.2 m/s (52.0 s) 1.28 10 m 2 2 x xv vx x t + +? ? ? ?? = = = ×? ? ? ?? ? ? ? . EVALUATE: When the acceleration is constant the instantaneous acceleration throughout the time interval equals the average acceleration for that time interval. We could have calculated the distance in part (a) as 2 2 21 1 0 0 2 2 (5.59 m/s )(8.00 s) 179 mx xx x v t a t? = + = = , which agrees with our previous calculation. 2.28. IDENTIFY: Apply the constant acceleration kinematic equations to the motion of the car. SET UP: 0.250 mi 1320 ft= . 60.0 mph 88.0 ft/s= . Let x+ be the direction the car is traveling. EXECUTE: (a) braking: 0 88.0 ft/sxv = , 0 146 ftx x? = , 0xv = . 2 20 02 ( )x x xv v a x x= + ? gives 2 2 2 20 0 0 (88.0 ft/s) 26.5 ft/s 2( ) 2(146 ft) x x x v v a x x ? ?= = = ?? Speeding up: 0 0xv = , 0 1320 ftx x? = , 19.9 st = . 210 0 2x xx x v t a t? = + gives 20 2 2 2( ) 2(1320 ft) 6.67 ft/s (19.9 s)x x x a t ?= = = 2-10 Chapter 2 (b) 20 0 (6.67 ft/s )(19.9 s) 133 ft/s 90.5 mphx x xv v a t= + = + = = (c) 0 2 0 88.0 ft/s 3.32 s 26.5 ft/s x x x v v t a ? ?= = =? EVALUATE: The magnitude of the acceleration while braking is much larger than when speeding up. That is why it takes much longer to go from 0 to 60 mph than to go from 60 mph to 0. 2.29. IDENTIFY: The acceleration xa is the slope of the graph of xv versus t. SET UP: The signs of xv and of xa indicate their directions. EXECUTE: (a) Reading from the graph, at 4.0 st = , 2.7 cm/sxv = , to the right and at 7.0 st = , 1.3 cm/sxv = , to the left. (b) xv versus t is a straight line with slope 28.0 cm/s 1.3 cm/s 6.0 s ? = ? . The acceleration is constant and equal to 21.3 cm/s , to the left. It has this value at all times. (c) Since the acceleration is constant, 210 0 2x xx x v t a t? = + . For 0t = to 4.5 s, 2 21 0 2(8.0 cm/s)(4.5 s) ( 1.3 cm/s )(4.5 s) 22.8 cmx x? = + ? = . For 0t = to 7.5 s, 2 21 0 2(8.0 cm/s)(7.5 s) ( 1.3 cm/s )(7.5 s) 23.4 cmx x? = + ? = (d) The graphs of xa and x versus t are given in Fig. 2.29. EVALUATE: In part (c) we could have instead used 00 2 x xv vx x t +? ?? = ? ?? ? . Figure 2.29 2.30. IDENTIFY: Use the constant acceleration equations to find x, 0xv , xv and xa for each constant-acceleration segment of the motion. SET UP: Let x+ be the direction of motion of the car and let 0x = at the first traffic light. EXECUTE: (a) For 0t = to 8 st = : 0 0 20 m/s (8 s) 80 m 2 2 x xv vx t + +? ? ? ?= = =? ? ? ?? ? ? ? . 20 20 m/s 2.50 m/s 8 s x x x v v a t ?= = = + . The car moves from 0x = to 80 mx = . The velocity xv increases linearly from zero to 20 m/s. The acceleration is a constant 22.50 m/s . Constant speed for 60 m: The car moves from 80 mx = to 140 mx = . xv is a constant 20 m/s. 0xa = . This interval starts at 8 st = and continues until 60 m 8 s 11 s 20 m/s t = + = . Slowing from 20 m/s until stopped: The car moves from 140 mx = to 180 mx = . The velocity decreases linearly from 20 m/s to zero. 00 2 x xv vx x t +? ?? = ? ?? ? gives 2(40 m) 4 s 20 m/s 0 t = =+ . 2 2 0 02 ( )x x xv v a x x= + ? gives 2 2(20.0 m/s) 5.00 m/s 2(40 m)x a ?= = ? This segment is from 11 st = to 15 st = . The acceleration is a constant 25.00 m/s? . The graphs are drawn in Figure 2.30a. (b) The motion diagram is sketched in Figure 2.30b. Motion Along a Straight Line 2-11 EVALUATE: When a! and v! are in the same direction, the speed increases ( 0t = to 8 st = ). When a! and v! are in opposite directions, the speed decreases ( 11 st = to 15 st = ). When 0a = the speed is constant 8 st = to 11 st = . Figure 2.30a-b 2.31. (a) IDENTIFY and SET UP: The acceleration xa at time t is the slope of the tangent to the xv versus t curve at time t. EXECUTE: At 3 s,t = the xv versus t curve is a horizontal straight line, with zero slope. Thus 0.xa = At 7 s,t = the xv versus t curve is a straight-line segment with slope 245 m/s 20 m/s 6.3 m/s .9 s 5 s ? =? Thus 26.3 m/s .xa = At 11 st = the curve is again a straight-line segment, now with slope 20 45 m/s 11.2 m/s . 13 s 9 s ? ? = ?? Thus 211.2 m/s .xa = ? EVALUATE: 0xa = when xv is constant, 0xa > when xv is positive and the speed is increasing, and 0xa < when xv is positive and the speed is decreasing. (b) IDENTIFY: Calculate the displacement during the specified time interval. SET UP: We can use the constant acceleration equations only for time intervals during which the acceleration is constant. If necessary, break the motion up into constant acceleration segments and apply the constant acceleration equations for each segment. For the time interval 0t = to 5 st = the acceleration is constant and equal to zero. For the time interval 5 st = to 9 st = the acceleration is constant and equal to 26.25 m/s . For the interval 9 st = to 13 st = the acceleration is constant and equal to 211.2 m/s .? EXECUTE: During the first 5 seconds the acceleration is constant, so the constant acceleration kinematic formulas can be used. 0 20 m/sxv = 0xa = 5 st = 0 ?x x? = 0 0xx x v t? = ( 0xa = so no 212 xa t term) 0 (20 m/s)(5 s) 100 m;x x? = = this is the distance the officer travels in the first 5 seconds. During the interval 5 st = to 9 s the acceleration is again constant. The constant acceleration formulas can be applied to this 4 second interval. It is convenient to restart our clock so the interval starts at time 0t = and ends at time 5 s.t = (Note that the acceleration is not constant over the entire 0t = to 9 st = interval.) 0 20 m/sxv = 26.25 m/sxa = 4 st = 0 100 mx = 0 ?x x? = 21 0 0 2x xx x v t a t? = + 2 21 0 2(20 m/s)(4 s) (6.25 m/s )(4 s) 80 m 50 m 130 m.x x? = + = + = Thus 0 130 m 100 m 130 m 230 m.x x? + = + = 2-12 Chapter 2 At 9 st = the officer is at 230 m,x = so she has traveled 230 m in the first 9 seconds. During the interval 9 st = to 13 st = the acceleration is again constant. The constant acceleration formulas can be applied for this 4 second interval but not for the whole 0t = to 13 st = interval. To use the equations restart our clock so this interval begins at time 0t = and ends at time 4 s.t = 0 45 m/sxv = (at the start of this time interval) 211.2 m/sxa = ? 4 st = 0 230 mx = 0 ?x x? = 21 0 0 2x xx x v t a t? = + 2 21 0 2(45 m/s)(4 s) ( 11.2 m/s )(4 s) 180 m 89.6 m 90.4 m.x x? = + ? = ? = Thus 0 90.4 m 230 m 90.4 m 320 m.x x= + = + = At 13 st = the officer is at 320 m,x = so she has traveled 320 m in the first 13 seconds. EVALUATE: The velocity xv is always positive so the displacement is always positive and displacement and distance traveled are the same. The average velocity for time interval t? is av- / .xv x t= ? ? For 0t = to 5 s, av- 20 m/s.xv = For 0t = to 9 s, av- 26 m/s.xv = For 0t = to 13 s, av- 25 m/s.xv = These results are consistent with Fig. 2.33. 2.32. IDENTIFY: In each constant acceleration interval, the constant acceleration equations apply. SET UP: When xa is constant, the graph of xv versus t is a straight line and the graph of x versus t is a parabola. When 0xa = , xv is constant and x versus t is a straight line. EXECUTE: The graphs are given in Figure 2.32. EVALUATE: The slope of the x versus t graph is ( )xv t and the slope of the xv versus t graph is ( )xa t . Figure 2.32 2.33. (a) IDENTIFY: The maximum speed occurs at the end of the initial acceleration period. SET UP: 220.0 m/sxa = 15.0 min 900 st = = 0 0xv = ?xv = 0x x xv v a t= + EXECUTE: 2 40 (20.0 m/s )(900 s) 1.80 10 m/sxv = + = × (b) IDENTIFY: Use constant acceleration formulas to find the displacement .x? The motion consists of three constant acceleration intervals. In the middle segment of the trip 0xa = and 41.80 10 m/s,xv = × but we can?t directly find the distance traveled during this part of the trip because we don?t know the time. Instead, find the distance traveled in the first part of the trip (where 220.0 m/sxa = + ) and in the last part of the trip (where 220.0 m/sxa = ? ). Subtract these two distances from the total distance of 83.84 10 m× to find the distance traveled in the middle part of the trip (where 0).xa = first segment SET UP: 0 ?x x? = 15.0 min 900 st = = 220.0 m/sxa = + 0 0xv = 21 0 0 2x xx x v t a t? = + EXECUTE: 2 2 6 310 20 (20.0 m/s )(900 s) 8.10 10 m 8.10 10 kmx x? = + = × = × second segment SET UP: 0 ?x x? = 15.0 min 900 st = = 220.0 m/sxa = ? 4 0 1.80 10 m/sxv = × 21 0 0 2x xx x v t a t? = + EXECUTE: 2 2 6 310 2(1.80 10 s)(900 s) ( 20.0 m/s )(900 s) 8.10 10 m 8.10 10 kmx x 4? = × + ? = × = × (The same distance as traveled as in the first segment.) Motion Along a Straight Line 2-13 Therefore, the distance traveled at constant speed is 8 6 6 8 53.84 10 m 8.10 10 m 8.10 10 m 3.678 10 m 3.678 10 km.× ? × ? × = × = × The fraction this is of the total distance is 8 8 3.678 10 m 0.958. 3.84 10 m × =× (c) IDENTIFY: We know the time for each acceleration period, so find the time for the constant speed segment. SET UP: 80 3.678 10 mx x? = × 41.80 10 m/sxv = × 0xa = ?t = 21 0 0 2x xx x v t a t? = + EXECUTE: 8 40 4 0 3.678 10 m 2.043 10 s 340.5 min. 1.80 10 m/sx x x t v ? ×= = = × =× The total time for the whole trip is thus 15.0 min 340.5 min 15.0 min 370min.+ + = EVALUATE: If the speed was a constant 41.80 10 m/s× for the entire trip, the trip would take 8 4(3.84 10 m)/(1.80 10 m/s) 356 min.× × = The trip actually takes a bit longer than this since the average velocity is less than 81.80 10 m/s× during the relatively brief acceleration phases. 2.34. IDENTIFY: Use constant acceleration equations to find 0x x? for each segment of the motion. SET UP: Let x+ be the direction the train is traveling. EXECUTE: 0t = to 14.0 s: 2 2 21 10 0 2 2 (1.60 m/s )(14.0 s) 157 mx xx x v t a t? = + = = . At 14.0 st = , the speed is 20 (1.60 m/s )(14.0 s) 22.4 m/sx x xv v a t= + = = . In the next 70.0 s, 0xa = and 0 0 (22.4 m/s)(70.0 s) 1568 mxx x v t? = = = . For the interval during which the train is slowing down, 0 22.4 m/sxv = , 23.50 m/sxa = ? and 0xv = . 2 2 0 02 ( )x x xv v a x x= + ? gives 2 2 2 0 0 2 0 (22.4 m/s) 72 m 2 2( 3.50 m/s ) x x x v v x x a ? ?? = = =? . The total distance traveled is 157 m 1568 m 72 m 1800 m+ + = . EVALUATE: The acceleration is not constant for the entire motion but it does consist of constant acceleration segments and we can use constant acceleration equations for each segment. 2.35 IDENTIFY: ( )xv t is the slope of the x versus t graph. Car B moves with constant speed and zero acceleration. Car A moves with positive acceleration; assume the acceleration is constant. SET UP: For car B, xv is positive and 0xa = . For car A, xa is positive and xv increases with t. EXECUTE: (a) The motion diagrams for the cars are given in Figure 2.35a. (b) The two cars have the same position at times when their x-t graphs cross. The figure in the problem shows this occurs at approximately 1 st = and 3 st = . (c) The graphs of xv versus t for each car are sketched in Figure 2.35b. (d) The cars have the same velocity when their x-t graphs have the same slope. This occurs at approximately 2 st = . (e) Car A passes car B when Ax moves above Bx in the x-t graph. This happens at 3 st = . (f) Car B passes car A when Bx moves above Ax in the x-t graph. This happens at 1 st = . EVALUATE: When 0xa = , the graph of xv versus t is a horizontal line. When xa is positive, the graph of xv versus t is a straight line with positive slope. Figure 2.35a-b 2.36. IDENTIFY: Apply the constant acceleration equations to the motion of each vehicle. The truck passes the car when they are at the same x at the same 0t > . 2-14 Chapter 2 SET UP: The truck has 0xa = . The car has 0 0xv = . Let x+ be in the direction of motion of the vehicles. Both vehicles start at 0 0x = . The car has 2C 3.20 m/sa = . The truck has 20.0 m/sxv = . EXECUTE: (a) 210 0 2x xx x v t a t? = + gives T 0Tx v t= and 21C C2x a t= . Setting T Cx x= gives 0t = and 10T C2v a t= , so 0T 2 C 2 2(20.0 m/s) 12.5 s 3.20 m/s v t a = = = . At this t, T (20.0 m/s)(12.5 s) 250 mx = = and 2 212 (3.20 m/s )(12.5 s) 250 mx = = . The car and truck have each traveled 250 m. (b) At 12.5 st = , the car has 20 (3.20 m/s )(12.5 s) 40 m/sx x xv v a t= + = = . (c) T 0Tx v t= and 21C C2x a t= . The x-t graph of the motion for each vehicle is sketched in Figure 2.36a. (d) T 0Tv v= . C Cv a t= . The -xv t graph for each vehicle is sketched in Figure 2.36b. EVALUATE: When the car overtakes the truck its speed is twice that of the truck. Figure 2.36a-b 2.37. IDENTIFY: For constant acceleration, Eqs. (2.8), (2.12), (2.13) and (2.14) apply. SET UP: Take y+ to be downward, so the motion is in the y+ direction. 19,300 km/h 5361 m/s= , 1600 km/h 444.4 m/s= , and 321 km/h 89.2 m/s= . 4.0 min 240 s= . EXECUTE: (a) Stage A: 240 st = , 0 5361 m/syv = , 444.4 m/syv = . 0y y yv v a t= + gives 0 2444.4 m/s 5361 m/s 20.5 m/s 240 s y y y v v a t ? ?= = = ? . Stage B: 94 st = , 0 444.4 m/syv = , 89.2 m/syv = . 0y y yv v a t= + gives 0 289.2 m/s 444.4 m/s 3.8 m/s 94 s y y y v v a t ? ?= = = ? . Stage C: 0 75 my y? = , 0 89.2 m/syv = , 0yv = . 2 20 02 ( )y y yv v a y y= + ? gives 2 2 2 0 2 0 0 (89.2 m/s) 53.0 m/s 2( ) 2(75 m) y y y v v a y y ? ?= = = ?? . In each case the negative sign means that the acceleration is upward. (b) Stage A: 00 5361 m/s 444.4 m/s (240 s) 697 km 2 2 y yv vy y t +? ? +? ?? = = =? ? ? ?? ?? ? . Stage B: 0 444.4 m/s 89.2 m/s (94 s) 25 km 2 y y +? ?? = =? ?? ? . Stage C: The problem states that 0 75 m = 0.075 kmy y? = . The total distance traveled during all three stages is 697 km 25 km 0.075 km 722 km+ + = . EVALUATE: The upward acceleration produced by friction in stage A is calculated to be greater than the upward acceleration due to the parachute in stage B. The effects of air resistance increase with increasing speed and in reality the acceleration was probably not constant during stages A and B. 2.38. IDENTIFY: Assume an initial height of 200 m and a constant acceleration of 29.80 m/s . SET UP: Let y+ be downward. 1 km/h 0.2778 m/s= and 1 mi/h 0.4470 m/s= . Motion Along a Straight Line 2-15 EXECUTE: (a) 0 200 my y? = , 29.80 m/sya = , 0 0yv = . 2 20 02 ( )y y yv v a y y= + ? gives 22(9.80 m/s )(200 m) 60 m/s 200 km/h 140 mi/hyv = = = = . (b) Raindrops actually have a speed of about 1 m/s as they strike the ground. (c) The actual speed at the ground is much less than the speed calculated assuming free-fall, so neglect of air resistance is a very poor approximation for falling raindrops. EVALUATE: In the absence of air resistance raindrops would land with speeds that would make them very dangerous. 2.39. IDENTIFY: Apply the constant acceleration equations to the motion of the flea. After the flea leaves the ground, ,ya g= downward. Take the origin at the ground and the positive direction to be upward. (a) SET UP: At the maximum height 0.yv = 0yv = 0 0.440 my y? = 29.80 m/sya = ? 0 ?yv = 2 2 0 02 ( )y y yv v a y y= + ? EXECUTE: 20 02 ( ) 2( 9.80 m/s )(0.440 m) 2.94 m/sy yv a y y= ? ? = ? ? = (b) SET UP: When the flea has returned to the ground 0 0.y y? = 0 0y y? = 0 2.94 m/syv = + 29.80 m/sya = ? ?t = 21 0 0 2y yy y v t a t? = + EXECUTE: With 0 0y y? = this gives 0 2 2 2(2.94 m/s) 0.600 s. 9.80 m/s y y v t a = ? = ? =? EVALUATE: We can use 0y y yv v a t= + to show that with 0 2.94 m/s,yv = 0yv = after 0.300 s. 2.40. IDENTIFY: Apply constant acceleration equations to the motion of the lander. SET UP: Let y+ be positive. Since the lander is in free-fall, 21.6 m/sya = + . EXECUTE: 0 0.8 m/syv = , 0 5.0 my y? = , 21.6 m/sya = + in 2 20 02 ( )y y yv v a y y= + ? gives 2 2 2 0 02 ( ) (0.8 m/s) 2(1.6 m/s )(5.0 m) 4.1 m/sy y yv v a y y= + ? = + = . EVALUATE: The same descent on earth would result in a final speed of 9.9 m/s, since the acceleration due to gravity on earth is much larger than on the moon. 2.41. IDENTIFY: Apply constant acceleration equations to the motion of the meterstick. The time the meterstick falls is your reaction time. SET UP: Let y+ be downward. The meter stick has 0 0yv = and 29.80 m/sya = . Let d be the distance the meterstick falls. EXECUTE: (a) 210 0 2y yy y v t a t? = + gives 2 2(4.90 m/s )d t= and 24.90 m/s d t = . (b) 2 0.176 m 0.190 s 4.90 m/s t = = EVALUATE: The reaction time is proportional to the square of the distance the stick falls. 2.42. IDENTIFY: Apply constant acceleration equations to the vertical motion of the brick. SET UP: Let y+ be downward. 29.80 m/sya = EXECUTE: (a) 0 0yv = , 2.50 st = , 29.80 m/sya = . 2 2 21 10 0 2 2 (9.80 m/s )(2.50 s) 30.6 my yy y v t a t? = + = = . The building is 30.6 m tall. (b) 20 0 (9.80 m/s )(2.50 s) 24.5 m/sy y yv v a t= + = + = (c) The graphs of ya , yv and y versus t are given in Fig. 2.42. Take 0y = at the ground. 2-16 Chapter 2 EVALUATE: We could use either 00 2 y yv vy y t +? ?? = ? ?? ? or 2 20 02 ( )y y yv v a y y= + ? to check our results. Figure 2.42 2.43. IDENTIFY: When the only force is gravity the acceleration is 29.80 m/s , downward. There are two intervals of constant acceleration and the constant acceleration equations apply during each of these intervals. SET UP: Let y+ be upward. Let 0y = at the launch pad. The final velocity for the first phase of the motion is the initial velocity for the free-fall phase. EXECUTE: (a) Find the velocity when the engines cut off. 0 525 my y? = , 22.25 m/sya = + , 0 0yv = . 2 2 0 02 ( )y y yv v a y y= + ? gives 22(2.25 m/s )(525 m) 48.6 m/syv = = . Now consider the motion from engine cut off to maximum height: 0 525 my = , 0 48.6 m/syv = + , 0yv = (at the maximum height), 29.80 m/sya = ? . 2 20 02 ( )y y yv v a y y= + ? gives 2 2 2 0 0 2 0 (48.6 m/s) 121 m 2 2( 9.80 m/s ) y y y v v y y a ? ?? = = =? and 121 m 525 m 646 my = + = . (b) Consider the motion from engine failure until just before the rocket strikes the ground: 0 525 my y? = ? , 29.80 m/sya = ? , 0 48.6 m/syv = + . 2 20 02 ( )y y yv v a y y= + ? gives 2 2(48.6 m/s) 2( 9.80 m/s )( 525 m) 112 m/syv = ? + ? ? = ? . Then 0y y yv v a t= + gives 0 2 112 m/s 48.6 m/s 16.4 s 9.80 m/s y y y v v t a ? ? ?= = =? . (c) Find the time from blast-off until engine failure: 0 525 my y? = , 0 0yv = , 22.25 m/sya = + . 21 0 0 2y yy y v t a t? = + gives 0 22( ) 2(525 m) 21.6 s2.25 m/sy y y t a ?= = = . The rocket strikes the launch pad 21.6 s 16.4 s 38.0 s+ = after blast off. The acceleration ya is 22.25 m/s+ from 0t = to 21.6 st = . It is 29.80 m/s? from 21.6 st = to 38.0 s . 0y y yv v a t= + applies during each constant acceleration segment, so the graph of yv versus t is a straight line with positive slope of 22.25 m/s during the blast-off phase and with negative slope of 29.80 m/s? after engine failure. During each phase 210 0 2y yy y v t a t? = + . The sign of ya determines the curvature of ( )y t . At 38.0 st = the rocket has returned to 0y = . The graphs are sketched in Figure 2.43. EVALUATE: In part (b) we could have found the time from 210 0 2y yy y v t a t? = + , finding yv first allows us to avoid solving for t from a quadratic equation. Figure 2.43 Motion Along a Straight Line 2-17 2.44. IDENTIFY: Apply constant acceleration equations to the vertical motion of the sandbag. SET UP: Take y+ upward. 29.80 m/sya = ? . The initial velocity of the sandbag equals the velocity of the balloon, so 0 5.00 m/syv = + . When the balloon reaches the ground, 0 40.0 my y? = ? . At its maximum height the sandbag has 0yv = . EXECUTE: (a) 0.250 st = : 2 2 21 10 0 2 2(5.00 m/s)(0.250 s) ( 9.80 m/s )(0.250 s) 0.94 my yy y v t a t? = + = + ? = . The sandbag is 40.9 m above the ground. 20 5.00 m/s ( 9.80 m/s )(0.250 s) 2.55 m/sy y yv v a t= + = + + ? = . 1.00 st = : 2 210 2(5.00 m/s)(1.00 s) ( 9.80 m/s )(1.00 s) 0.10 my y? = + ? = . The sandbag is 40.1 m above the ground. 20 5.00 m/s ( 9.80 m/s )(1.00 s) 4.80 m/sy y yv v a t= + = + + ? = ? . (b) 0 40.0 my y? = ? , 0 5.00 m/syv = , 29.80 m/sya = ? . 210 0 2y yy y v t a t? = + gives 2 240.0 m (5.00 m/s) (4.90 m/s )t t? = ? . 2 2(4.90 m/s ) (5.00 m/s) 40.0 m 0t t? ? = and ( )21 5.00 ( 5.00) 4(4.90)( 40.0) s (0.51 2.90) s9.80t = ± ? ? ? = ± . t must be positive, so 3.41 st = . (c) 20 5.00 m/s ( 9.80 m/s )(3.41 s) 28.4 m/sy y yv v a t= + = + + ? = ? (d) 0 5.00 m/syv = , 29.80 m/sya = ? , 0yv = . 2 20 02 ( )y y yv v a y y= + ? gives 2 2 2 0 0 2 0 (5.00 m/s) 1.28 m 2 2( 9.80 m/s ) y y y v v y y a ? ?? = = =? . The maximum height is 41.3 m above the ground. (e) The graphs of ya , yv , and y versus t are given in Fig. 2.44. Take 0y = at the ground . EVALUATE: The sandbag initially travels upward with decreasing velocity and then moves downward with increasing speed. Figure 2.44 2.45. IDENTIFY: The balloon has constant acceleration ,ya g= downward. (a) SET UP: Take the y+ direction to be upward. 2.00 s,t = 0 6.00 m/s,yv = ? 29.80 m/s ,ya = ? ?yv = EXECUTE: 20 6.00 m/s ( 9.80 m/s )(2.00 s) 25.5 m/sy y yv v a t= + = ? + ? = ? (b) SET UP: 0 ?y y? = EXECUTE: 2 21 10 0 2 2( 6.00 m/s)(2.00 s) ( 9.80 m/s )(2.00 s) 31.6 my yy y v t a t 2? = + = ? + ? = ? (c) SET UP: 0 10.0 m,y y? = ? 0 6.00 m/s,yv = ? 29.80 m/s ,ya = ? ?yv = 2 2 0 02 ( )y y yv v a y y= + ? EXECUTE: 2 2 20 02 ( ) ( 6.00 m/s) 2( 9.80 m/s )( 10.0 m) 15.2 m/sy y yv v a y y= ? + ? = ? ? + ? ? = ? (d) The graphs are sketched in Figure 2.45. Figure 2.45 EVALUATE: The speed of the balloon increases steadily since the acceleration and velocity are in the same direction. 25.5 m/syv = when 0 31.6 m,y y? = so yv is less than this (15.2 m/s) when 0y y? is less (10.0 m). 2-18 Chapter 2 2.46. IDENTIFY: Since air resistance is ignored, the egg is in free-fall and has a constant downward acceleration of magnitude 29.80 m/s . Apply the constant acceleration equations to the motion of the egg. SET UP: Take y+ to be upward. At the maximum height, 0yv = . EXECUTE: (a) 0 50.0 my y? = ? , 5.00 st = , 29.80 m/sya = ? . 210 0 2y yy y v t a t? = + gives 20 1 1 0 2 2 50.0 m ( 9.80 m/s )(5.00 s) 14.5 m/s 5.00 sy y y y v a t t ? ?= ? = ? ? = + . (b) 0 14.5 m/syv = + , 0yv = (at the maximum height), 29.80 m/sya = ? . 2 20 02 ( )y y yv v a y y= + ? gives 2 2 2 0 0 2 0 (14.5 m/s) 10.7 m 2 2( 9.80 m/s ) y y y v v y y a ? ?? = = =? . (c) At the maximum height 0yv = . (d) The acceleration is constant and equal to 29.80 m/s , downward, at all points in the motion, including at the maximum height. (e) The graphs are sketched in Figure 2.46. EVALUATE: The time for the egg to reach its maximum height is 0 214.5 m/s 1.48 s9.8 m/s y y y v v t a ? ?= = =? . The egg has returned to the level of the cornice after 2.96 s and after 5.00 s it has traveled downward from the cornice for 2.04 s. Figure 2.46 2.47. IDENTIFY: Use the constant acceleration equations to calculate xa and 0.x x? (a) SET UP: 224 m/s,xv = 0 0,xv = 0.900 s,t = ?xa = 0x x xv v a t= + EXECUTE: 20 224 m/s 0 249 m/s 0.900 s x x x v v a t ? ?= = = (b) 2 2/ (249 m/s ) /(9.80 m/s ) 25.4xa g = = (c) 2 2 21 10 0 2 20 (249 m/s )(0.900 s) 101 mx xx x v t a t? = + = + = (d) SET UP: Calculate the acceleration, assuming it is constant: 1.40 s,t = 0 283 m/s,xv = 0xv = (stops), ?xa = 0x x xv v a t= + EXECUTE: 20 0 283 m/s 202 m/s 1.40 s x x x v v a t ? ?= = = ? 2 2/ ( 202 m/s ) /(9.80 m/s ) 20.6;xa g = ? = ? 20.6xa g= ? If the acceleration while the sled is stopping is constant then the magnitude of the acceleration is only 20.6g . But if the acceleration is not constant it is certainly possible that at some point the instantaneous acceleration could be as large as 40g . EVALUATE: It is reasonable that for this motion the acceleration is much larger than g . 2.48. IDENTIFY: Since air resistance is ignored, the boulder is in free-fall and has a constant downward acceleration of magnitude 29.80 m/s . Apply the constant acceleration equations to the motion of the boulder. SET UP: Take y+ to be upward. EXECUTE: (a) 0 40.0 m/syv = + , 20.0 m/syv = + , 29.80 m/sya = ? . 0y y yv v a t= + gives 0 2 20.0 m/s 40.0 m/s 2.04 s 9.80 m/s y y y v v t a ? ?= = = +? . Motion Along a Straight Line 2-19 (b) 20.0 m/syv = ? . 0 220.0 m/s 40.0 m/s 6.12 s9.80 m/s y y y v v t a ? ? ?= = = +? . (c) 0 0y y? = , 0 40.0 m/syv = + , 29.80 m/sya = ? . 210 0 2y yy y v t a t? = + gives 0t = and 0 2 2 2(40.0 m/s) 8.16 s 9.80 m/s y y v t a = ? = ? = +? . (d) 0yv = , 0 40.0 m/syv = + , 29.80 m/sya = ? . 0y y yv v a t= + gives 0 20 40.0 m/s 4.08 s9.80 m/s y y y v v t a ? ?= = =? . (e) The acceleration is 29.80 m/s , downward, at all points in the motion. (f) The graphs are sketched in Figure 2.48. EVALUATE: 0yv = at the maximum height. The time to reach the maximum height is half the total time in the air, so the answer in part (d) is half the answer in part (c). Also note that 2.04 s 4.08 s 6.12 s< < . The boulder is going upward until it reaches its maximum height and after the maximum height it is traveling downward. Figure 2.48 2.49. IDENTIFY: We can avoid solving for the common height by considering the relation between height, time of fall and acceleration due to gravity and setting up a ratio involving time of fall and acceleration due to gravity. SET UP: Let Eng be the acceleration due to gravity on Enceladus and let g be this quantity on earth. Let h be the common height from which the object is dropped. Let y+ be downward, so 0y y h? = . 0 0yv = EXECUTE: 210 0 2y yy y v t a t? = + gives 21 E2h gt= and 21 En En2h g t= . Combining these two equations gives 2 2 E En Eng t g t= and 2 2 2 2E En En 1.75 s (9.80 m/s ) 0.0868 m/s 18.6 s t g g t ? ? ? ?= = =? ? ? ?? ?? ? . EVALUATE: The acceleration due to gravity is inversely proportional to the square of the time of fall. 2.50. IDENTIFY: The acceleration is not constant so the constant acceleration equations cannot be used. Instead, use Eqs.(2.17) and (2.18). Use the values of xv and of x at 1.0 st = to evaluate 0xv and 0x . SET UP: 11 1 n nt dt t n += +? , for 0n ? . EXECUTE: (a) 2 3 210 0 020 (0.60 m/s ) t x x x xv v tdt v t v t? ?= + = + = +? . 5.0 m/sxv = when 1.0 st = gives 0 4.4 m/sxv = . Then, at 2.0 st = , 3 24.4 m/s (0.60 m/s )(2.0 s) 6.8 m/sxv = + = . (b) 2 310 0 0 020 1 ( ) 6 t x xx x v t dt x v t t? ?= + + = + +? . 6.0 mx = at 1.0 st = gives 0 1.4 mx = . Then, at 2.0 st = , 3 311.4 m (4.4 m/s)(2.0 s) (1.24 m/s )(2.0 s) 11.8 m 6 x = + + = . (c) 3 3( ) 1.4 m (4.4 m/s) (0.20 m/s )x t t t= + + . 3 2( ) 4.4 m/s (0.60 m/s )xv t t= + . 3( ) (1.20m/s )xa t t= . The graphs are sketched in Figure 2.50. 2-20 Chapter 2 EVALUATE: We can verify that xx dva dt= and x dx v dt = . Figure 2.50 2.51. 2xa At Bt= ? with 31.50 m/sA = and 40.120 m/sB = (a) IDENTIFY: Integrate ( )xa t to find ( )xv t and then integrate ( )xv t to find ( ).x t SET UP: 0 0 t x x xv v a dt= + ? EXECUTE: 2 2 31 10 0 2 30 ( ) t x x xv v At Bt dt v At Bt= + ? = + ?? At rest at 0t = says that 0 0,xv = so 2 3 3 2 41 1 1 1 2 3 2 3(1.50 m/s ) (0.120 m/s )xv At Bt t t 3= ? = ? 3 2 4 3(0.75 m/s ) (0.040 m/s )xv t t= ? SET UP: 0 0 t xx x v dt? + ? EXECUTE: ( )2 3 3 41 1 1 10 02 3 6 120 tx x At Bt dt x At Bt= + ? = + ?? At the origin at 0t = says that 0 0,x = so 3 4 3 3 4 41 1 1 1 6 12 6 12(1.50 m/s ) (0.120 m/s )x At Bt t t= ? = ? 3 3 4 4(0.25 m/s ) (0.010 m/s )x t t= ? EVALUATE: We can check our results by using them to verify that ( )x dxv t dt= and ( ) . x x dv a t dt = (b) IDENTIFY and SET UP: At time t, when xv is a maximum, 0.xdvdt = (Since , x x dv a dt = the maximum velocity is when 0.xa = For earlier times xa is positive so xv is still increasing. For later times xa is negative and xv is decreasing.) EXECUTE: 0xx dva dt= = so 2 0At Bt? = One root is 0,t = but at this time 0xv = and not a maximum. The other root is 3 4 1.50 m/s 12.5 s 0.120 m/s A t B = = = At this time 3 2 4 3(0.75 m/s ) (0.040 m/s )xv t t= ? gives 3 2 4 3(0.75 m/s )(12.5 s) (0.040 m/s )(12.5 s) 117.2 m/s 78.1 m/s 39.1 m/s.xv = ? = ? = EVALUATE: For 12.5 s,t < 0xa > and xv is increasing. For 12.5 s,t > 0xa < and xv is decreasing. 2.52. IDENTIFY: ( )a t is the slope of the v versus t graph and the distance traveled is the area under the v versus t graph. SET UP: The v versus t graph can be approximated by the graph sketched in Figure 2.52. EXECUTE: (a) Slope 0 for 1.3 msa t= = ? . (b) max Area under - graphh v t= Triangle RectangleA A? + ( )1 (1.3 ms) 133 cm/s (2.5 ms 1.3 ms)(133 cm s)2? + ? 0.25 cm? (c) slopea = of v-t graph. 25133cm s(0.5 ms) (1.0 ms) 1.0 10 cm s 1.3ms a a? ? = × . (1.5 ms) 0 because the slope is zero.a = Motion Along a Straight Line 2-21 (d) areah = under v-t graph. ( ) 3Triangle 1(0.5 ms) (0.5 ms) 33 cm/s 8.3 10 cm2h A ?? = = × . 2 Triangle 1 (1.0 ms) (1.0 ms)(100 cm s) 5.0 10 cm 2 h A ?? = = × . ( )Triangle Rectangle 1(1.5 ms) (1.3 ms) 133 cm/s (0.2 ms)(1.33) 0.11 cm2h A A? + = = EVALUATE: The acceleration is constant until 1.3 mst = , and then it is zero. 2980 cm/sg = . The acceleration during the first 1.3 ms is much larger than this and gravity can be neglected for the portion of the jump that we are considering. Figure 2.52 2.53. (a) IDENTIFY and SET UP: The change in speed is the area under the xa versus t curve between vertical lines at 2.5 st = and 7.5 s.t = EXECUTE: This area is 2 212 (4.00 cm/s 8.00 cm/s )(7.5 s 2.5 s) 30.0 cm/s+ ? = This acceleration is positive so the change in velocity is positive. (b) Slope of xv versus t is positive and increasing with t. The graph is sketched in Figure 2.53. Figure 2.53 EVALUATE: The calculation in part (a) is equivalent to av-( ) .x xv a t? = ? Since xa is linear in t, av- 0( ) / 2.x x xa a a= + Thus 2 21av- 2 (4.00 cm/s 8.00 cm/s )xa = + for the time interval 2.5 st = to 7.5 s.t = 2.54. IDENTIFY: The average speed is the total distance traveled divided by the total time. The elapsed time is the distance traveled divided by the average speed. SET UP: The total distance traveled is 20 mi. With an average speed of 8 mi/h for 10 mi, the time for that first 10 miles is 10 mi 1.25 h 8 mi/h = . EXECUTE: (a) An average speed of 4 mi/h for 20 mi gives a total time of 20 mi 5.0 h 4 mi/h = . The second 10 mi must be covered in 5.0 h 1.25 h 3.75 h? = . This corresponds to an average speed of 10 mi 2.7 mi/h 3.75 h = . (b) An average speed of 12 mi/h for 20 mi gives a total time of 20 mi 1.67 h 12 mi/h = . The second 10 mi must be covered in 1.67 h 1.25 h 0.42 h? = . This corresponds to an average speed of 10 mi 24 mi/h 0.42 h = . (c) An average speed of 16 mi/h for 20 mi gives a total time of 20 mi 1.25 h 16 mi/h = . But 1.25 h was already spent during the first 10 miles and the second 10 miles would have to be covered in zero time. This is not possible and an average speed of 16 mi/h for the 20-mile ride is not possible. EVALUATE: The average speed for the total trip is not the average of the average speeds for each 10-mile segment. The rider spends a different amount of time traveling at each of the two average speeds. 2.55. IDENTIFY: ( )x dxv t dt= and x x dv a dt = . SET UP: 1( )n nd t nt dt ?= , for 1n ? . EXECUTE: (a) 3 2 2( ) (9.00 m/s ) (20.0 m/s ) 9.00 m/sxv t t t= ? + . 3 2( ) (18.0 m/s ) 20.0 m/sxa t t= ? . The graphs are sketched in Figure 2.55. 2-22 Chapter 2 (b) The particle is instantaneously at rest when ( ) 0xv t = . 0 0xv = and the quadratic formula gives 21 (20.0 (20.0) 4(9.00)(9.00)) s 1.11 s 0.48 s 18.0 t = ± ? = ± . 0.63 st = and 1.59 st = . These results agree with the -xv t graphs in part (a). (c) For 0.63 st = , 3 2 2(18.0 m/s )(0.63 s) 20.0 m/s 8.7 m/sxa = ? = ? . For 1.59 st = , 28.6 m/sxa = + . At 0.63 st = the slope of the -xv t graph is negative and at 1.59 st = it is positive, so the same answer is deduced from the ( )xv t graph as from the expression for ( )xa t . (d) ( )xv t is instantaneously not changing when 0xa = . This occurs at 2 3 20.0 m/s 1.11 s 18.0m/s t = = . (e) When the particle is at its greatest distance from the origin, 0xv = and 0xa < (so the particle is starting to move back toward the origin). This is the case for 0.63 st = , which agrees with the x-t graph in part (a) . At 0.63 st = , 2.45 mx = . (f) The particle?s speed is changing at its greatest rate when xa has its maximum magnitude. The -xa t graph in part (a) shows this occurs at 0t = and at 2.00 st = . Since xv is always positive in this time interval, the particle is speeding up at its greatest rate when xa is positive, and this is for 2.00 st = . The particle is slowing down at its greatest rate when xa is negative and this is for 0t = . EVALUATE: Since ( )xa t is linear in t, ( )xv t is a parabola and is symmetric around the point where ( )xv t has its minimum value ( 1.11 st = ). For this reason, the answer to part (d) is midway between the two times in part (c). Figure 2.55 2.56. IDENTIFY: The average velocity is av-x xv t ?= ? . The average speed is the distance traveled divided by the elapsed time. SET UP: Let x+ be in the direction of the first leg of the race. For the round trip, 0x? ? and the total distance traveled is 50.0 m. For each leg of the race both the magnitude of the displacement and the distance traveled are 25.0 m. EXECUTE: (a) av- 25.0 m 1.25 m/s20.0 sx x v t ?= = =? . This is the same as the average speed for this leg of the race. (b) av- 25.0 m 1.67 m/s 15.0 sx x v t ?= = =? . This is the same as the average speed for this leg of the race. (c) 0x? = so av- 0xv = . (d) The average speed is 50.0 m 1.43 m/s 35.0 s = . EVALUATE: Note that the average speed for the round trip is not equal to the arithmetic average of the average speeds for each leg. 2.57. IDENTIFY: Use information about displacement and time to calculate average speed and average velocity. Take the origin to be at Seward and the positive direction to be west. (a) SET UP: distance traveledaverage speed time = EXECUTE: The distance traveled (different from the net displacement 0( )x x? ) is 76 km 34 km 110 km.+ = Find the total elapsed time by using 0av-x x x x v t t ? ?= =? to find t for each leg of the journey. Seward to Auora: 0 av- 76 km 0.8636 h 88 km/hx x x t v ?= = = Motion Along a Straight Line 2-23 Auora to York: 0 av- 34 km 0.4722 h 72 km/hx x x t v ? ?= = =? Total 0.8636 h 0.4722 h 1.336 h.t = + = Then 110 km average speed 82 km/h. 1.336 h = = (b) SET UP: av- ,x xv t ?= ? where x? is the displacement, not the total distance traveled. For the whole trip he ends up 76 km 34 km 42 km? = west of his starting point. av- 42 km 31 km/h.l.336 hxv = = EVALUATE: The motion is not uniformly in the same direction so the displacement is less than the distance traveled and the magnitude of the average velocity is less than the average speed. 2.58. IDENTIFY: The vehicles are assumed to move at constant speed. The speed (mi/h) divided by the frequency with which vehicles pass a given point (vehicles/h) is the total space per vehicle (the length of the vehicle plus space to the next vehicle). SET UP: 396 km/h 96 10 m/h= × EXECUTE: (a) The total space per vehicle is 396 10 m/h 40 m/vehicle 2400 vehicles/h × = . Since the average length of a vehicle is 4.6 m, the average space between vehicles is 40 m 4.6 m 35 m? = . (b) The frequency of vehicles (vehicles/h) is 396 10 m/h 7000 vehicles/h (4.6 9.2) m/vehicle × =+ . EVALUATE: The traffic flow rate per lane would nearly triple. Note that the traffic flow rate is directly proportional to the traffic speed. 2.59. (a) IDENTIFY: Calculate the average acceleration using 0av- x x xx v v va t t ? ?= =? Use the information about the time and total distance to find his maximum speed. SET UP: 0 0xv = since the runner starts from rest. 4.0 s,t = but we need to calculate ,xv the speed of the runner at the end of the acceleration period. EXECUTE: For the last 9.1 s 4.0 s 5.1 s? = the acceleration is zero and the runner travels a distance of 1 (5.1 s) xd v= (obtained using 210 0 2 )x xx x v t a t? = + During the acceleration phase of 4.0 s, where the velocity goes from 0 to ,xv the runner travels a distance 0 2 (4.0 s) (2.0 s)2 2 x x x x v v v d t v +? ?= = =? ?? ? The total distance traveled is 100 m, so 1 2 100 m.d d+ = This gives (5.1 s) (2.0 s) 100 m.x xv v+ = 100 m 14.08 m/s. 7.1 sx v = = Now we can calculate av- :xa 20 av- 14.08 s 0 3.5 m/s . 4.0 s x x x v v a t ? ?= = = (b) For this time interval the velocity is constant, so av 0.xa ? = EVALUATE: Now that we have xv we can calculate 1 (5.1 s)(14.08 m/s) 71.9 md = = and 2 (2.0 s)(14.08 m/s) 28.2 m.d = = So, 1 2 100 m,d d+ = which checks. (c) IDENTIFY and SET UP: 0av- ,x xx v va t ?= where now the time interval is the full 9.1 s of the race. We have calculated the final speed to be 14.08 m/s, so 2 av- 14.08 m/s 1.5 m/s . 9.1 sx a = = EVALUATE: The acceleration is zero for the last 5.1 s, so it makes sense for the answer in part (c) to be less than half the answer in part (a). (d) The runner spends different times moving with the average accelerations of parts (a) and (b). 2.60. IDENTIFY: Apply the constant acceleration equations to the motion of the sled. The average velocity for a time interval t? is av-x xv t ?= ? . 2-24 Chapter 2 SET UP: Let x+ be parallel to the incline and directed down the incline. The problem doesn?t state how much time it takes the sled to go from the top to 14.4 m from the top. EXECUTE: (a) 14.4 m to 25.6 m: av- 25.6 m 14.4 m 5.60 m/s2.00 sxv ?= = . 25.6 to 40.0 m: av- 40.0 m 25.6 m 7.20 m/s 2.00 sx v ?= = . 40.0 m to 57.6 m: av- 57.6 m 40.0 m 8.80 m/s2.00 sxv ?= = . (b) For each segment we know 0x x? and t but we don?t know 0xv or xv . Let 1 14.4 mx = and 2 25.6 mx = . For this interval 1 2 2 1 2 v v x x t + ?? ? =? ?? ? and 2 1at v v= ? . Solving for 2v gives 2 11 2 2 x x v at t ?= + . Let 2 25.6 mx = and 3 40.0 mx = . For this second interval, 2 3 3 22 v v x x t + ?? ? =? ?? ? and 3 2at v v= ? . Solving for 2v gives 3 21 2 2 x x v at t ?= ? + . Setting these two expressions for 2v equal to each other and solving for a gives 2 3 2 2 12 2 1 1 [( ) ( )] [(40.0 m 25.6 m) (25.6 m 14.4 m)] 0.80 m/s (2.00 s) a x x x x t = ? ? ? = ? ? ? = . Note that this expression for a says av-23 av-12 v v a t ?= , where av-12v and av-23v are the average speeds for successive 2.00 s intervals. (c) For the motion from 14.4 mx = to 25.6 mx = , 0 11.2 mx x? = , 20.80 m/sxa = and 2.00 st = . 21 0 0 2x xx x v t a t? = + gives 20 10 2 11.2 m 1 (0.80 m/s )(2.00 s) 4.80 m/s2.00 s 2x x x x v a t t ?= ? = ? = . (d) For the motion from 0x = to 14.4 mx = , 0 14.4 mx x? = , 0 0xv = , and 4.8 m/sxv = . 0 0 2 x xv vx x t +? ?? = ? ?? ? gives 0 0 2( ) 2(14.4 m) 6.0 s 4.8 m/sx x x x t v v ?= = =+ . (e) For this 1.00 s time interval, 1.00 st = , 0 4.8 m/sxv = , 20.80 m/sxa = . 2 2 21 1 0 0 2 2(4.8 m/s)(1.00 s) (0.80 m/s )(1.00 s) 5.2 mx xx x v t a t? = + = + = . EVALUATE: With 0x = at the top of the hill, 2 2 210 2( ) (0.40 m/s )x xx t v t a t t= + = . We can verify that 6.0 st = gives 14.4 mx = , 8.0 st = gives 25.6 m, 10.0 st = gives 40.0 m, and 12.0 st = gives 57.6 m. 2.61. IDENTIFY: When the graph of xv versus t is a straight line the acceleration is constant, so this motion consists of two constant acceleration segments and the constant acceleration equations can be used for each segment. Since xv is always positive the motion is always in the x+ direction and the total distance moved equals the magnitude of the displacement. The acceleration xa is the slope of the xv versus t graph. SET UP: For the 0t = to 10.0 st = segment, 0 4.00 m/sxv = and 12.0 m/sxv = . For the 10.0 st = to 12.0 s segment, 0 12.0 m/sxv = and 0xv = . EXECUTE: (a) For 0t = to 10.0 st = , 00 4.00 m/s 12.0 m/s (10.0 s) 80.0 m2 2 x xv vx x t + +? ? ? ?? = = =? ? ? ?? ? ? ? . For 10.0 st = to 12.0 st = , 0 12.0 m/s 0 (2.00 s) 12.0 m2x x +? ?? = =? ?? ? . The total distance traveled is 92.0 m. (b) 0 80.0 m 12.0 m 92.0 mx x? = + = (c) For 0t = to 10.0 s, 212.0 m/s 4.00 m/s 0.800 m/s 10.0 sx a ?= = . For 10.0 st = to 10.2 s, 20 12.0 m/s 6.00 m/s 2.00 sx a ?= = ? . The graph of xa versus t is given in Figure 2.61. Motion Along a Straight Line 2-25 EVALUATE: When xv and xa are both positive, the speed increases. When xv is positive and xa is negative, the speed decreases. Figure 2.61 2.62. IDENTIFY: Since light travels at constant speed, d ct= SET UP: The distance from the earth to the sun is 111.50 10 m× . The distance from the earth to the moon is 83.84 10 m× . 186,000 mi/sc = . EXECUTE: (a) 18 154365 d 24 h 3600 s(3.0 10 m/s)(1 y) 9.5 10 m 1 y 1 d 1 h d ct ? ?? ?? ?= = × = ×? ?? ?? ?? ?? ?? ? (b) 8 9(3.0 10 m/s)(10 s) 0.30 md ct ?= = × = (c) 11 8 1.5 10 m 500 s 8 33 min 3.0 10 m s d t . c ×= = = =× (d) 8 8 2(3.84 10 m) 2.6 s 3.0 10 m s d t c ×= = =× (e) 93 10 mi 16,100 s 4 5 h 186,000 mi s d t . c ×= = = = EVALUATE: The speed of light is very large but it still takes light a measurable length of time to travel a large distance. 2.63. IDENTIFY: Speed is distance d divided by time t. The distance around a circular path is 2d R?= , where R is the radius of the circular path. SET UP: The radius of the earth is 6E 6.38 10 mR = × . The earth rotates once in 1 day 86,400 s= . The radius of the earth?s orbit around the sun is 111.50 10 m× and the earth completes this orbit in 71 year 3.156 10 s= × . The speed of light in vacuum is 83.00 10 m/sc = × . EXECUTE: (a) 6 E2 2 (6.38 10 m) 464 m/s 86,400 s d R v t t ? ? ×= = = = . (b) 11 4 7 2 2 (1.50 10 m) 2.99 10 m/s 3.156 10 s R v t ? ? ×= = = ×× . (c) The time for light to go around once is 6 E 8 2 2 (6.38 10 m) 0.1336 s c 3.00 10 m/s d R t c ? ? ×= = = =× . In 1.00 s light would go around the earth 1.00 s 7.49 times 0.1336 s = . EVALUATE: All these speeds are large compared to speeds of objects in our everyday experience. 2.64. IDENTIFY: When the graph of xv versus t is a straight line the acceleration is constant, so this motion consists of two constant acceleration segments and the constant acceleration equations can be used for each segment. For 0t = to 5.0 s, xv is positive and the ball moves in the x+ direction. For 5.0 st = to 20.0 s, xv is negative and the ball moves in the x? direction. The acceleration xa is the slope of the xv versus t graph. SET UP: For the 0t = to 5.0 st = segment, 0 0xv = and 30.0 m/sxv = . For the 5.0 st = to 20.0 st = segment, 0 20.0 m/sxv = ? and 0xv = . 2-26 Chapter 2 EXECUTE: (a) For 0t = to 5.0 s, 00 0 30.0 m/s (5.0 m/s) 75.0 m2 2 x xv vx x t + +? ? ? ?? = = =? ? ? ?? ? ? ? . The ball travels a distance of 75.0 m. For 5.0 st = to 20.0 s, 0 20.0 m/s 0 (15.0 m/s) 150.0 m2x x ? +? ?? = = ?? ?? ? . The total distance traveled is 75.0 m 150.0 m 225.0 m+ = . (b) The total displacement is 0 75.0 m +( 150.0 m) 75.0 mx x? = ? = ? . The ball ends up 75.0 m in the negative x- direction from where it started. (c) For 0t = to 5.0 s, 230.0 m/s 0 6.00 m/s 5.0 sx a ?= = . For 5.0 st = to 20.0 s, 20 ( 20.0 m/s) 1.33 m/s 15.0 sx a ? ?= = + . The graph of xa versus t is given in Figure 2.64. (d) The ball is in contact with the floor for a small but nonzero period of time and the direction of the velocity doesn't change instantaneously. So, no, the actual graph of ( )xv t is not really vertical at 5.00 s. EVALUATE: For 0t = to 5.0 s, both xv and xa are positive and the speed increases. For 5.0 st = to 20.0 s, xv is negative and xa is positive and the speed decreases. Since the direction of motion is not the same throughout, the displacement is not equal to the distance traveled. Figure 2.64 2.65. IDENTIFY and SET UP: Apply constant acceleration equations. Find the velocity at the start of the second 5.0 s; this is the velocity at the end of the first 5.0 s. Then find 0x x? for the first 5.0 s. EXECUTE: For the first 5.0 s of the motion, 0 0,xv = 5.0 s.t = 0x x xv v a t= + gives (5.0 s).x xv a= This is the initial speed for the second 5.0 s of the motion. For the second 5.0 s: 0 (5.0 s),x xv a= 5.0 s,t = 0 150 m.x x? = 21 0 0 2x xx x v t a t? = + gives 2 2150 m (25 s ) (12.5 s )x xa a= + and 24.0 m/sxa = Use this xa and consider the first 5.0 s of the motion: 2 2 21 1 0 0 2 20 (4.0 m/s )(5.0 s) 50.0 m.x xx x v t a t? = + = + = EVALUATE: The ball is speeding up so it travels farther in the second 5.0 s interval than in the first. In fact, 0x x? is proportional to t 2 since it starts from rest. If it goes 50.0 m in 5.0 s, in twice the time (10.0 s) it should go four times as far. In 10.0 s we calculated it went 50 m 150 m 200 m,+ = which is four times 50 m. 2.66. IDENTIFY: Apply 210 0 2x xx x v t a t? = + to the motion of each train. A collision means the front of the passenger train is at the same location as the caboose of the freight train at some common time. SET UP: Let P be the passenger train and F be the freight train. For the front of the passenger train 0 0x = and for the caboose of the freight train 0 200 mx = . For the freight train F 15.0 m/sv = and F 0a = . For the passenger train P 25.0 m/sv = and 2P 0.100 m/sa = ? . EXECUTE: (a) 210 0 2x xx x v t a t? = + for each object gives 21P P P2x v t a t= + and F F200 mx v t= + . Setting P Fx x= gives 21P P F2 200 mv t a t v t+ = + . 2 2(0.0500 m/s ) (10.0 m/s) 200 m 0t t? + = . The quadratic formula gives ( )21 10.0 (10.0) 4(0.0500)(200) s (100 77.5) s0.100t = + ± ? = ± . The collision occurs at 100 s 77.5 s 22.5 st = ? = . The equations that specify a collision have a physical solution (real, positive t), so a collision does occur. Motion Along a Straight Line 2-27 (b) 2 21P 2(25.0 m/s)(22.5 s) ( 0.100 m/s )(22.5 s) 537 mx = + ? = . The passenger train moves 537 m before the collision. The freight train moves (15.0 m/s)(22.5 s) 337 m= . (c) The graphs of Fx and Px versus t are sketched in Figure 2.66. EVALUATE: The second root for the equation for t, 177.5 st = is the time the trains would meet again if they were on parallel tracks and continued their motion after the first meeting. Figure 2.66 2.67. IDENTIFY: Apply constant acceleration equations to the motion of the two objects, you and the cockroach. You catch up with the roach when both objects are at the same place at the same time. Let T be the time when you catch up with the cockroach. SET UP: Take 0x = to be at the 0t = location of the roach and positive x to be in the direction of motion of the two objects. roach: 0 1.50 m/s,xv = 0,xa = 0 0,x = 1.20 m,x = t T= you: 0 0.80 m/s,xv = 0 0.90 m,x = ? 1.20 m,x = ,t T= ?xa = Apply 210 0 2x xx x v t a t? = + to both objects: EXECUTE: roach: 1.20 m (1.50 m/s) ,T= so 0.800 s.T = you: 2121.20 m ( 0.90 m) (0.80 m/s) xT a T? ? = + 21 22.10 m (0.80 m/s)(0.800 s) (0.800 s)xa= + 22.10 m 0.64 m (0.320 s ) xa= + 24.6 m/s .xa = EVALUATE: Your final velocity is 0 4.48 m/s.x x xv v a t= + = Then 00 2.10 m,2 x xv vx x t +? ?? = =? ?? ? which checks. You have to accelerate to a speed greater than that of the roach so you will travel the extra 0.90 m you are initially behind. 2.68. IDENTIFY: The insect has constant speed 15 m/s during the time it takes the cars to come together. SET UP: Each car has moved 100 m when they hit. EXECUTE: The time until the cars hit is 100 m 10 s 10 m/s = . During this time the grasshopper travels a distance of (15 m/s)(10 s) 150 m= . EVALUATE: The grasshopper ends up 100 m from where it started, so the magnitude of his final displacement is 100 m. This is less than the total distance he travels since he spends part of the time moving in the opposite direction. 2.69. IDENTIFY: Apply constant acceleration equations to each object. Take the origin of coordinates to be at the initial position of the truck, as shown in Figure 2.69a Let d be the distance that the auto initially is behind the truck, so 0(auto)x d= ? and 0(truck) 0.x = Let T be the time it takes the auto to catch the truck. Thus at time T the truck has undergone a displacement 0 40.0 m,x x? = so is at 0 40.0 m 40.0 m.x x= + = The auto has caught the truck so at time T is also at 40.0 m.x = Figure 2.69a 2-28 Chapter 2 (a) SET UP: Use the motion of the truck to calculate T : 0 40.0 m,x x? = 0 0xv = (starts from rest), 22.10 m/s ,xa = t T= 21 0 0 2x xx x v t a t? = + Since 0 0,xv = this gives 02( ) x x x t a ?= EXECUTE: 22(40.0 m) 6.17 s2.10 m/sT = = (b) SET UP: Use the motion of the auto to calculate d: 0 40.0 m ,x x d? = + 0 0,xv = 23.40 m/s ,xa = 6.17 st = 21 0 0 2x xx x v t a t? = + EXECUTE: 2 21240.0 m (3.40 m/s )(6.17 s)d + = 64.8 m 40.0 m 24.8 md = ? = (c) auto: 20 0 (3.40 m/s )(6.17 s) 21.0 m/sx x xv v a t= + = + = truck: 20 0 (2.10 m/s )(6.17 s) 13.0 m/sx x xv v a t= + = + = (d) The graph is sketched in Figure 2.69b. Figure 2.69b EVALUATE: In part (c) we found that the auto was traveling faster than the truck when they come abreast. The graph in part (d) agrees with this: at the intersection of the two curves the slope of the x-t curve for the auto is greater than that of the truck. The auto must have an average velocity greater than that of the truck since it must travel farther in the same time interval. 2.70. IDENTIFY: Apply the constant acceleration equations to the motion of each car. The collision occurs when the cars are at the same place at the same time. SET UP: Let x+ be to the right. Let 0x = at the initial location of car 1, so 01 0x = and 02x D= . The cars collide when 1 2x x= . 0 1 0xv = , 1x xa a= , 0 2 0xv v= ? and 2 0xa = . EXECUTE: (a) 210 0 2x xx x v t a t? = + gives 211 2 xx a t= and 2 0x D v t= ? . 1 2x x= gives 21 02 xa t D v t= ? . 21 02 0xa t v t D+ ? = . The quadratic formula gives ( )20 01 2 x x t v v a D a = ? ± + . Only the positive root is physical, so ( )20 01 2 x x t v v a D a = ? + + . (b) 21 0 02x xv a t v a D v= = + ? (c) The x-t and -xv t graphs for the two cars are sketched in Figure 2.70. Motion Along a Straight Line 2-29 EVALUATE: In the limit that 0xa = , 0 0D v t? = and 0/t D v= , the time it takes car 2 to travel distance D. In the limit that 0 0v = , 2 x D t a = , the time it takes car 1 to travel distance D. Figure 2.70 2.71. IDENTIFY: The average speed is the distance traveled divided by the time. The average velocity is av-x xv t ?= ? . SET UP: The distance the ball travels is half the circumference of a circle of diameter 50.0 cm so is 1 1 2 2 (50.0 cm) 78.5 cmd? ?= = . Let x+ be horizontally from the starting point toward the ending point, so x? equals the diameter of the bowl. EXECUTE: (a) The average speed is 12 78.5 cm 7.85 cm/s 10.0 s d t ? = = . (b) The average velocity is av- 50.0 cm 5.00 cm/s 10.0 sx x v t ?= = =? . EVALUATE: The average speed is greater than the magnitude of the average velocity, since the distance traveled is greater than the magnitude of the displacement. 2.72. IDENTIFY: xa is the slope of the xv versus t graph. x is the area under the xv versus t graph. SET UP: The slope of xv is positive and decreasing in magnitude. As xv increases, the displacement in a given amount of time increases. EXECUTE: The -xa t and x-t graphs are sketched in Figure 2.72. EVALUATE: xv is the slope of the x versus t graph. The ( )x t graph we sketch has zero slope at 0t = , the slope is always positive, and the slope initially increases and then approaches a constant. This behavior agrees with the ( )xv t that is given in the graph in the problem. Figure 2.72 2.73. IDENTIFY: Apply constant acceleration equations to each vehicle. SET UP: (a) It is very convenient to work in coordinates attached to the truck. Note that these coordinates move at constant velocity relative to the earth. In these coordinates the truck is at rest, and the initial velocity of the car is 0 0.xv = Also, the car?s acceleration in these coordinates is the same as in coordinates fixed to the earth. EXECUTE: First, let?s calculate how far the car must travel relative to the truck: The situation is sketched in Figure 2.73. Figure 2.73 2-30 Chapter 2 The car goes from 0 24.0 mx = ? to 51.5 m.x = So 0 75.5 mx x? = for the car. Calculate the time it takes the car to travel this distance: 20.600 m/s ,xa = 0 0,xv = 0 75.5 m,x x? = ?t = 21 0 0 2x xx x v t a t? = + 0 2 2( ) 2(75.5 m) 15.86 s 0.600 m/sx x x t a ?= = = It takes the car 15.9 s to pass the truck. (b) Need how far the car travels relative to the earth, so go now to coordinates fixed to the earth. In these coordinates 0 20.0 m/sxv = for the car. Take the origin to be at the initial position of the car. 0 20.0 m/s,xv = 20.600 m/s ,xa = 15.86 s,t = 0 ?x x? = 2 2 21 1 0 0 2 2(20.0 m/s)(15.86 s) (0.600 m/s )(15.86 s)x xx x v t a t? = + = + 0 317.2 m 75.5 m 393 m.x x? = + = (c) In coordinates fixed to the earth: 2 0 20.0 m/s (0.600 m/s )(15.86 s) 29.5 m/sx x xv v a t= + = + = EVALUATE: In 15.9 s the truck travels 0 (20.0 m/s)(15.86 s) 317.2 m.x x? = = The car travels 392.7 m 317.2 m 75 m? = farther than the truck, which checks with part (a). In coordinates attached to the truck, for the car 0 0,xv = 9.5 m/sxv = and in 15.86 s the car travels 00 75 m,2 x xv vx x t +? ?? = =? ?? ? which checks with part (a). 2.74. IDENTIFY: The acceleration is not constant so the constant acceleration equations cannot be used. Instead, use ( ) xx dv a t dt = and 0 0 ( ) t xx x v t dt= + ? . SET UP: 11 1 n nt dt t n += +? for 0n ? . EXECUTE: (a) 2 310 0 30( ) [ ] t x t x t dt x t t? ? ? ?= + ? = + ?? . 0x = at 0t = gives 0 0x = and 3 3 31 3( ) (4.00 m/s) (0.667 m/s )x t t t t t? ?= ? = ? . 3( ) 2 (4.00 m/s )xx dva t t tdt ?= = ? = ? . (b) The maximum positive x is when 0xv = and 0xa < . 0xv = gives 2 0t? ?? = and 3 4.00 m/s 1.41 s 2.00 m/s t ? ?= = = . At this t, xa is negative. For 1.41 st = , 3 3(4.00 m/s)(1.41 s) (0.667 m/s )(1.41 s) 3.77 mx = ? = . EVALUATE: After 1.41 st = the object starts to move in the x? direction and goes to x = ?? as t ? ? . 2.75. ( ) ,a t t? ?= + with 22.00 m/s? = ? and 33.00 m/s? = (a) IDENTIFY and SET UP: Integrage ( )xa t to find ( )xv t and then integrate ( )xv t to find ( ).x t EXECUTE: 210 0 0 20 0 ( ) t t x x x x xv v a dt v dt v t t? ? ? ?= + = + + = + +? ? 2 2 31 1 1 0 0 0 0 02 2 60 0 ( ) t t x x xx x v dt x v t t dt x v t t t? ? ? ?= + = + + + = + + +? ? At 0,t = 0.x x= To have 0x x= at 1 4.00 st = requires that 2 31 10 1 1 12 6 0.xv t t t? ?+ + = Thus 2 3 2 21 1 1 10 1 16 2 6 2(3.00 m/s )(4.00 s) ( 2.00 m/s )(4.00 s) 4.00 m/s.xv t t? ?= ? ? = ? ? ? = ? (b) With 0xv as calculated in part (a) and 4.00 s,t = 2 2 3 21 1 0 0 2 24.00 s ( 2.00 m/s )(4.00 s) (3.00 m/s )(4.00 s) 12.0 m/s.xv v t t? ?= + + = ? + ? + = + EVALUATE: 0xa = at 0.67 s.t = For 0.67 s,t > 0.xa > At 0,t = the particle is moving in the -directionx? and is speeding up. After 0.67 s,t = when the acceleration is positive, the object slows down and then starts to move in the -directionx+ with increasing speed. Motion Along a Straight Line 2-31 2.76. IDENTIFY: Find the distance the professor walks during the time t it takes the egg to fall to the height of his head. SET UP: Let y+ be downward. The egg has 0 0yv = and 29.80 m/sya = . At the height of the professor?s head, the egg has 0 44.2 my y? = . EXECUTE: 210 0 2y yy y v t a t? = + gives 0 22( ) 2(44.2 m) 3.00 s9.80 m/sy y y t a ?= = = . The professor walks a distance 0 0 (1.20 m/s)(3.00 s) 3.60 mxx x v t? = = = . Release the egg when your professor is 3.60 m from the point directly below you. EVALUATE: Just before the egg lands its speed is 2(9.80 m/s )(3.00s) 29.4 m/s= . It is traveling much faster than the professor. 2.77. IDENTIFY: Use the constant acceleration equations to establish a relationship between maximum height and acceleration due to gravity and between time in the air and acceleration due to gravity. SET UP: Let y+ be upward. At the maximum height, 0yv = . When the rock returns to the surface, 0 0y y? = . EXECUTE: (a) 2 20 02 ( )y y yv v a y y= + ? gives 21 02y ya H v= ? , which is constant, so E E M Ma H a H= . 2 E M E 2 M 9.80 m/s 2.64 3.71 m/s a H H H H a ? ? ? ?= = =? ? ? ?? ?? ? . (b) 210 0 2y yy y v t a t? = + with 0 0y y? = gives 02y ya t v= ? , which is constant, so E E M Ma T a T= . E M E M 2.64 a T T T a ? ?= =? ?? ? . EVALUATE: On Mars, where the acceleration due to gravity is smaller, the rocks reach a greater height and are in the air for a longer time. 2.78. IDENTIFY: Calculate the time it takes her to run to the table and return. This is the time in the air for the thrown ball. The thrown ball is in free-fall after it is thrown. Assume air resistance can be neglected. SET UP: For the thrown ball, let y+ be upward. 29.80 m/sya = ? . 0 0y y? = when the ball returns to its original position. EXECUTE: (a) It takes her 5.50 m 2.20 s 2.50 m/s = to reach the table and an equal time to return. For the ball, 0 0y y? = , 4.40 st = and 29.80 m/sya = ? . 210 0 2y yy y v t a t? = + gives 21 1 0 2 2 ( 9.80 m/s )(4.40 s) 21.6 m/sy yv a t= ? = ? ? = . (b) Find 0y y? when 2.20 st = . 2 2 21 10 0 2 2(21.6 m/s)(2.20 s) ( 9.80 m/s )(2.20 s) 23.8 my yy y v t a t? = + = + ? = EVALUATE: It takes the ball the same amount of time to reach its maximum height as to return from its maximum height, so when she is at the table the ball is at its maximum height. Note that this large maximum height requires that the act either be done outdoors, or in a building with a very high ceiling. 2.79. (a) IDENTIFY: Use constant acceleration equations, with ,ya g= downward, to calculate the speed of the diver when she reaches the water. SET UP: Take the origin of coordinates to be at the platform, and take the -directiony+ to be downward. 0 21.3 m,y y? = + 29.80 m/s ,ya = + 0 0yv = (since diver just steps off), ?yv = 2 2 0 02 ( )y y yv v a y y= + ? EXECUTE: 202 ( ) 2(9.80 m/s )(31.3 m) 20.4 m/s.y yv a y y= + ? = + = + We know that yv is positive because the diver is traveling downward when she reaches the water. The announcer has exaggerated the speed of the diver. EVALUATE: We could also use 210 0 2y yy y v t a t? = + to find 2.085 s.t = The diver gains 9.80 m/s of speed each second, so has 2(9.80 m/s )(2.085 s) 20.4 m/syv = = when she reaches the water, which checks. (b) IDENTIFY: Calculate the initial upward velocity needed to give the diver a speed of 25.0 m/s when she reaches the water. Use the same coordinates as in part (a). SET UP: 0 ?,yv = 25.0 m/s,yv = + 29.80 m/s ,ya = + 0 21.3 my y? = + 2 2 0 02 ( )y y yv v a y y= + ? 2-32 Chapter 2 EXECUTE: 2 2 20 02 ( ) (25.0 m/s) 2(9.80 m/s )(21.3 m) 14.4 m/sy y yv v a y y= ? ? ? = ? ? = ? 0( yv is negative since the direction of the initial velocity is upward.) EVALUATE: One way to decide if this speed is reasonable is to calculate the maximum height above the platform it would produce: 0 14.4 m/s,yv = ? 0yv = (at maximum height), 29.80 m/s ,ya = + 0 ?y y? = 2 2 0 02 ( )y y yv v a y y= + ? 2 2 2 0 0 0 ( 14.4 s) 10.6 m 2 2( 9.80 m/s) y y y v v y y a ? ? ?? = = = ?+ This is not physically attainable; a vertical leap of 10.6 m upward is not possible. 2.80. IDENTIFY: The flowerpot is in free-fall. Apply the constant acceleration equations. Use the motion past the window to find the speed of the flowerpot as it reaches the top of the window. Then consider the motion from the windowsill to the top of the window. SET UP: Let y+ be downward. Throughout the motion 29.80 m/sya = + . EXECUTE: Motion past the window: 0 1.90 my y? = , 0.420 st = , 29.80 m/sya = + . 210 0 2y yy y v t a t? = + gives 20 1 1 0 2 2 1.90 m (9.80 m/s )(0.420 s) 2.466 m/s 0.420 sy y y y v a t t ?= ? = ? = . This is the velocity of the flowerpot when it is at the top of the window. Motion from the windowsill to the top of the window: 0 0yv = , 2.466 m/syv = , 29.80 m/sya = + . 2 2 0 02 ( )y y yv v a y y= + ? gives 2 2 2 0 0 2 (2.466 m/s) 0 0.310 m 2 2(9.80 m/s ) y y y v v y y a ? ?? = = = . The top of the window is 0.310 m below the windowsill. EVALUATE: It takes the flowerpot 0 22.466 m/s 0.252 s9.80 m/s y y y v v t a ?= = = to fall from the sill to the top of the window. Our result says that from the windowsill the pot falls 0.310 m 1.90 m 2.21 m+ = in 0.252 s 0.420 s 0.672 s+ = . 2 2 21 10 0 2 2 (9.80 m/s )(0.672 s) 2.21 my yy y v t a t? = + = = , which checks. 2.81. IDENTIFY: For parts (a) and (b) apply the constant acceleration equations to the motion of the bullet. In part (c) neglect air resistance, so the bullet is free-fall. Use the constant acceleration equations to establish a relation between initial speed 0v and maximum height H. SET UP: For parts (a) and (b) let x+ be in the direction of motion of the bullet. For part (c) let y+ be upward, so ya g= ? . At the maximum height, 0yv = . EXECUTE: (a) 0 0.700 mx x? = , 0 0xv = , 965 m/sxv = . 2 20 02 ( )x x xv v a x x= + ? gives 2 2 2 5 20 0 (965 m/s) 0 6.65 10 m/s 2( ) 2(0.700 m) x x x v v a x x ? ?= = = ×? . 46.79 10x a g = × , so 4(6.79 10 )xa g= × . (b) 00 2 x xv vx x t +? ?? = ? ?? ? gives 0 0 2( ) 2(0.700 m) 1.45 ms 0 965 m/sx x x x t v v ?= = =+ + . (c) 2 20 02 ( )y y yv v a y y= + ? and 0yv = gives 2 0 0 2y y v a y y = ?? , which is constant. 2 2 01 02 1 2 v v H H = . 22 1 0102 2 2 1 2 01 01 / 4 vv H H H H v v ? ? ? ?= = =? ? ? ?? ? ? ? . EVALUATE: 2 2 2 0 2 (965 m/s) 47.5 km 2 2( 9.80 m/s ) y y y v v H a ? ?= = =? . Rifle bullets fired vertically don't actually reach such a large height; it is not an accurate approximation to ignore air resistance. 2.82. IDENTIFY: Assume the firing of the second stage lasts a very short time, so the rocket is in free-fall after 25.0 s. The motion consists of two constant acceleration segments. SET UP: Let y+ be upward. After 25.0 st = , 29.80 m/sya = ? . EXECUTE: (a) Find the height of the rocket at 25.0 st = : 0 0yv = , 23.50 m/sya = + , 25.0 st = . 2 2 31 1 0 0 2 2 (3.50 m/s)(25.0 s) 1.0938 10 my yy y v t a t? = + = = × . Find the displacement of the rocket from firing of the Motion Along a Straight Line 2-33 second stage until the maximum height is reached: 0 132.5 m/syv = , 0yv = (at maximum height), 29.80 m/sya = ? . 2 2 0 02 ( )y y yv v a y y= + ? gives 2 2 2 0 0 2 0 (132.5 m/s) 896 m 2 2( 9.80 m/s ) y y y v v y y a ? ?? = = =? . The total height is 1094 m 896 m 1990 m+ = . (b) 0 132.5 m/syv = + , 29.80 m/sya = ? , 0 1094 my y? = ? . 210 0 2y yy y v t a t? = + gives 2 21093.8 m (132.5 m/s) (4.90 m/s )t t? = ? . The quadratic formula gives 33.7 st = as the positive root. The rocket returns to the launch pad 33.7 s after the second stage fires. (c) 20 132.5 m/s ( 9.80 m/s )(33.7 s) 198 m/sy y yv v a t= + = + + ? = ? . The rocket has speed 198 m/s as it reaches the launch pad. EVALUATE: The speed when the rocket returns to the launch pad is greater than 132.5 m/s. When the rocket returns to the height where the second stage fired, its velocity is 132.5 m/s downward and it continues to speed up during the rest of the descent. 2.83. Take positive y to be upward. (a) IDENTIFY: Consider the motion from when he applies the acceleration to when the shot leaves his hand. SET UP: 0 0,yv = ?,yv = 245.0 m/s ,ya = 0 0.640 my y? = 2 2 0 02 ( )y y yv v a y y= + ? EXECUTE: 202 ( ) 2(45.0 m/s )(0.640 m) 7.59 m/sy yv a y y= ? = = (b) IDENTIFY: Consider the motion of the shot from the point where he releases it to its maximum height, where 0.v = Take 0y = at the ground. SET UP: 0 2.20 m,y = ?,y = 29.80 m/sya = ? (free fall), 0 7.59 m/syv = (from part (a), 0yv = at maximum height) 2 2 0 02 ( )y y yv v a y y= + ? EXECUTE: 2 2 2 0 0 2 0 (7.59 m/s) 2.94 m 2 2( 9.80 m/s ) y y y v v y y a ? ?? = = =? 2.20 m 2.94 m 5.14 m.y = + = (c) IDENTIFY: Consider the motion of the shot from the point where he releases it to when it returns to the height of his head. Take 0y = at the ground. SET UP: 0 2.20 m,y = 1.83 m,y = 29.80 m/s ,ya = ? 0 7.59 m/s,yv = + ?t = 10 0 2y yy y v t a t 2? = + EXECUTE: 2 2121.83 m 2.20 m (7.59 m/s) ( 9.80 m/s )t t? = + ? 2 2(7.59 m/s) (4.90 m/s )t t= ? 24.90 7.59 0.37 0,t t? ? = with t in seconds. Use the quadratic formula to solve for t: ( )21 7.59 (7.59) 4(4.90)( 0.37) 0.774 0.8229.80t = ± ? ? = ± t must be positive, so 0.774 s 0.822 s 1.60 st = + = EVALUATE: Calculate the time to the maximum height: 0 ,y y yv v a t= + so 2 0( ) / (7.59 m/s)/( 9.80 m/s ) 0.77 s.y y yt v v a= ? = ? ? = It also takes 0.77 s to return to 2.2 m above the ground, for a total time of 1.54 s. His head is a little lower than 2.20 m, so it is reasonable for the shot to reach the level of his head a little later than 1.54 s after being thrown; the answer of 1.60 s in part (c) makes sense. 2.84. IDENTIFY: The teacher is in free-fall and falls with constant acceleration 29.80 m/s , downward. The sound from her shout travels at constant speed. The sound travels from the top of the cliff, reflects from the ground and then travels upward to her present location. If the height of the cliff is h and she falls a distance y in 3.0 s, the sound must travel a distance ( )h h y+ ? in 3.0 s. SET UP: Let y+ be downward, so for the teacher 29.80 m/sya = and 0 0yv = . Let 0y = at the top of the cliff. EXECUTE: (a) For the teacher, 2 212 (9.80 m/s )(3.0 s) 44.1 my = = . For the sound, s( )h h y v t+ ? = . 1 1 s2 2( ) ([340 m/s][3.0 s] 44.1 m) 532 mh v t y= + = + = , which rounds to 530 m. (b) 2 20 02 ( )y y yv v a y y= + ? gives 202 ( ) 2(9.80 m/s )(532 m) 102 m/sy yv a y y= ? = = . 2-34 Chapter 2 EVALUATE: She is in the air for 0 2102 m/s 10.4 s9.80 m/s y y y v v t a ?= = = and strikes the ground at high speed. 2.85. IDENTIFY and SET UP: Let y+ be upward. Each ball moves with constant acceleration 29.80 m/s .ya = ? In parts (c) and (d) require that the two balls be at the same height at the same time. EXECUTE: (a) At ceiling, 0,yv = 0 3.0 m,y y? = 29.80 m/s .ya = ? Solve for 0 .yv 2 2 0 02 ( )y y yv v a y y= + ? gives 0 7.7 m/s.yv = (b) 0y y yv v a t= + with the information from part (a) gives 0.78 s.t = (c) Let the first ball travel downward a distance d in time t. It starts from its maximum height, so 0 0.yv = 21 0 0 2y y y y v t a t? = = gives 2 2(4.9 m/s )d t= The second ball has 20 3 (7.7 m/s) 5.1 m/s.yv = = In time t it must travel upward 3.0 m d? to be at the same place as the first ball. 1 0 0 2y yy y v t a t 2? = + gives 2 23.0 m (5.1 m/s) (4.9 m/s ) .d t t? = ? We have two equations in two unknowns, d and t. Solving gives 0.59 st = and 1.7 m.d = (d) 3.0 m 1.3 md? = EVALUATE: In 0.59 s the first ball falls 2 2(4.9 m/s )(0.59 s) 1.7 m,d = = so is at the same height as the second ball. 2.86. IDENTIFY: The helicopter has two segments of motion with constant acceleration: upward acceleration for 10.0 s and then free-fall until it returns to the ground. Powers has three segments of motion with constant acceleration: upward acceleration for 10.0 s, free-fall for 7.0 s and then downward acceleration of 22.0 m/s . SET UP: Let y+ be upward. Let 0y = at the ground. EXECUTE: (a) When the engine shuts off both objects have upward velocity 2 0 (5.0 m/s )(10.0 s) 50.0 m/sy y yv v a t= + = = and are at 2 2 21 10 2 2 (5.0 m/s )(10.0 s) 250 my yy v t a t= + = = . For the helicopter, 0yv = (at the maximum height), 0 50.0 m/syv = + , 0 250 my = , and 29.80 m/sya = ? . 2 2 0 02 ( )y y yv v a y y= + ? gives 2 2 2 0 0 2 0 (50.0 m/s) 250 m 378 m 2 2( 9.80 m/s ) y y y v v y y a ? ?= + = + =? , which rounds to 380 m. (b) The time for the helicopter to crash from the height of 250 m where the engines shut off can be found using 0 50.0 m/syv = + , 29.80 m/sya = ? , and 0 250 my y? = ? . 210 0 2y yy y v t a t? = + gives 2 2250 m (50.0 m/s) (4.90 m/s )t t? = ? . 2 2(4.90 m/s ) (50.0 m/s) 250 m 0t t? ? = . The quadratic formula gives ( )21 50.0 (50.0) 4(4.90)(250) s9.80t = ± + . Only the positive solution is physical, so 13.9 st = . Powers therefore has free-fall for 7.0 s and then downward acceleration of 22.0 m/s for 13.9 s 7.0 s 6.9 s? = . After 7.0 s of free-fall he is at 2 2 21 10 0 2 2250 m (50.0 m/s)(7.0 s) ( 9.80 m/s )(7.0 s) 360 my yy y v t a t? = + = + + ? = and has velocity 2 0 50.0 m/s ( 9.80 m/s )(7.0 s) 18.6 m/sx x xv v a t= + = + ? = ? . After the next 6.9 s he is at 2 2 21 1 0 0 2 2360 m ( 18.6 m/s)(6.9 s) ( 2.00 m/s )(6.9 s) 184 my yy y v t a t? = + = + ? + ? = . Powers is 184 m above the ground when the helicopter crashes. EVALUATE: When Powers steps out of the helicopter he retains the initial velocity he had in the helicopter but his acceleration changes abruptly from 25.0 m/s upward to 29.80 m/s downward. Without the jet pack he would have crashed into the ground at the same time as the helicopter. The jet pack slows his descent so he is above the ground when the helicopter crashes. 2.87. IDENTIFY: Apply the constant acceleration equations to his motion. Consider two segments of the motion: the last 1.0 s and the motion prior to that. The final velocity for the first segment is the initial velocity for the second segment. SET UP: Let y+ be downward, so 29.80 m/sya = + . EXECUTE: Motion from the roof to a height of / 4h above ground: 0 3 / 4y y h? = , 29.80 m/sya = + , 0 0yv = . 2 2 0 02 ( )y y yv v a y y= + ? gives 02 ( ) 3.834 m /sy yv a y y h= ? = . Motion from height of / 4h to the ground: 0 / 4y y h? = , 29.80 m/sya = + , 0 3.834 m /syv h= , 1.00 st = . 210 0 2y yy y v t a t? = + gives Motion Along a Straight Line 2-35 3.834 m 4.90 m 4 h h= + . Let 2h u= and solve for u . 214 3.834 m 4.90 m 0u u? ? = . ( )22 3.834 ( 3.834) 4.90 mu = ± ? + . Only the positive root is physical, so 16.52 mu = and 2 273 mh u= = , which rounds to 270 m. The building is 270 m tall. EVALUATE: With 273 mh = the total time of fall is 2 7.46 s y h t a = = . In 7.47 s 1.00 s 6.46 s? = Spider-Man falls a distance 2 210 2 (9.80 m/s )(6.46 s) 204 my y? = = . This leaves 69 m for the last 1.0 s of fall, which is / 4h . 2.88. IDENTIFY: Apply constant acceleration equations to the motion of the rock. Sound travels at constant speed. SET UP: Let fallt be the time for the rock to fall to the ground and let st be the time it takes the sound to travel from the impact point back to you. fall s 10.0 st t+ = . Both the rock and sound travel a distance d that is equal to the height of the cliff. Take y+ downward for the motion of the rock. The rock has 0 0yv = and 29.80 m/sya = . EXECUTE: (a) For the rock, 210 0 2y yy y v t a t? = + gives fall 229.80 m/s d t = . For the sound, s 10.0 s330 m/s d t = = . Let 2 d? = . 20.00303 0.4518 10.0 0? ?+ ? = . 19.6? = and 384 md = . (b) You would have calculated 2 212 (9.80 m/s )(10.0 s) 490 md = = . You would have overestimated the height of the cliff. It actually takes the rock less time than 10.0 s to fall to the ground. EVALUATE: Once we know d we can calculate that fall 8.8 st = and s 1.2 st = . The time for the sound of impact to travel back to you is 12% of the total time and cannot be neglected. The rock has speed 86 m/s just before it strikes the ground. 2.89. (a) IDENTIFY: Let y+ be upward. The can has constant acceleration .ya g= ? The initial upward velocity of the can equals the upward velocity of the scaffolding; first find this speed. SET UP: 0 15.0 m,y y? = ? 3.25 s,t = 29.80 m/s ,ya = ? 0 ?yv = EXECUTE: 210 0 2y yy y v t a t? = + gives 0 11.31 m/syv = Use this 0 yv in 0y y yv v a t= + to solve for :yv 20.5 m/syv = ? (b) IDENTIFY: Find the maximum height of the can, above the point where it falls from the scaffolding: SET UP: 0,yv = 0 11.31 m/s,yv = + 29.80 m/s ,ya = ? 0 ?y y? = EXECUTE: 2 20 02 ( )y y yv v a y y= + ? gives 0 6.53 my y? = The can will pass the location of the other painter. Yes, he gets a chance. EVALUATE: Relative to the ground the can is initially traveling upward, so it moves upward before stopping momentarily and starting to fall back down. 2.90. IDENTIFY: Both objects are in free-fall. Apply the constant acceleration equations to the motion of each person. SET UP: Let y+ be downward, so 29.80 m/sya = + for each object. EXECUTE: (a) Find the time it takes the student to reach the ground: 0 180 my y? = , 0 0yv = , 29.80 m/sya = . 21 0 0 2y yy y v t a t? = + gives 0 22( ) 2(180 m) 6.06 s9.80 m/sy y y t a ?= = = . Superman must reach the ground in 6.06 s 5.00 s 1.06 s? = : 1.06 st = , 0 180 my y? = , 29.80 m/sya = + . 210 0 2y yy y v t a t? = + gives 20 1 1 0 2 2 180 m (9.80 m/s )(1.06 s) 165 m/s 1.06 sy y y y v a t t ?= ? = ? = . Superman must have initial speed 0 165 m/sv = . (b) The graphs of y-t for Superman and for the student are sketched in Figure 2.90. (c) The minimum height of the building is the height for which the student reaches the ground in 5.00 s, before Superman jumps. 2 2 21 10 0 2 2 (9.80 m/s )(5.00 s) 122 my yy y v t a t? = + = = . The skyscraper must be at least 122 m high. 2-36 Chapter 2 EVALUATE: 165 m/s 369 mi/h= , so only Superman could jump downward with this initial speed. Figure 2.90 2.91. IDENTIFY: Apply constant acceleration equations to the motion of the rocket and to the motion of the canister after it is released. Find the time it takes the canister to reach the ground after it is released and find the height of the rocket after this time has elapsed. The canister travels up to its maximum height and then returns to the ground. SET UP: Let y+ be upward. At the instant that the canister is released, it has the same velocity as the rocket. After it is released, the canister has 29.80 m/sya = ? . At its maximum height the canister has 0yv = . EXECUTE: (a) Find the speed of the rocket when the canister is released: 0 0yv = , 23.30 m/sya = , 0 235 my y? = . 2 20 02 ( )y y yv v a y y= + ? gives 202 ( ) 2(3.30 m/s )(235 m) 39.4 m/sy yv a y y= ? = = . For the motion of the canister after it is released, 0 39.4 m/syv = + , 29.80 m/sya = ? , 0 235 my y? = ? . 21 0 0 2y yy y v t a t? = + gives 2 2235 m (39.4 m/s) (4.90 m/s )t t? = ? . The quadratic formula gives 12.0 st = as the positive solution. Then for the motion of the rocket during this 12.0 s, 2 2 21 1 0 0 2 2235 m (39.4 m/s)(12.0 s) (3.30 m/s )(12.0 s) 945 my yy y v t a t? = + = + + = . (b) Find the maximum height of the canister above its release point: 0 39.4 m/syv = + , 0yv = , 29.80 m/sya = ? . 2 2 0 02 ( )y y yv v a y y= + ? gives 2 2 2 0 0 2 0 (39.4 m/s) 79.2 m 2 2( 9.80 m/s ) y y y v v y y a ? ?? = = =? . After its release the canister travels upward 79.2 m to its maximum height and then back down 79.2 m 235 m+ to the ground. The total distance it travels is 393 m. EVALUATE: The speed of the rocket at the instant that the canister returns to the launch pad is 2 0 39.4 m/s (3.30 m/s )(12.0 s) 79.0 m/sy y yv v a t= + = + = . We can calculate its height at this instant by 2 2 0 02 ( )y y yv v a y y= + ? with 0 0yv = and 79.0 m/syv = . 2 2 2 0 0 2 (79.0 m/s) 946 m 2 2(3.30 m/s ) y y y v v y y a ?? = = = , which agrees with our previous calculation. 2.92. IDENTIFY: Both objects are in free-fall and move with constant acceleration 29.80 m/s , downward. The two balls collide when they are at the same height at the same time. SET UP: Let y+ be upward, so 29.80 m/sya = ? for each ball. Let 0y = at the ground. Let ball A be the one thrown straight up and ball B be the one dropped from rest at height H. 0 0Ay = , 0By H= . EXECUTE: (a) 210 0 2y yy y v t a t? = + applied to each ball gives 210 2Ay v t gt= ? and 212By H gt= ? . A By y= gives 2 21 1 0 2 2v t gt H gt? = ? and 0 H t v = . (b) For ball A at its highest point, 0yAv = and 0y y yv v a t= + gives 0vt g= . Setting this equal to the time in part (a) gives 0 0 H v v g = and 2 0vH g = . EVALUATE: In part (a), using 0 H t v = in the expressions for Ay and By gives 2 0 1 2A B g H y y H v ? ?= = ?? ?? ? . H must be less than 2 02v g in order for the balls to collide before ball A returns to the ground. This is because it takes ball A Motion Along a Straight Line 2-37 time 0 2v t g = to return to the ground and ball B falls a distance 2 2 01 2 2v gt g = during this time. When 2 02vH g = the two balls collide just as ball A reaches the ground and for H greater than this ball A reaches the ground before they collide. 2.93. IDENTIFY and SET UP: Use /xv dx dt= and /x xa dv dt= to calculate ( )xv t and ( )xa t for each car. Use these equations to answer the questions about the motion. EXECUTE: 2 ,Ax t t? ?= + 2 ,AAx dxv tdt ? ?= = + 2 Ax Ax dv a dt ?= = 2 3,Bx t t? ?= ? 22 3 ,BBx dxv t tdt ? ?= = ? 2 6 Bx Bx dv a t dt ? ?= ? ? (a) IDENTIFY and SET UP: The car that initially moves ahead is the one that has the larger 0 .xv EXECUTE: At 0,t = Axv ?= and 0.Bxv = So initially car A moves ahead. (b) IDENTIFY and SET UP: Cars at the same point implies .A Bx x= 2 2 3t t t t? ? ? ?+ = ? EXECUTE: One solution is 0,t = which says that they start from the same point. To find the other solutions, divide by t: 2t t t? ? ? ?+ = ? 2 ( ) 0t t? ? ? ?+ ? + = ( ) ( )2 21 1( ) ( ) 4 1.60 (1.60) 4(0.20)(2.60) 4.00 s 1.73 s2 0.40t ? ? ? ? ???= ? ? ± ? ? = + ± ? = ± So A Bx x= for 0,t = 2.27 st = and 5.73 s.t = EVALUATE: Car A has constant, positive .xa Its xv is positive and increasing. Car B has 0 0xv = and xa that is initially positive but then becomes negative. Car B initially moves in the -directionx+ but then slows down and finally reverses direction. At 2.27 st = car B has overtaken car A and then passes it. At 5.73 s,t = car B is moving in the -directionx? as it passes car A again. (c) IDENTIFY: The distance from A to B is .B Ax x? The rate of change of this distance is ( ) .B Ad x xdt ? If this distance is not changing, ( ) 0.B A d x x dt ? = But this says 0.Bx Axv v? = (The distance between A and B is neither decreasing nor increasing at the instant when they have the same velocity.) SET UP: Ax Bxv v= requires 22 2 3t t t? ? ? ?+ = ? EXECUTE: 23 2( ) 0t t? ? ? ?+ ? + = ( ) ( )2 21 12( ) 4( ) 12 3.20 4( 1.60) 12(0.20)(2.60)6 1.20t ? ? ? ? ???= ? ? ± ? ? = ± ? ? 2.667 s 1.667 s,t = ± so Ax Bxv v= for 1.00 st = and 4.33 s.t = EVALUATE: At 1.00 s,t = 5.00 m/s.Ax Bxv v= = At 4.33 s,t = 13.0 m/s.Ax Bxv v= = Now car B is slowing down while A continues to speed up, so their velocities aren?t ever equal again. (d) IDENTIFY and SET UP: Ax Bxa a= requires 2 2 6 t? ? ?= ? EXECUTE: 2 2 3 2.80 m/s 1.20 m/s 2.67 s. 3 3(0.20 m/s ) t ? ? ? ? ?= = = EVALUATE: At 0,t = ,Bx Axa a> but Bxa is decreasing while Axa is constant. They are equal at 2.67 st = but for all times after that .Bx Axa a< 2.94. IDENTIFY: The apple has two segments of motion with constant acceleration. For the motion from the tree to the top of the grass the acceleration is g , downward and the apple falls a distance H h? . For the motion from the top of the grass to the ground the acceleration is a, upward, the apple travels downward a distance h, and the final speed is zero. SET UP: Let y+ be upward and let 0y = at the ground. The apple is initially a height H h+ above the ground. EXECUTE: (a) Motion from 0y H h= + to y H= : 0y y H? = ? , 0 0yv = , ya g= ? . 2 20 02 ( )y y yv v a y y= + ? gives 2yv gH= ? . The speed of the apple is 2 g H as it enters the grass. 2-38 Chapter 2 (b) Motion from 0y h= to 0y = : 0y y h? = ? , 0 2yv gH= ? . 2 20 02 ( )y y yv v a y y= + ? gives 2 2 0 0 2 2( ) 2( ) y y y v v g H gH a y y h h ? ?= = =? ? . The acceleration of the apple while it is in the grass is /g H h , upward. (c) Graphs of y-t, -yv t and -ya t are sketched in Figure 2.94. EVALUATE: The acceleration a produced by the grass increases when H increases and decreases when h increases. Figure 2.94 2.95. IDENTIFY: Apply constant acceleration equations to the motion of the two objects, the student and the bus. SET UP: For convenience, let the student's (constant) speed be 0v and the bus's initial position be 0.x Note that these quantities are for separate objects, the student and the bus. The initial position of the student is taken to be zero, and the initial velocity of the bus is taken to be zero. The positions of the student 1x and the bus 2x as functions of time are then 1 0x v t= and 22 0 (1 2) .x x at= + EXECUTE: (a) Setting 1 2x x= and solving for the times t gives ( )20 0 01 2t v v axa= ± ? . ( )2 221 (5.0 m s) (5.0 m s) 2(0.170 m s )(40.0 m) 9.55 s and 49 3 s(0.170 m s )t .= ± ? = . The student will be likely to hop on the bus the first time she passes it (see part (d) for a discussion of the later time). During this time, the student has run a distance 0 (5 m s)(9.55 s) 47 8 m.v t .= = (b) The speed of the bus is 2(0.170 m/s )(9.55 s) 1.62 m/s= . (c) The results can be verified by noting that the x lines for the student and the bus intersect at two points, as shown in Figure 2.95a. (d) At the later time, the student has passed the bus, maintaining her constant speed, but the accelerating bus then catches up to her. At this later time the bus's velocity is ( )( )20.170 m s 49.3 s 8.38 m s.= (e) No; 20 02v ax< , and the roots of the quadratic are imaginary. When the student runs at 3.5 m s, Figure 2.95b shows that the two lines do not intersect: (f) For the student to catch the bus, 20 02 .v ax> and so the minimum speed is ( )( )22 0.170 m s 40 m s 3.688 m s.= She would be running for a time 23.69 m s 21.7 s,0.170 m/s = and covers a distance ( ) ( )3.688 m s 21.7 s 80.0 m.= However, when the student runs at 3.688 m s, the lines intersect at one point, at 80 mx = , as shown in Figure 2.95c. EVALUATE: The graph in part (c) shows that the student is traveling faster than the bus the first time they meet but at the second time they meet the bus is traveling faster. 2 tot 1t t t= ? Figure 2.95 Motion Along a Straight Line 2-39 2.96. IDENTIFY: Apply 210 0 2y yy y v t a t? = + to the motion from the maximum height, where 0 0yv = . The time spent above max / 2y on the way down equals the time spent above max / 2y on the way up. SET UP: Let y+ be downward. ya g= . 0 max / 2y y y? = when he is a distance max / 2y above the floor. EXECUTE: The time from the maximum height to max / 2y above the floor is given by 21max 12/ 2y gt= . The time from the maximum height to the floor is given by 21max tot2y gt= and the time from a height of max / 2y to the floor is . max1 2 max max / 2 1 2.4 / 2 2 1 yt t y y = = =? ? . EVALUATE: The person spends over twice as long above max / 2y as below max / 2y . His average speed is less above max / 2y than it is when he is below this height. 2.97. IDENTIFY: Apply constant acceleration equations to both objects. SET UP: Let y+ be upward, so each ball has ya g= ? . For the purpose of doing all four parts with the least repetition of algebra, quantities will be denoted symbolically. That is, let ( )221 0 2 01 1, .2 2y h v t gt y h g t t= + ? = ? ? In this case, 0 1.00 st = . EXECUTE: (a) Setting 1 2 0,y y= = expanding the binomial ( )20t t? and eliminating the common term 2 21 1 0 0 02 2 yields g t v t gt t gt= ? . Solving for t: 21 0 02 0 0 0 0 1 2 1 /( ) gt t t g t v v gt ? ?= = ? ?? ?? ? . Substitution of this into the expression for 1y and setting 1 0y = and solving for h as a function of 0v yields, after some algebra, ( ) ( ) 21 0 02 21 02 2 0 0 . gt v h gt gt v ?= ? Using the given value 2 0 1.00 s and 9.80 m s ,t g= = ( ) 0 0 2 4.9 m s 20.0 m 4.9 m . 9.8 m s v h v ? ??= = ? ??? ? This has two solutions, one of which is unphysical (the first ball is still going up when the second is released; see part (c)). The physical solution involves taking the negative square root before solving for 0v , and yields 8.2 m s. The graph of y versus t for each ball is given in Figure 2.97. (b) The above expression gives for (i), 0.411 m and for (ii) 1.15 km. (c) As 0v approaches 9.8 m s , the height h becomes infinite, corresponding to a relative velocity at the time the second ball is thrown that approaches zero. If 0 9.8 m s,v > the first ball can never catch the second ball. (d) As 0v approaches 4.9 m/s, the height approaches zero. This corresponds to the first ball being closer and closer (on its way down) to the top of the roof when the second ball is released. If 0 4.9 m s,v < the first ball will already have passed the roof on the way down before the second ball is released, and the second ball can never catch up. EVALUATE: Note that the values of 0v in parts (a) and (b) are all greater than minv and less than maxv . Figure 2.97 2.98. IDENTIFY: Apply constant acceleration equations to the motion of the boulder. SET UP: Let y+ be downward, so ya g= + . 2-40 Chapter 2 EXECUTE: (a) Let the height be h and denote the 1.30-s interval as ;t? the simultaneous equations 2 21 2 1 2 3 2, ( )h gt h g t t= = ? ? can be solved for t. Eliminating h and taking the square root, 3 ,2 t t t =? ? and , 1 2/3 t t ?= ? and substitution into 21 2h gt= gives 246 m.h = (b) The above method assumed that 0t > when the square root was taken. The negative root (with 0)t? = gives an answer of 2.51 m, clearly not a ?cliff?. This would co rrespond to an object that was initially near the bottom of this ?cliff? being thrown upward and taking 1.30 s to rise to the top and fall to the bottom. Although physically possible, the conditions of the problem preclude this answer. EVALUATE: For the first two-thirds of the distance, 0 164 my y? = , 0 0yv = , and 29.80 m/sya = . 02 ( ) 56.7 m/sy yv a y y= ? = . Then for the last third of the distance, 0 82.0 my y? = , 0 56.7 m/syv = and 29.80 m/sya = . 210 0 2y yy y v t a t? = + gives 2 2(4.90 m/s ) (56.7 m/s) 82.0 m 0t t+ ? = . ( )21 56.7 (56.7) 4(4.9)(82.0) s 1.30 s9.8t = ? + + = , as required. 3-1 M OTION IN TWO OR THREE DIMENSIONS 3.1. IDENTIFY and SET UP: Use Eq.(3.2), in component form. EXECUTE: ( ) 2 1av 2 1 5.3 m 1.1 m 1.4 m/s 3.0 s 0x x x x v t t t ? ? ?= = = =? ? ? ( ) 2 1av 2 1 0.5 m 3.4 m 1.3 m/s 3.0 s 0y y y y v t t t ? ? ? ?= = = = ?? ? ? EVALUATE: Our calculation gives that avv! is in the 4th quadrant. This corresponds to increasing x and decreasing y. 3.2. IDENTIFY: Use Eq.(3.2), written in component form. The distance from the origin is the magnitude of r! . SET UP: At time 1t , 1 1 0x y= = . EXECUTE: (a) av-( )? ( 3.8m s)(12.0 s) 45.6 mxx v t= = ? = ? and av-( )? (4.9m s)(12.0 s) 58.8 myy v t= = = . (b) 2 2 2 2( 45.6 m) (58.8 m) 74.4 m.r x y= + = ? + = EVALUATE: ?r! is in the direction of avv! . Therefore, x? is negative since av-xv is negative and y? is positive since av-yv is positive. 3.3. (a) IDENTIFY and SET UP: From r! we can calculate x and y for any t. Then use Eq.(3.2), in component form. EXECUTE: ( ) ( )2 2 ? ?4.0 cm 2.5 cm/s 5.0 cm/st t? ?= + +? ?r i j! At 0,t = ( ) ?4.0 cm .=r i! At 2.0 s,t = ( ) ( )? ?14.0 cm 10.0 cm .= +r i j! ( )av 10.0 cm 5.0 cm/s.2.0 sx x v t ?= = =? ( )av 10.0 cm 5.0 cm/s.2.0 sy y v t ?= = =? ( ) ( ) av av 1.3 m/s tan 0.9286 1.4 m/s y x v v ? ?= = = ? 360 42.9 317? = ° ? ° = ° ( ) ( )2 2av av avx yv v v= + 2 2 av (1.4 m/s) ( 1.3 m/s) 1.9 m/sv = + ? = Figure 3.1 3 3-2 Chapter 3 ( ) ( )2 2av av av 7.1 cm/sx yv v v= + = ( ) ( ) av av tan 1.00y x v v ? = = 45 .? = ° Figure 3.3a EVALUATE: Both x and y increase, so avv! is in the 1st quadrant. (b) IDENTIFY and SET UP: Calculate r! by taking the time derivative of ( ).tr! EXECUTE: ( ) ( )2 ? ?5.0 cm/s 5.0 cm/sd t dt ? ?= = +? ? r v i j !! 0 :t = 0,xv = 5.0 cm/s;yv = 5.0 cm/sv = and 90? = ° 1.0 s:t = 5.0 cm/s,xv = 5.0 cm/s;yv = 7.1 cm/sv = and 45? = ° 2.0 s:t = 10.0 cm/s,xv = 5.0 cm/s;yv = 11 cm/sv = and 27? = ° (c) The trajectory is a graph of y versus x. 2 24.0 cm (2.5 cm/s ) ,x t= + (5.0 cm/s)y t= For values of t between 0 and 2.0 s, calculate x and y and plot y versus x. Figure 3.3b EVALUATE: The sketch shows that the instantaneous velocity at any t is tangent to the trajectory. 3.4. IDENTIFY: d dtv = r/! ! . This vector will make a 45° -angle with both axes when its x- and y-components are equal. SET UP: 1( ) n nd t nt dt ?= . EXECUTE: 2? ?2 3bt ctv = i + j! . x yv v= gives 2 3t b c= . EVALUATE: Both components of v! change with t. 3.5. IDENTIFY and SET UP: Use Eq.(3.8) in component form to calculate ( )av xa and ( )av .ya Motion in Two or Three Dimensions 3-3 EXECUTE: (a) The velocity vectors at 1 0t = and 2 30.0 st = are shown in Figure 3.5a. Figure 3.5a (b) ( ) 22 1av 2 1 170 m/s 90 m/s 8.67 m/s 30.0 s x x x x v v v a t t t ? ? ? ?= = = = ?? ? ( ) 2 1 2av 2 1 40 m/s 110 m/s 2.33 m/s 30.0 s y y y y v v v a t t t ? ? ?= = = = ?? ? (c) ( ) ( )2 2 2av av 8.98 m/sx ya a a= + = ( ) ( ) 2 av 2 av 2.33 m/s tan 0.269 8.67 m/s y x a a ? ?= = =? 15 180 195? = ° + ° = ° Figure 3.5b EVALUATE: The changes in xv and yv are both in the negative x or y direction, so both components of ava! are in the 3rd quadrant. 3.6. IDENTIFY: Use Eq.(3.8), written in component form. SET UP: 2 2 2 2(0.45m s )cos31.0 0.39m s , (0.45m s )sin31.0 0.23m sx ya a= ° = = ° = EXECUTE: (a) av- xx va t ?= ? and 22.6 m s (0.39 m s )(10.0 s) 6.5 m sxv = + = . av- yy v a t ?= ? and 21.8 m s (0.23 m s )(10.0 s) 0.52 m syv = ? + = . (b) 2 2(6.5m s) (0.52m s) 6.48m sv = + = , at an angle of 0.52arctan 4.6 6.5 ? ? = °? ?? ? above the horizontal. (c) The velocity vectors 1v ! and 2v ! are sketched in Figure 3.6. The two velocity vectors differ in magnitude and direction. EVALUATE: 1v! is at an angle of 35° below the -axisx+ and has magnitude 1 3.2 m/sv = , so 2 1v v> and the direction of 2v ! is rotated counterclockwise from the direction of 1v ! . Figure 3.6 3.7. IDENTIFY and SET UP: Use Eqs.(3.4) and (3.12) to find ,xv ,yv ,xa and ya as functions of time. The magnitude and direction of r ! and a ! can be found once we know their components. 3-4 Chapter 3 EXECUTE: (a) Calculate x and y for t values in the range 0 to 2.0 s and plot y versus x. The results are given in Figure 3.7a. Figure 3.7a (b) x dx v dt ?= = 2y dyv tdt ?= = ? 0xy dv a dt = = 2yy dv a dt ?= = ? Thus ? ?2a t?= ?v i j! ?2?= ?a j! (c) velocity: At 2.0 s,t = 2.4 m/s,xv = 22(1.2 m/s )(2.0 s) 4.8 m/syv = ? = ? 2 2 5.4 m/sx yv v v= + = 4.8 m/s tan 2.00 2.4 m/s y x v v ? ?= = = ? 63.4 360 297? = ? ° + ° = ° Figure 3.7b acceleration: At 2.0 s,t = 0,xa = 2 22(1.2 m/s ) 2.4 m/sya = ? = ? 2 2 22.4 m/sx ya a a= + = 22.4 m/s tan 0 y x a a ? ?= = = ?? 270? = ° Figure 3.7c EVALUATE: (d) a! has a component a" in the same direction as ,v ! so we know that v is increasing (the bird is speeding up.) a ! also has a component a? perpendicular to ,v ! so that the direction of v ! is changing; the bird is turning toward the -directiony? (toward the right) Figure 3.7d Motion in Two or Three Dimensions 3-5 v ! is always tangent to the path; v ! at 2.0 st = shown in part (c) is tangent to the path at this t, conforming to this general rule. a ! is constant and in the -direction;y? the direction of v! is turning toward the -direction.y? 3.8. IDENTIFY: The component ?a! of a! perpendicular to the path is related to the change in direction of v! and the component a " ! of a ! parallel to the path is related to the change in the magnitude of v ! . SET UP: When the speed is increasing, a "! is in the direction of v! and when the speed is decreasing, a "! is opposite to the direction of v ! . When v is constant, a" is zero and when the path is a straight line, a? is zero. EXECUTE: The acceleration vectors in each case are sketched in Figure 3.8a-c. EVALUATE: ?a! is toward the center of curvature of the path. Figure 3.8a-c 3.9. IDENTIFY: The book moves in projectile motion once it leaves the table top. Its initial velocity is horizontal. SET UP: Take the positive y-direction to be upward. Take the origin of coordinates at the initial position of the book, at the point where it leaves the table top. x-component: 0,xa = 0 1.10 m/s,xv = 0.350 st = y-component: 29.80 m/s ,ya = ? 0 0,yv = 0.350 st = Figure 3.9a Use constant acceleration equations for the x and y components of the motion, with 0xa = and .ya g= ? EXECUTE: (a) 0 ?y y? = 2 2 21 1 0 0 2 20 ( 9.80 m/s )(0.350 s) 0.600 m.y yy y v t a t? = + = + ? = ? The table top is 0.600 m above the floor. (b) 0 ?x x? = 21 0 0 2 (1.10 m/s)(0.350 s) 0 0.358 m.x xx x v t a t? = + = + = (c) 0 1.10 m/sx x xv v a t= + = (The x-component of the velocity is constant, since 0.)xa = 2 0 0 ( 9.80 m/s )(0.350 s) 3.43 m/sy y yv v a t= + = + ? = ? 2 2 3.60 m/sx yv v v= + = 3.43 m/s tan 3.118 1.10 m/s y x v v ? ?= = = ? 72.2? = ? ° Direction of v ! is 72.2° below the horizontal Figure 3.9b 3-6 Chapter 3 (d) The graphs are given in Figure 3.9c Figure 3.9c EVALUATE: In the x-direction, 0xa = and xv is constant. In the y-direction, 29.80 m/sya = ? and yv is downward and increasing in magnitude since ya and yv are in the same directions. The x and y motions occur independently, connected only by the time. The time it takes the book to fall 0.600 m is the time it travels horizontally. 3.10. IDENTIFY: The bomb moves in projectile motion. Treat the horizontal and vertical components of the motion separately. The vertical motion determines the time in the air. SET UP: The initial velocity of the bomb is the same as that of the helicopter. Take y+ downward, so 0xa = , 29.80 m/sya = + , 0 60.0 m/sxv = and 0 0yv = . EXECUTE: (a) 210 0 2y yy y v t a t? = + with 0 300 my y? = gives 0 22( ) 2(300 m) 7.82 s9.80 m/sy y y t a ?= = = . (b) The bomb travels a horizontal distance 210 0 2 (60.0 m/s)(7.82 s) 470 mx xx x v t a t? = + = = . (c) 0 60.0 m/sx xv v= = . 20 (9.80 m/s )(7.82 s) 76.6 m/sy y yv v a t= + = = . (d) The graphs are given in Figure 3.10. (e) Because the airplane and the bomb always have the same x-component of velocity and position, the plane will be 300 m directly above the bomb at impact. EVALUATE: The initial horizontal velocity of the bomb doesn?t affect its vertical motion. Figure 3.10 3.11. IDENTIFY: Each object moves in projectile motion. SET UP: Take y+ to be downward. For each cricket, 0xa = and 29.80 m/sya = + . For Chirpy, 0 0 0x yv v= = . For Milada, 0 0.950 m/sxv = , 0 0yv = EXECUTE: Milada's horizontal component of velocity has no effect on her vertical motion. She also reaches the ground in 3.50 s. 210 0 2 (0.950 m/s)(3.50 s) 3.32 mx xx x v t a t? = + = = EVALUATE: The x and y components of motion are totally separate and are connected only by the fact that the time is the same for both. 3.12. IDENTIFY: The person moves in projectile motion. She must travel 1.75 m horizontally during the time she falls 9.00 m vertically. SET UP: Take y+ downward. 0xa = , 29.80 m/sya = + . 0 0xv v= , 0 0yv = . EXECUTE: Time to fall 9.00 m: 210 0 2y yy y v t a t? = + gives 0 22( ) 2(9.00 m) 1.36 s9.80 m/sy y y t a ?= = = . Speed needed to travel 1.75 m horizontally during this time: 210 0 2x xx x v t a t? = + gives 0 0 0 1.75 m 1.29 m/s 1.36 sx x x v v t ?= = = = . EVALUATE: If she increases her initial speed she still takes 1.36 s to reach the level of the ledge, but has traveled horizontally farther than 1.75 m. 3.13. IDENTIFY: The car moves in projectile motion. The car travels 21.3 m 1.80 m 19.5 m? = downward during the time it travels 61.0 m horizontally. SET UP: Take y+ to be downward. 0xa = , 29.80 m/sya = + . 0 0xv v= , 0 0yv = . Motion in Two or Three Dimensions 3-7 EXECUTE: Use the vertical motion to find the time in the air: 21 0 0 2y yy y v t a t? = + gives 0 22( ) 2(19.5 m) 1.995 s9.80 m/sy y y t a ?= = = Then 210 0 2x xx x v t a t? = + gives 00 0 61.0 m 30.6 m/s1.995 sx x x v v t ?= = = = . (b) 30.6m sxv = since 0xa = . 0 19.6m sy y yv v a t= + = ? . 2 2 36.3m sx yv v v= + = . EVALUATE: We calculate the final velocity by calculating its x and y components. 3.14. IDENTIFY: The marble moves with projectile motion, with initial velocity that is horizontal and has magnitude 0v . Treat the horizontal and vertical motions separately. If 0v is too small the marble will land to the left of the hole and if 0v is too large the marble will land to the right of the hole. SET UP: Let x+ be horizontal to the right and let y+ be upward. 0 0xv v= , 0 0yv = , 0xa = , 29.80 m/sya = ? EXECUTE: Use the vertical motion to find the time it takes the marble to reach the height of the level ground; 0 2.75 my y? = ? . 210 0 2y yy y v t a t? = + gives 0 22( ) 2( 2.75 m) 0.749 s9.80 m/sy y y t a ? ?= = =? . The time does not depend on 0v . Minimum 0 :v 0 2.00 mx x? = , 0.749 st = . 210 0 2x xx x v t a t? = + gives 00 2.00 m 2.67 m/s0.749 s x x v t ?= = = . Maximum 0v : 0 3.50 mx x? = and 0 3.50 m 4.67 m/s0.749 sv = = . EVALUATE: The horizontal and vertical motions are independent and are treated separately. Their only connection is that the time is the same for both. 3.15. IDENTIFY: The ball moves with projectile motion with an initial velocity that is horizontal and has magnitude 0v . The height h of the table and 0v are the same; the acceleration due to gravity changes from 2 E 9.80 m/sg = on earth to Xg on planet X. SET UP: Let x+ be horizontal and in the direction of the initial velocity of the marble and let y+ be upward. 0 0xv v= , 0 0yv = , 0xa = , ya g= ? , where g is either Eg or Xg . EXECUTE: Use the vertical motion to find the time in the air: 0y y h? = ? . 210 0 2y yy y v t a t? = + gives 2ht g= . Then 210 0 2x xx x v t a t? = + gives 0 0 0 2x hx x v t v g? = = . 0x x D? = on earth and 2.76D on Planet X. 0 0( ) 2x x g v h? = , which is constant, so E X2.76D g D g= . 2EX E2 0.131 1.28 m/s(2.76) g g g= = = . EVALUATE: On Planet X the acceleration due to gravity is less, it takes the ball longer to reach the floor, and it travels farther horizontally. 3.16. IDENTIFY: The football moves in projectile motion. SET UP: Let y+ be upward. 0xa = , ya g= ? . At the highest point in the trajectory, 0yv = . EXECUTE: (a) 0y y yv v a t= + . The time t is 0 216.0m s 1.63 s9.80m s yv g = = . (b) Different constant acceleration equations give different expressions but the same numerical result: 2 021 1 02 2 13.1 m2 y y v gt v t g = = = . (c) Regardless of how the algebra is done, the time will be twice that found in part (a), or 3.27 s (d) 0xa = , so 0 0 (20.0 m s)(3.27 s) 65 3 mxx x v t .? = = = . (e) The graphs are sketched in Figure 3.16. 3-8 Chapter 3 EVALUATE: When the football returns to its original level, 20.0 m/sxv = and 16.0 m/syv = ? . Figure 3.16 3.17. IDENTIFY: The shell moves in projectile motion. SET UP: Let x+ be horizontal, along the direction of the shell's motion, and let y+ be upward. 0xa = , 29.80 m/sya = ? . EXECUTE: (a) 0 0 0cos (80.0 m/s)cos60.0 40.0 m/sxv v ?= = =° , 0 0 0sin (80.0 m/s)sin60.0 69.3 m/syv v ?= = =° . (b) At the maximum height 0yv = . 0y y yv v a t= + gives 0 20 69.3 m/s 7.07 s9.80 m/s y y y v v t a ? ?= = =? . (c) 2 20 02 ( )y y yv v a y y= + ? gives 2 2 2 0 0 2 0 (69.3 m/s) 245 m 2 2( 9.80 m/s ) y y y v v y y a ? ?? = = =? . (d) The total time in the air is twice the time to the maximum height, so 21 0 0 2 (40.0 m/s)(14.14 s) 566 mx xx x v t a t? = + = = . (e) At the maximum height, 0 40.0 m/sx xv v= = and 0yv = . At all points in the motion, 0xa = and 29.80 m/sya = ? . EVALUATE: The equation for the horizontal range R derived in Example 3.8 is 2 0 0sin 2vR g ?= . This gives 2 2 (80.0 m/s) sin(120.0 ) 566 m 9.80 m/s R = =° , which agrees with our result in part (d). 3.18. IDENTIFY: The flare moves with projectile motion. The equations derived in Example 3.8 can be used to find the maximum height h and range R. SET UP: From Example 3.8, 2 2 0 0sin 2 v h g ?= and 2 0 0sin 2vR g ?= . EXECUTE: (a) 2 2 2 (125 m/s) (sin55.0 ) 535 m 2(9.80 m/s ) h = =° . 2 2 (125 m/s) (sin110.0 ) 1500 m 9.80 m/s R = =° . (b) h and R are proportional to 1/ g , so on the Moon, 2 2 9.80 m/s (535 m) 3140 m 1.67 m/s h ? ?= =? ?? ? and 2 2 9.80 m/s (1500 m) 8800 m 1.67 m/s R ? ?= =? ?? ? . EVALUATE: The projectile travels on a parabolic trajectory. It is incorrect to say that 0( / 2) tanh R ?= . 3.19. IDENTIFY: The baseball moves in projectile motion. In part (c) first calculate the components of the velocity at this point and then get the resultant velocity from its components. SET UP: First find the x- and y-components of the initial velocity. Use coordinates where the -directiony+ is upward, the -directionx+ is to the right and the origin is at the point where the baseball leaves the bat. 0 0 0cos (30.0 m/s)cos36.9 24.0 m/sxv v ?= = ° = 0 0 0sin (30.0 m/s)sin36.9 18.0 m/syv v ?= = ° = Figure 3.19a Use constant acceleration equations for the x and y motions, with 0xa = and .ya g= ? Motion in Two or Three Dimensions 3-9 EXECUTE: (a) y-component (vertical motion): 0 10.0 m/s,y y? = + 0 18.0 m/s,yv = 29.80 m/s ,ya = ? ?t = 21 0 0 2y yy y v a t? = + 2 210.0 m (18.0 m/s) (4.90 m/s )t t= ? 2 2(4.90 m/s ) (18.0 m/s) 10.0 m 0t t? + = Apply the quadratic formula: ( ) ( )( ) ( )219.80 18.0 18.0 4 4.90 10.0 s 1.837 1.154 st ? ?= ± ? ? = ±? ?? ? The ball is at a height of 10.0 above the point where it left the bat at 1 0.683 st = and at 2 2.99 s.t = At the earlier time the ball passes through a height of 10.0 m as its way up and at the later time it passes through 10.0 m on its way down. (b) 0 24.0 m/s,x xv v= = + at all times since 0.xa = 0y y yv v a t= + 1 0.683 s:t = 218.0 m/s ( 9.80 m/s )(0.683 s) 11.3 m/s.yv = + + ? = + ( yv is positive means that the ball is traveling upward at this point. 2 2.99 s:t = 218.0 m/s ( 9.80 m/s )(2.99 s) 11.3 m/s.yv = + + ? = ? ( yv is negative means that the ball is traveling downward at this point.) (c) 0 24.0 m/sx xv v= = Solve for :yv ?,yv = 0 0y y? = (when ball returns to height where motion started), 29.80 m/s ,ya = ? 0 18.0 m/syv = + 2 2 0 02 ( )y y yv v a y y= + ? 0 18.0 m/sy yv v= ? = ? (negative, since the baseball must be traveling downward at this point) Now that have the components can solve for the magnitude and direction of .v ! 2 2 x yv v v= + ( ) ( )2 224.0 m/s 18.0 m/s 30.0 m/sv = + ? = 18.0 m/s tan 24.0 m/s y x v v ? ?= = 36.9 ,? = ? ° 36.9° below the horizontal Figure 3.19b The velocity of the ball when it returns to the level where it left the bat has magnitude 30.0 m/s and is directed at an angle of 36.9° below the horizontal. EVALUATE: The discussion in parts (a) and (b) explains the significance of two values of t for which 0 10.0 m.y y? = + When the ball returns to its initial height, our results give that its speed is the same as its initial speed and the angle of its velocity below the horizontal is equal to the angle of its initial velocity above the horizontal; both of these are general results. 3.20. IDENTIFY: The shot moves in projectile motion. SET UP: Let y+ be upward. EXECUTE: (a) If air resistance is to be ignored, the components of acceleration are 0 horizontally and 29.80 m sg? = ? vertically downward. (b) The x-component of velocity is constant at (12.0 m s)cos51.0 7.55 m sxv = ° = . The y-component is 0 (12.0 m s)sin51.0 9.32 m syv = ° = at release and 20 (10.57 m s) (9.80 m s )(2.08 s) 11.06 m sy yv v gt= ? = ? = ? when the shot hits. (c) 0 0 (7.55 m s)(2.08 s) 15 7 mxx x v t .? = = = . (d) The initial and final heights are not the same. (e) With 0y = and 0 yv as found above, Eq.(3.18) gives 0 1.81my = . (f) The graphs are sketched in Figure 3.20. 3-10 Chapter 3 EVALUATE: When the shot returns to its initial height, 9.32 m/syv = ? . The shot continues to accelerate downward as it travels downward 1.81 m to the ground and the magnitude of yv at the ground is larger than 9.32 m/s. Figure 3.20 3.21. IDENTIFY: Take the origin of coordinates at the point where the quarter leaves your hand and take positive y to be upward. The quarter moves in projectile motion, with 0,xa = and .ya g= ? It travels vertically for the time it takes it to travel horizontally 2.1 m. Figure 3.21 (a) SET UP: Use the horizontal (x-component) of motion to solve for t, the time the quarter travels through the air: ?,t = 0 2.1 m,x x? = 0 3.2 m/s,xv = 0xa = 21 0 0 02 ,x x xx x v t a t v t? = + = since 0xa = EXECUTE: 0 0 2.1 m 0.656 s 3.2 m/sx x x t v ?= = = SET UP: Now find the vertical displacement of the quarter after this time: 0 ?,y y? = 29.80 m/s ,ya = ? 0 5.54 m/s,yv = + 0.656 st = 21 0 0 2y yy y v t a t? + + EXECUTE: 2 210 2(5.54 m/s)(0.656 s) ( 9.80 m/s )(0.656 s) 3.63 m 2.11 m 1.5 m.y y? = + ? = ? = (b) SET UP: ?,yv = 0.656 s,t = 29.80 m/s ,ya = ? 0 5.54 m/syv = + 0y y yv v a t= + EXECUTE: 25.54 m/s ( 9.80 m/s )(0.656 s) 0.89 m/s.yv = + ? = ? EVALUATE: The minus sign for yv indicates that the y-component of v! is downward. At this point the quarter has passed through the highest point in its path and is on its way down. The horizontal range if it returned to its original height (it doesn?t!) would be 3.6 m. It reaches its maximum height after traveling horizontally 1.8 m, so at 0 2.1 mx x? = it is on its way down. 3.22. IDENTIFY: Use the analysis of Example 3.10. SET UP: From Example 3.10, 0 0cos d t v ?= and 21 dart 0 0 2( sin )y v t gt?= ? . EXECUTE: Substituting for t in terms of d in the expression for darty gives dart 0 2 2 0 0 tan . 2 cos gd y d v ? ? ? ?= ?? ?? ? Using the given values for d and 0? to express this as a function of 0v , 2 2 2 0 26.62 m s (3.00 m) 0.90 .y v ? ?= ?? ?? ? (a) 0 12.0 m/sv = gives 2.14 my = . (b) 0 8.0 m/sv = gives 1.45 my = . 0 0 0cos (6.4 m/s)cos60xv v ?= = ° 0 3.20 m/sxv = 0 0 0sin (6.4 m/s)sin 60yv v ?= = ° 0 5.54 m/syv = Motion in Two or Three Dimensions 3-11 (c) 0 4.0 m/sv = gives 2.29 my = ? . In this case, the dart was fired with so slow a speed that it hit the ground before traveling the 3-meter horizontal distance. EVALUATE: For (a) and (d) the trajectory of the dart has the shape shown in Figure 3.26 in the textbook. For (c) the dart moves in a parabola and returns to the ground before it reaches the x-coordinate of the monkey. 3.23. IDENTIFY: Take the origin of coordinates at the roof and let the -directiony+ be upward. The rock moves in projectile motion, with 0xa = and .ya g= ? Apply constant acceleration equations for the x and y components of the motion. SET UP: 0 0 0cos 25.2 m/sxv v ?= = 0 0 0sin 16.3 m/syv v ?= = Figure 3.23a (a) At the maximum height 0.yv = 29.80 m/s ,ya = ? 0,yv = 0 16.3 m/s,yv = + 0 ?y y? = 2 2 0 02 ( )y y yv v a y y= + ? EXECUTE: 2 2 2 0 0 2 0 (16.3 m/s) 13.6 m 2 2( 9.80 m/s ) y y y v v y y a ? ?? = = = +? (b) SET UP: Find the velocity by solving for its x and y components. 0 25.2 m/sx xv v= = (since 0xa = ) ?,yv = 29.80 m/s ,ya = ? 0 15.0 my y? = ? (negative because at the ground the rock is below its initial position), 0 16.3 m/syv = 2 2 0 02 ( )y y yv v a y y= + ? 2 0 02 ( )y y yv v a y y= ? + ? ( yv is negative because at the ground the rock is traveling downward.) EXECUTE: 2 2(16.3 m/s) 2( 9.80 m/s )( 15.0 m) 23.7 m/syv = ? + ? ? = ? Then 2 2 2 2(25.2 m/s) ( 23.7 m/s) 34.6 m/s.x yv v v= + = + ? = (c) SET UP: Use the vertical motion (y-component) to find the time the rock is in the air: ?,t = 23.7 m/syv = ? (from part (b)), 29.80 m/s ,ya = ? 0 16.3 m/syv = + EXECUTE: 0 223.7 m/s 16.3 m/s 4.08 s9.80 m/s y y y v v t a ? ? ?= = = +? SET UP: Can use this t to calculate the horizontal range: 4.08 s,t = 0 25.2 m/s,xv = 0,xa = 0 ?x x? = EXECUTE: 210 0 2 (25.2 m/s)(4.08 s) 0 103 mx xx x v t a t? = + = + = (d) Graphs of x versus t, y versus t, xv versus t, and yv versus t: Figure 3.23b 3-12 Chapter 3 EVALUATE: The time it takes the rock to travel vertically to the ground is the time it has to travel horizontally. With 0 16.3 m/syv = + the time it takes the rock to return to the level of the roof ( 0)y = is 02 / 3.33 s.yt v g= = The time in the air is greater than this because the rock travels an additional 15.0 m to the ground. 3.24. IDENTIFY: Consider the horizontal and vertical components of the projectile motion. The water travels 45.0 m horizontally in 3.00 s. SET UP: Let y+ be upward. 0xa = , 29.80 m/sya = ? . 0 0 0cosxv v ?= , 0 0 0sinyv v ?= . EXECUTE: (a) 210 0 2x xx x v t a t? = + gives 0 0 0(cos )x x v t?? = and 0 45.0 mcos 0.600(25.0 m/s)(3.00 s)? = = ; 0 53.1? = ° (b) At the highest point 0 (25.0 m/s)cos53.1 15.0 m/sx xv v= = ° = , 0yv = and 2 2 15.0 m/sx yv v v= + = . At all points in the motion, 29.80 m/sa = downward. (c) Find 0y y? when 3 00st = . : 2 2 21 1 0 0 2 2(25.0 m/s)(sin53.1 )(3.00 s) ( 9.80 m/s )(3.00 s) 15.9 my yy y v t a t? = + = ° + ? = 0 15.0 m/sx xv v= = , 20 (25.0 m/s)(sin53.1 ) (9.80m/s )(3.00 s) 9.41 m/sy y yv v a t= + = ° ? = ? , and 2 2 2 2(15.0 m/s) ( 9.41 m/s) 17.7 m/sx yv v v= + = + ? = EVALUATE: The acceleration is the same at all points of the motion. It takes the water 0 2 20.0 m/s 2.04 s 9.80 m/s y y v t a = ? = ? =? to reach its maximum height. When the water reaches the building it has passed its maximum height and its vertical component of velocity is downward. 3.25. IDENTIFY and SET UP: The stone moves in projectile motion. Its initial velocity is the same as that of the balloon. Use constant acceleration equations for the x and y components of its motion. Take y+ to be upward. EXECUTE: (a) Use the vertical motion of the rock to find the initial height. 6.00 s,t = 0 20.0 m/s,yv = + 29.80 m/s ,ya = + 0 ?y y? = 21 0 0 2y yy y v t a t? = + gives 0 296 my y? = (b) In 6.00 s the balloon travels downward a distance 0 (20.0 s)(6.00 s) 120 m.y y? = = So, its height above ground when the rock hits is 296 m 120 m 176 m.? = (c) The horizontal distance the rock travels in 6.00 s is 90.0 m. The vertical component of the distance between the rock and the basket is 176 m, so the rock is 2 2(176 m) (90 m) 198 m+ = from the basket when it hits the ground. (d) (i) The basket has no horizontal velocity, so the rock has horizontal velocity 15.0 m/s relative to the basket. Just before the rock hits the ground, its vertical component of velocity is 0y y yv v a t= + = 220.0 m/s (9.80 m/s )(6.00 s) 78.8 m/s,+ = downward, relative to the ground. The basket is moving downward at 20.0 m/s, so relative to the basket the rock has downward component of velocity 58.8 m/s. (e) horizontal: 15.0 m/s; vertical: 78.8 m/s EVALUATE: The rock has a constant horizontal velocity and accelerates downward 3.26. IDENTIFY: The shell moves as a projectile. To just clear the top of the cliff, the shell must have 0 25.0 my y? = when it has 0 60.0 mx x? = . SET UP: Let y+ be upward. 0xa = , ya g= ? . 0 0 cos43xv v= ° , 0 0 sin 43yv v= ° . EXECUTE: (a) horizontal motion: 0 0 0 60.0 m so ( cos43 )x x x v t t v t ? = = ° . vertical motion: 2 2 21 10 0 02 2 gives 25.0m ( sin 43.0 ) ( 9.80m/s )y yy y v t a t v t t? = + = ° + ? . Solving these two simultaneous equations for 0v and t gives 0 3.26 m/sv = and 2.51 st = . (b) yv when shell reaches cliff: 2 0 (32.6 m/s) sin 43.0 (9.80 m/s )(2.51 s) 2.4 m/s y y yv v a t= + = ° ? = ? The shell is traveling downward when it reaches the cliff, so it lands right at the edge of the cliff. EVALUATE: The shell reaches its maximum height at 0 2.27 sy y v t a = ? = , which confirms that at 2.51 st = it has passed its maximum height and is on its way down when it strikes the edge of the cliff. 3.27. IDENTIFY: The suitcase moves in projectile motion. The initial velocity of the suitcase equals the velocity of the airplane. Motion in Two or Three Dimensions 3-13 SET UP: Take y+ to be upward. 0xa = , ya g= ? . EXECUTE: Use the vertical motion to find the time it takes the suitcase to reach the ground: 2 0 0 0 sin23 , 9.80 m/s , 114 m, ?y yv v a y y t= ° = ? ? = ? = 210 0 2 gives 9.60 sy yy y v t a t t? = + = . The distance the suitcase travels horizontally is 0 0 0( cos23.0 ) 795 mxx x v v t? = = ° = . EVALUATE: An object released from rest at a height of 114 m strikes the ground at 02( ) 4.82 sy yt g ?= =? . The suitcase is in the air much longer than this since it initially has an upward component of velocity. 3.28. IDENTIFY: Determine how rada depends on the rotational period T . SET UP: 2 rad 2 4 R a T ?= . EXECUTE: For any item in the washer, the centripetal acceleration will be inversely proportional to the square of the rotational period; tripling the centripetal acceleration involves decreasing the period by a factor of 3 , so that the new period T ? is given in terms of the previous period T by / 3T T? = . EVALUATE: The rotational period must be decreased in order to increase the rate of rotation and therefore increase the centripetal acceleration. 3.29. IDENTIFY: Apply Eq. (3.30). SET UP: 24 hT = . EXECUTE: (a) 2 6 2 3 rad 2 4 (6.38 10 m) 0.034 m/s 3.4 10 . ((24 h)(3600 s/h)) a g ? ?×= = = × (b) Solving Eq. (3.30) for the period T with rada g= , 2 6 2 4 (6.38 10 m) 5070 s =1.4 h. 9.80 m/s T ? ×= = EVALUATE: rada is proportional to 21/ T , so to increase rada by a factor of 31 2943.4 10? =× requires that T be multiplied by a factor of 1 294 . 24 h 1.4 h 294 = . 3.30. IDENTIFY: Each blade tip moves in a circle of radius 3.40 mR = and therefore has radial acceleration 2 rad /a v R= . SET UP: 550 rev/min 9.17 rev/s= , corresponding to a period of 1 0.109 s 9.17 rev/s T = = . EXECUTE: (a) 2 196 m/sRv T ?= = . (b) 2 4 2 3 rad 1.13 10 m/s 1.15 10 v a g R = = × = × . EVALUATE: 2 rad 2 4 R a T ?= gives the same results for rada as in part (b). 3.31. IDENTIFY: Apply Eq.(3.30). SET UP: 7.0 mR = . 29.80 m/sg = . EXECUTE: (a) Solving Eq. (3.30) for T in terms of R and rada , 2 2 2 rad4 / 4 (7.0 m)/(3.0)(9.80 m/s ) 3.07 sT R a? ?= = = . (b) rad 10a g= gives 1.68 sT = . EVALUATE: When rada increases, T decreases. 3.32. IDENTIFY: Each planet moves in a circular orbit and therefore has acceleration 2rad /a v R= . SET UP: The radius of the earth?s orbit is 111.50 10 mr = × and its orbital period is 7365 days 3.16 10 sT = = × . For Mercury, 105.79 10 mr = × and 688.0 days 7.60 10 sT = = × . EXECUTE: (a) 42 2.98 10 m/srv T ?= = × (b) 2 3 2 rad 5.91 10 m/s v a r ?= = × . (c) 44.79 10 m/sv = × , and 2 2rad 3.96 10 m/sa ?= × . 3-14 Chapter 3 EVALUATE: Mercury has a larger orbital velocity and a larger radial acceleration than earth. 3.33. IDENTIFY: Uniform circular motion. SET UP: Since the magnitude of v! is constant. tan 0dv dt= = v ! and the resultant acceleration is equal to the radial component. At each point in the motion the radial component of the acceleration is directed in toward the center of the circular path and its magnitude is given by 2 / .v R EXECUTE: (a) 2 2 2 rad (7.00 m/s) 3.50 m/s , 14.0 m v a R = = = upward. (b) The radial acceleration has the same magnitude as in part (a), but now the direction toward the center of the circle is downward. The acceleration at this point in the motion is 23.50 m/s , downward. (c) SET UP: The time to make one rotation is the period T , and the speed v is the distance for one revolution divided by T . EXECUTE: 2 Rv T ?= so 2 2 (14.0 m) 12.6 s 7.00 m/s R T v ? ?= = = EVALUATE: The radial acceleration is constant in magnitude since v is constant and is at every point in the motion directed toward the center of the circular path. The acceleration is perpendicular to v ! and is nonzero because the direction of v ! changes. 3.34. IDENTIFY: The acceleration is the vector sum of the two perpendicular components, rada and tana . SET UP: tana is parallel to v! and hence is associated with the change in speed; 2tan 0.500 m/sa = . EXECUTE: (a) 2 2 2rad / (3 m/s) /(14 m) 0.643 m/sa v R= = = . 2 2 2 2 1/ 2 2((0.643 m/s ) (0.5 m/s ) ) 0.814 m/s , 37.9a = + = ° to the right of vertical. (b) The sketch is given in Figure 3.34. Figure 3.34 3.35. IDENTIFY: Each part of his body moves in uniform circular motion, with 2 rad v a R = . The speed in rev/s is 1/ T , where T is the period in seconds (time for 1 revolution). The speed v increases with R along the length of his body but all of him rotates with the same period T . SET UP: For his head 8.84 mR = and for his feet 6.84 mR = . EXECUTE: (a) 2rad (8.84 m)(12.5)(9.80 m/s ) 32.9 m/sv Ra= = = (b) Use 2 rad 2 4 R a T ?= . Since his head has rad 12.5a g= and 8.84 mR = , 2 rad 8.84 m 2 2 1.688 s 12.5(9.80 m/s ) R T a ? ?= = = . Then his feet have 2 2 rad 2 2 4 (6.84 m) 94.8 m/s 9.67 (1.688 s) R a g T ?= = = = . The difference between the acceleration of his head and his feet is 212.5 9.67 2.83 27.7 m/sg g g? = = . (c) 1 1 0.592 rev/s 35.5 rpm 1.69 sT = = = EVALUATE: His feet have speed 2rad (6.84 m)(94.8 m/s ) 25.5 m/sv Ra= = = 3.36. IDENTIFY: The relative velocities are S/Fv! , the velocity of the scooter relative to the flatcar, S/Gv! , the scooter relative to the ground and F/Gv ! , the flatcar relative to the ground. S/G S/F F/Gv = v + v ! ! ! . Carry out the vector addition by drawing a vector addition diagram. SET UP: S/F S/G F/G?v = v v! ! ! . F/Gv! is to the right, so F/G?v! is to the left. EXECUTE: In each case the vector addition diagram gives Motion in Two or Three Dimensions 3-15 (a) 5.0 m/s to the right (b) 16.0 m/s to the left (c) 13.0 m/s to the left. EVALUATE: The scooter has the largest speed relative to the ground when it is moving to the right relative to the flatcar, since in that case the two velocities S/Fv ! and F/Gv ! are in the same direction and their magnitudes add. 3.37. IDENTIFY: Relative velocity problem. The time to walk the length of the moving sidewalk is the length divided by the velocity of the woman relative to the ground. SET UP: Let W stand for the woman, G for the ground, and S for the sidewalk. Take the positive direction to be the direction in which the sidewalk is moving. The velocities are W/Gv (woman relative to the ground), W/Sv (woman relative to the sidewalk), and S/Gv (sidewalk relative to the ground). Eq.(3.33) becomes W/G W/S S/G .v v v= + The time to reach the other end is given by W/G distance traveled relative to ground t v = EXECUTE: (a) S/G 1.0 m/sv = W/S 1.5 m/sv = + W/G W/S S/G 1.5 m/s 1.0 m/s 2.5 m/s.v v v= + = + = W/G 35.0 m 35.0 m 14 s. 2.5 m/s t v = = = (b) S/G 1.0 m/sv = W/S 1.5 m/sv = ? W/G W/S S/G 1.5 m/s 1.0 m/s 0.5 m/s.v v v= + = ? + = ? (Since W/Gv now is negative, she must get on the moving sidewalk at the opposite end from in part (a).) W/G 35.0 m 35.0 m 70 s. 0.5 m/s t v ? ?= = =? EVALUATE: Her speed relative to the ground is much greater in part (a) when she walks with the motion of the sidewalk. 3.38. IDENTIFY: Calculate the rower?s speed relative to the shore for each segment of the round trip. SET UP: The boat?s speed relative to the shore is 6.8 km/h downstream and 1.2 km/h upstream. EXECUTE: The walker moves a total distance of 3.0 km at a speed of 4.0 km/h, and takes a time of three fourths of an hour (45.0 min). The total time the rower takes is 1.5 km 1.5 km 1.47 h 88.2 min. 6.8 km/h 1.2 km/h + = = EVALUATE: It takes the rower longer, even though for half the distance his speed is greater than 4.0 km/h. The rower spends more time at the slower speed. 3.39. IDENTIFY: Apply the relative velocity relation. SET UP: The relative velocities are C/Ev! , the canoe relative to the earth, R/Ev! , the velocity of the river relative to the earth and C/Rv ! , the velocity of the canoe relative to the river. EXECUTE: C/E C/R R/Ev = v + v! ! ! and therefore C/R C/E R/E?v = v v! ! ! . The velocity components of C/Rv! are 0.50 m/s (0.40 m/s)/ 2, east and (0.40 m/s)/ 2, south,? + for a velocity relative to the river of 0.36 m/s, at 52.5° south of west. EVALUATE: The velocity of the canoe relative to the river has a smaller magnitude than the velocity of the canoe relative to the earth. 3.40. IDENTIFY: Use the relation that relates the relative velocities. SET UP: The relative velocities are the velocity of the plane relative to the ground, P/Gv! , the velocity of the plane relative to the air, P/Av ! , and the velocity of the air relative to the ground, A/Gv ! . P/Gv ! must due west and A/Gv ! must be south. A/G 80 km/hv = and P/A 320 km/hv = . P/G P/A A/Gv = v + v! ! ! . The relative velocity addition diagram is given in Figure 3.40. EXECUTE: (a) A/G P/A 80 km/h sin 320 km/h v v ? = = and 14? = ° , north of west. (b) 2 2 2 2P/G P/A A/G (320 km/h) (80.0 km/h) 310 km/hv v v= ? = ? = . 3-16 Chapter 3 EVALUATE: To travel due west the velocity of the plane relative to the air must have a westward component and also a component that is northward, opposite to the wind direction. Figure 3.40 3.41. IDENTIFY: Relative velocity problem in two dimensions. His motion relative to the earth (time displacement) depends on his velocity relative to the earth so we must solve for this velocity. (a) SET UP: View the motion from above. The velocity vectors in the problem are: M/E ,v ! the velocity of the man relative to the earth W/E ,v ! the velocity of the water relative to the earth M/W ,v ! the velocity of the man relative to the water The rule for adding these velocities is M/E M/W W/Ev = v + v ! ! ! Figure 3.41a The problem tells us that W/Ev ! has magnitude 2.0 m/s and direction due south. It also tells us that M/Wv ! has magnitude 4.2 m/s and direction due east. The vector addition diagram is then as shown in Figure 3.41b This diagram shows the vector addition M/E M/W W/Ev = v + v ! ! ! and also has M/Wv ! and W/Ev ! in their specified directions. Note that the vector diagram forms a right triangle. Figure 3.41b The Pythagorean theorem applied to the vector addition diagram gives 2 2 2M/E M/W W/E.v v v= + EXECUTE: 2 2 2 2M/E M/W W/E (4.2 m/s) (2.0 m/s) 4.7 m/sv v v= + = + = M/W W/E 4.2 m/s tan 2.10; 2.0 m/s v v ? = = = 65 ;? = ° or 90 25 .? ?= ° ? = ° The velocity of the man relative to the earth has magnitude 4.7 m/s and direction 25 S° of E. (b) This requires careful thought. To cross the river the man must travel 800 m due east relative to the earth. The man?s velocity relative to the earth is M/E.v ! But, from the vector addition diagram the eastward component of M/Ev equals M/W 4.2 m/s.v = Thus 0 800 m 190 s. 4.2 m/sx x x t v ?= = = (c) The southward component of M/Ev ! equals W/E 2.0 m/s.v = Therefore, in the 190 s it takes him to cross the river the distance south the man travels relative to the earth is 0 (2.0 m/s)(190 s) 380 m.yy y v t? = = = EVALUATE: If there were no current he would cross in the same time, (800 m) /(4.2 m/s) 190 s.= The current carries him downstream but doesn?t affect his motion in the perpendicular direction, from bank to bank. 3.42. IDENTIFY: Use the relation that relates the relative velocities. SET UP: The relative velocities are the water relative to the earth, W/Ev! , the boat relative to the water, B/Wv! , and the boat relative to the earth, B/Ev ! . B/Ev ! is due east, W/Ev ! is due south and has magnitude 2.0 m/s. B/W 4.2 m/sv = . B/E B/W W/E= +v v v! ! ! . The velocity addition diagram is given in Figure 3.42. Motion in Two or Three Dimensions 3-17 EXECUTE: (a) Find the direction of B/Wv! . W/E B/W 2.0 m/s sin 4.2 m/s v v ? = = . 28.4? = ° , north of east. (b) 2 2 2 2B/E B/W W/E (4.2 m/s) (2.0 m/s) 3.7 m/sv v v= ? = ? = (c) B/E 800 m 800 m 216 s 3.7 m/s t v = = = . EVALUATE: It takes longer to cross the river in this problem than it did in Problem 3.41. In the direction straight across the river (east) the component of his velocity relative to the earth is lass than 4.2 m/s. Figure 3.42 3.43. IDENTIFY: Relative velocity problem in two dimensions. (a) SET UP: P/Av! is the velocity of the plane relative to the air. The problem states that P/Av! has magnitude 35 m/s and direction south. A/Ev ! is the velocity of the air relative to the earth. The problem states that A/Ev ! is to the southwest ( 45 S° of W) and has magnitude 10 m/s. The relative velocity equation is P/E P/A A/E.= +v v v! ! ! Figure 3.43a EXECUTE: (b) P/A( ) 0,xv = P/A( ) 35 m/syv = ? A/E( ) (10 m/s)cos45 7.07 m/s,xv = ? ° = ? A/E( ) (10 m/s)sin 45 7.07 m/syv = ? ° = ? P/E P/A A/E( ) ( ) ( ) 0 7.07 m/s 7.1 m/sx x xv v v= + = ? = ? P/E P/A A/E( ) ( ) ( ) 35 m/s 7.07 m/s 42 m/sy y yv v v= + = ? ? = ? (c) 2 2 P/E P/E P/E( ) ( )x yv v v= + 2 2 P/E ( 7.1 m/s) ( 42 m/s) 43 m/sv = ? + ? = P/E P/E ( ) 7.1 tan 0.169 ( ) 42 x y v v ? ?= = =? 9.6 ;? = ° ( 9.6° west of south) Figure 3.43b EVALUATE: The relative velocity addition diagram does not form a right triangle so the vector addition must be done using components. The wind adds both southward and westward components to the velocity of the plane relative to the ground. 3-18 Chapter 3 3.44. IDENTIFY: Use Eqs.(2.17) and (2.18). SET UP: At the maximum height 0yv = . EXECUTE: (a) 3 20 0, 3 2x x y yv v t v v t t ? ??= + = + ? , and 4 2 30 0, 12 2 6x yx v t t y v t t t ? ? ?= + = + ? . (b) Setting 0yv = yields a quadratic in 20, 0 2yt v t t ??= + ? , which has as the positive solution 2 0 1 2 13.59 st v? ? ?? ? ?= + + =? ? . Using this time in the expression for y(t) gives a maximum height of 341 m. (c) The path of the rocket is sketched in Figure 3.44. (d) 0y = gives 2 300 2 6yv t t t ? ?= + ? and 2 0 06 2 yt t v ? ?? ? = . The positive solution is 20.73 st = . For this t, 43.85 10 mx = × . EVALUATE: The graph in part (c) shows the path is not symmetric about the highest point and the time to return to the ground is less than twice the time to the maximum height. Figure 3.44 3.45. IDENTIFY: d dt r v = !! and d dt v a = !! SET UP: 1( )n nd t nt dt ?= . At 1.00 st = , 24.00 m/sxa = and 23.00 m/sya = . At 0t = , 0x = and 50.0 my = . EXECUTE: (a) 2x dxv Btdt= = . 2 x x dv a B dt = = , which is independent of t. 24.00 m/sxa = gives 22.00 m/sB = . 23y dy v Dt dt = = . 6yy dva Dtdt= = . 23.00 m/sya = gives 20.500 m/sD = . 0x = at 0t = gives 0A = . 50.0 my = at 0t = gives 50.0 mC = . (b) At 0t = , 0xv = and 0yv = , so 0v =! . At 0t = , 22 4.00 m/sxa B= = and 0ya = , so 2 ?(4.00 m/s )a = i! . (c) At 10.0 st = , 22(2.00 m/s )(10.0 s) 40.0 m/sxv = = and 3 23(0.500 m/s )(10.0 s) 150 m/syv = = . 2 2 155 m/sx yv v v= + = . (d) 2 2(2.00 m/s )(10.0 s) 200 mx = = , 3 350.0 m (0.500 m/s )(10.0 s) 550 my = + = . ? ?(200 m) (550 m)r = i + j! . EVALUATE: The velocity and acceleration vectors as functions of time are 2? ?( ) (2 ) (3 )t Bt Dtv = i + j ! and ? ?( ) (2 ) (6 )t B Dta = i + j ! . The acceleration is not constant. 3.46. IDENTIFY: 0 0 ( ) t t dt?r = r + v! ! ! and ddtva = ! . SET UP: At 0t = , 0 0x = and 0 0y = . EXECUTE: (a) Integrating, 3 2? ?( ) ( ) 3 2 t t t ? ?? ?r = i + j! . Differentiating, ? ?( 2 )t? ??a = i + j! . (b) The positive time at which 0x = is given by 2 3t ? ?= . At this time, the y-coordinate is 2 2 3 3 3(2.4 m/s)(4.0 m/s ) 9.0 m 2 2 2(1.6 m/s ) y t ? ?? ?= = = = . EVALUATE: The acceleration is not constant. Motion in Two or Three Dimensions 3-19 3.47. IDENTIFY: Once the rocket leaves the incline it moves in projectile motion. The acceleration along the incline determines the initial velocity and initial position for the projectile motion. SET UP: For motion along the incline let x+ be directed up the incline. 2 20 02 ( )x x xv v a x x= + ? gives 22(1.25 m/s )(200 m) 22.36 m/sxv = = . When the projectile motion begins the rocket has 0 22.36 m/sv = at 35.0° above the horizontal and is at a vertical height of (200.0 m)sin35.0 114.7 m=° . For the projectile motion let x+ be horizontal to the right and let y+ be upward. Let 0y = at the ground. Then 0 114.7 my = , 0 0 cos35.0 18.32 m/sxv v= =° , 0 0 sin35.0 12.83 m/syv v= =° , 0xa = , 29.80 m/sya = ? . Let 0x = at point A, so 0 (200.0 m)cos35.0 163.8 mx = =° . EXECUTE: (a) At the maximum height 0yv = . 2 20 02 ( )y y yv v a y y= + ? gives 2 2 2 0 0 2 0 (12.83 m/s) 8.40 m 2 2( 9.80 m/s ) y y y v v y y a ? ?? = = =? and 114.7 m 8.40 m 123 my = + = . The maximum height above ground is 123 m. (b) The time in the air can be calculated from the vertical component of the projectile motion: 0 114.7 my y? = ? , 0 12.83 m/syv = , 29.80 m/sya = ? . 210 0 2y yy y v t a t? = + gives 2 2(4.90 m/s ) (12.83 m/s) 114.7 mt t? ? . The quadratic formula gives ( )21 12.83 (12.83) 4(4.90)(114.7) s9.80t = ± + . The positive root is 6.32 st = . Then 21 0 0 2 (18.32 m/s)(6.32 s) 115.8 mx xx x v t a t? = + = = and 163.8 m 115.8 m 280 mx = + = . The horizontal range of the rocket is 280 m. EVALUATE: The expressions for h and R derived in Example 3.8 do not apply here. They are only for a projectile fired on level ground. 3.48. IDENTIFY: The person moves in projectile motion. Use the results in Example 3.8 to determine how T , h and D depend on g and set up a ratio. SET UP: From Example 3.8, the time in the air is 0 02 sinvt g ?= , the maximum height is 2 2 0 0sin 2 v h g ?= and the horizontal range (called D in the problem) is 2 0 0sin 2vD g ?= . The person has the same 0v and 0? on Mars as on the earth. EXECUTE: 0 02 sintg v ?= , which is constant, so E E M Mt g t g= . E EM E E E M E 2.64 0.379 g g t t t t g g ? ? ? ?= = =? ? ? ?? ? ? ? . 2 2 0 0sin 2 v hg ?= , which is constant, so E E M Mh g h g= . EM E E M 2.64 g h h h g ? ?= =? ?? ? . 20 0sin 2Dg v ?= , which is constant, so E E M MD g D g= . EM E E M 2.64 g D D D g ? ?= =? ?? ? . EVALUATE: All three quantities are proportional to 1/ g so all increase by the same factor of E M/ 2.64g g = . 3.49. IDENTIFY: The range for a projectile that lands at the same height from which it was launched is 2 0 sin 2v ?R g = . SET UP: The maximum range is for 45? = ° . EXECUTE: Assuming 45? = ° , and 50 mR = , 0 22 m/sv gR= = . EVALUATE: We have assumed that debris was launched at all angles, including the angle of 45° that gives maximum range. 3.50. IDENTIFY: The velocity has a horizontal tangential component and a vertical component. The vertical component of acceleration is zero and the horizontal component is 2 rad xva R = SET UP: Let y+ be upward and x+ be in the direction of the tangential velocity at the instant we are considering. 3-20 Chapter 3 EXECUTE: (a) The bird?s tangential velocity can be found from circumference 2 (8.00 m) 50.27 m 10.05 m/s time of rotation 5.00 s 5.00 sx v ?= = = = Thus its velocity consists of the components 10.05 m/sxv = and 3.00 m/syv = . The speed relative to the ground is then 2 2 10.5 m/sx yv v v= + = . (b) The bird?s speed is constant, so its acceleration is strictly centripetal?entirely in the horizontal direction, toward the center of its spiral path?and has magnitude 2 2 2 rad (10.05 m/s) 12.6 m/s 8.00 m xva r = = = . (c) Using the vertical and horizontal velocity components 1 3.00 m/s tan 16.6 10.05 m/s ? ?= = ° . EVALUATE: The angle between the bird?s velocity and the horizontal remains constant as the bird rises. 3.51. IDENTIFY: Take y+ to be downward. Both objects have the same vertical motion, with 0 yv and .ya g= + Use constant acceleration equations for the x and y components of the motion. SET UP: Use the vertical motion to find the time in the air: 0 0,yv = 9.80 m/s ,ya 2= 0 25 m,y y? = ?t = EXECUTE: 210 0 2y yy y v t a t? = + gives 2.259 st = During this time the dart must travel 90 m, so the horizontal component of its velocity must be 0 0 90 m 40 m/s 2.25 sx x x v t ?= = = EVALUATE: Both objects hit the ground at the same time. The dart hits the monkey for any muzzle velocity greater than 40 m/s. 3.52. IDENTIFY: The person moves in projectile motion. Her vertical motion determines her time in the air. SET UP: Take y+ upward. 0 15.0 m/sxv = , 0 10.0 m/syv = + , 0xa = , 29.80 m/sya = ? . EXECUTE: (a) Use the vertical motion to find the time in the air: 210 0 2y yy y v t a t? = + with 0 30.0 my y? = ? gives 2 230.0 m (10.0 m/s) (4.90 m/s )t t? = ? . The quadratic formula gives ( )21 10.0 ( 10.0) 4(4.9)( 30) s2(4.9)t = + ± ? ? ? . The positive solution is 3.70 st = . During this time she travels a horizontal distance 210 0 2 (15.0 m/s)(3.70 s) 55.5 mx xx x v t a t? = + = = . She will land 55.5 m south of the point where she drops from the helicopter and this is where the mats should have been placed. (b) The x-t, y-t, xv -t and yv -t graphs are sketched in Figure 3.52. EVALUATE: If she had dropped from rest at a height of 30.0 m it would have taken her 22(30.0 m) 2.47 s9.80 m/st = = . She is in the air longer than this because she has an initial vertical component of velocity that is upward. Figure 3.52 3.53. IDENTIFY: The cannister moves in projectile motion. Its initial velocity is horizontal. Apply constant acceleration equations for the x and y components of motion. Motion in Two or Three Dimensions 3-21 SET UP: Take the origin of coordinates at the point where the canister is released. Take y+ to be upward. The initial velocity of the canister is the velocity of the plane, 64.0 m/s in the -direction.x+ Figure 3.53 Use the vertical motion to find the time of fall: ?,t = 0 0,yv = 29.80 m/s ,ya = ? 0 90.0 my y? = ? (When the canister reaches the ground it is 90.0 m below the origin.) 21 0 0 2y yy y v t a t? = + EXECUTE: Since 0 0,yv = 0 22( ) 2( 90.0 m) 4.286 s.9.80 m/sy y y t a ? ?= = =? SET UP: Then use the horizontal component of the motion to calculate how far the canister falls in this time: 0 ?,x x? = 0,xa ? 0 64.0 m/s,xv = EXECUTE: 210 0 2 (64.0 m/s)(4.286 s) 0 274 m.x x v t at? = + = + = EVALUATE: The time it takes the cannister to fall 90.0 m, starting from rest, is the time it travels horizontally at constant speed. 3.54. IDENTIFY: The equipment moves in projectile motion. The distance D is the horizontal range of the equipment plus the distance the ship moves while the equipment is in the air. SET UP: For the motion of the equipment take x+ to be to the right and y+ to be upwards. Then 0xa = , 29.80 m/sya = ? , 0 0 0cos 7.50 m/sxv v ?= = and 0 0 0sin 13.0 m/syv v ?= = . When the equipment lands in the front of the ship, 0 8.75 my y? = ? . EXECUTE: Use the vertical motion of the equipment to find its time in the air: 210 0 2y yy y v t a t? = + gives ( )21 13.0 ( 13.0) 4(4.90)(8.75) s9.80t = ± ? + . The positive root is 3.21 st = . The horizontal range of the equipment is 210 0 2 (7.50 m/s)(3.21 s) 24.1 mx xx x v t a t? = + = = . In 3.21 s the ship moves a horizontal distance (0.450 m/s)(3.21 s) 1.44 m= , so 24.1 m 1.44 m 25.5 mD = + = . EVALUATE: The equation 2 0 0sin 2vR g ?= from Example 3.8 can't be used because the starting and ending points of the projectile motion are at different heights. 3.55. IDENTIFY: Projectile motion problem. Take the origin of coordinates at the point where the ball leaves the bat, and take y+ to be upward. 0 0 0cosxv v ?= 0 0 0sin ,yv v ?= but we don?t know 0.v Figure 3.55 Write down the equation for the horizontal displacement when the ball hits the ground and the corresponding equation for the vertical displacement. The time t is the same for both components, so this will give us two equations in two unknowns ( 0v and t). 3-22 Chapter 3 (a) SET UP: y-component: 29.80 m/s ,ya = ? 0 0.9 m,y y? = ? 0 0 sin 45yv v= ° 21 0 0 2y yy y v t a t? = + EXECUTE: 2 210 20.9 m ( sin 45 ) ( 9.80 m/s )v t t? = ° + ? SET UP: x-component: 0,xa = 0 188 m,x x? = 0 0 cos45xv v= ° 21 0 0 2x xx x v t a t? = + EXECUTE: 0 0 0 188 m cos45x x x t v v ?= = ° Put the expression for t from the x-component motion into the y-component equation and solve for 0.v (Note that sin 45 cos45 .° = ° ) 2 2 0 0 0 188 m 188 m 0.9 m ( sin 45 ) (4.90 m/s ) cos45 cos45 v v v ? ? ? ?? = ° ?? ? ? ?° °? ? ? ? 2 2 0 188 m 4.90 m/s 188 m 0.9 m 188.9 m cos45v ? ? = + =? ?°? ? 2 2 0 cos45 4.90 m/s , 188 m 188.9 m v °? ? =? ?? ? 2 0 188 m 4.90 m/s 42.8 m/s cos45 188.9 m v ? ?= =? ?°? ? (b) Use the horizontal motion to find the time it takes the ball to reach the fence: SET UP: x-component: 0 116 m,x x? = 0,xa = 0 0 cos45 (42.8 m/s)cos45 30.3 m/s,xv v= ° = ° = ?t = 21 0 0 2x xx x v t a t? = + EXECUTE: 0 0 116 m 3.83 s 30.3 m/sx x x t v ?= = = SET UP: Find the vertical displacement of the ball at this t: y-component: 0 ?,y y? = 29.80 m/s ,ya = ? 0 0 sin 45 30.3 m/s,yv v= ° = 3.83 st = 21 0 0 2y yy y v t a t? = + EXECUTE: 210 2(30.3 s)(3.83 s) ( 9.80 m/s )(3.83 s)y y 2? = + ? 0 116.0 m 71.9 m 44.1 m,y y? = ? = + above the point where the ball was hit. The height of the ball above the ground is 44.1 m 0.90 m 45.0 m.+ = It?s height then above the top of the fence is 45.0 m 3.0 m 42.0 m.? = EVALUATE: With 0 42.8 m/s,v = 0 30.3 m/syv = and it takes the ball 6.18 s to return to the height where it was hit and only slightly longer to reach a point 0.9 m below this height. 0(188 m) /( cos45 )t v= ° gives 6.21 s,t = which agrees with this estimate. The ball reaches its maximum height approximately (188 m) / 2 94 m= from home plate, so at the fence the ball is not far past its maximum height of 47.6 m, so a height of 45.0 m at the fence is reasonable. 3.56. IDENTIFY: The water moves in projectile motion. SET UP: Let 0 0 0x y= = and take y+ to be positive. 0xa = , ya g= ? . EXECUTE: The equations of motions are 210 2( sin )y v ? t gt= ? and 0( cos )x v ? t= . When the water goes in the tank for the minimum velocity, 2y D= and 6x D= . When the water goes in the tank for the maximum velocity, 2y D= and 7x D= . In both cases, sin cos 2 / 2.? ?= = To reach the minimum distance: 0 2 6 2 D v t= , and 210 222 2D v t gt= ? . Solving the first equation for t gives 0 6 2D t v = . Substituting this into the second equation gives 2 1 2 0 6 2 2 6 D D D g v ? ?= ? ? ?? ?? ? . Solving this for 0v gives 0 3v gD= . Motion in Two or Three Dimensions 3-23 To reach the maximum distance: 0 2 7 2 D v t= , and 210 222 2D v t gt= ? . Solving the first equation for t gives 0 7 2D t v = . Substituting this into the second equation gives 2 1 2 0 7 2 2 7 D D D g v ? ?= ? ? ?? ?? ? . Solving this for 0v gives 0 49 /5 3.13v gD gD= = , which, as expected, is larger than the previous result. EVALUATE: A launch speed of 0 6 2.45v gD gD= = is required for a horizontal range of 6D. The minimum speed required is greater than this, because the water must be at a height of at least 2D when it reaches the front of the tank. 3.57. IDENTIFY: The equations for h and R from Example 3.8 can be used. SET UP: 2 2 0 0sin 2 v h g ?= and 2 0 0sin 2vR g ?= . If the projectile is launched straight up, 0 90? = ° . EXECUTE: (a) 2 0 2 v h g = and 0 2v gh= . (b) Calculate 0? that gives a maximum height of h when 0 2 2v gh= . 2 20 0 8 sin 4 sin 2 gh h h g ? ?= = . 10 2sin? = and 0 30.0? = ° . (c) ( )22 2 sin 60.0 6.93 gh R h g = =° . EVALUATE: 2 0 2 0 2 sin v h g ?= so 0 2 0 2 sin(2 ) sin h R ? ?= . For a given 0? , R increases when h increases. For 0 90? = ° , 0R = and for 0 0? = ° , 0h = and 0R = . For 0 45? = ° , 4R h= . 3.58. IDENTIFY: To clear the bar the ball must have a height of 10.0 ft when it has a horizontal displacement of 36.0 ft. The ball moves as a projectile. When 0v is very large, the ball reaches the goal posts in a very short time and the acceleration due to gravity causes negligible downward displacement. SET UP: 36.0 ft 10.97 m= ; 10.0 ft 3.048 m= . Let x+ be to the right and y+ be upward, so 0xa = , ya g= ? , 0 0 0cosxv v ?= and 0 0 0sinyv v ?= EXECUTE: (a) The ball cannot be aimed lower than directly at the bar. 0 10.0 fttan 36.0 ft? = and 0 15.5? = ° . (b) 210 0 2x xx x v t a t? = + gives 0 0 0 0 0cosx x x x x t v v ? ? ?= = . Then 210 0 2y yy y v t a t? = + gives 2 2 0 0 0 0 0 0 0 02 2 2 2 0 0 0 0 0 0 1 ( ) 1 ( ) ( sin ) ( ) tan cos 2 cos 2 cos x x x x x x y y v g x x g v v v ? ?? ? ? ? ?? ? ?? = ? = ? ?? ?? ? . 2 0 0 0 0 0 0 ( ) 10.97 m 9.80 m/s 12.2 m/s cos 2[( ) tan ( )] cos45.0 2[10.97 m 3.048 m] x x g v x x y y? ? ?= = =? ? ? ?° EVALUATE: With the 0v in part (b) the horizontal range of the ball is 2 0 0sin 2 15.2 m 49.9 ft v R g ?= = = . The ball reaches the highest point in its trajectory when 0 / 2x x R? = , so when it reaches the goal posts it is on its way down. 3.59. IDENTIFY: Apply Eq.(3.27) and solve for x. SET UP: The change in height is y h= ? . EXECUTE: (a) We get a quadratic equation in x, the solution to which is 2 2 2 20 0 0 0 0 0 0 0 02 0 0 cos 2 cos tan sin sin 2 cos v gh v x v v gh g v g ? ?? ? ?? ? ? ? ?= + = + +? ? ? ?? ? . If 0h = , the square root reduces to 0 0sin v ? , and x R= . 3-24 Chapter 3 (b) The expression for x becomes 2 20 0 0(10.2 m)cos [sin sin 0.98]x ? ? ?= + + + . The graph of x as a function of 0? is sketched in Figure 3.59. The angle 0 90? = ° corresponds to the projectile being launched straight up, and there is no horizontal motion. If 0 0? = , the projectile moves horizontally until it has fallen the distance h. (d) The graph shows that the maximum horizontal distance is for an angle less than 45° . EVALUATE: For 0 45? = ° the x and y components of the initial velocity are equal. For 0 45? < ° the x component of the initial velocity is less than the y component. Height comes from the initial position and less vertical component of initial velocity is needed for the maximum range. Figure 3.59 3.60. IDENTIFY: The snowball moves in projectile motion. In part (a) the vertical motion determines the time in the air. In part (c), find the height of the snowball above the ground after it has traveled horizontally 4.0 m. SET UP: Let y+ be downward. 0xa = , 29.80 m/sya = + . 0 0 0cos 5.36 m/sxv v ?= = , 0 0 0sin 4.50 m/syv v ?= = . EXECUTE: (a) Use the vertical motion to find the time in the air: 210 0 2y yy y v t a t? = + with 0 14.0 my y? = gives 2 214.0 m (4.50 m/s) (4.9 m/s )t t= + . The quadratic formula gives ( )21 4.50 (4.50) 4(4.9)( 14.0) s2(4.9)t = ? ± ? ? . The positive root is 1.29 st = . Then 210 0 2 (5.36 m/s)(1.29 s) 6.91 mx xx x v t a t? = + = = . (b) The x-t, y-t, xv -t and yv -t graphs are sketched in Figure 3.60. (c) 210 0 2x xx x v t a t? = + gives 0 0 4.0 m 0.746 s 5.36 m/sx x x t v ?= = = . In this time the snowball travels downward a distance 210 0 2 6.08 my yy y v t a t? = + = and is therefore 14.0 m 6.08 m 7.9 m? = above the ground. The snowball passes well above the man and doesn?t hit him. EVALUATE: If the snowball had been released from rest at a height of 14.0 m it would have reached the ground in 2 2(14.0 m) 1.69 s 9.80 m/s t = = . The snowball reaches the ground in a shorter time than this because of its initial downward component of velocity. Figure 3.60 3.61. (a) IDENTIFY and SET UP: Use the equation derived in Example 3.8: 0 0 0 0 2 sin ( cos ) v R v g ?? ? ?= ? ?? ? Motion in Two or Three Dimensions 3-25 Call the range 1R when the angle is 0? and 2R when the angle is 90 .?° ? 0 0 1 0 0 2 sin ( cos ) v R v g ?? ? ?= ? ?? ? 0 0 2 0 0 2 sin(90 ) ( cos(90 )) v R v g ?? ? ?° ?= ° ? ? ?? ? The problem asks us to show that 1 2.R R= EXECUTE: We can use the trig identities in Appendix B to show: 0 0 0cos(90 ) cos( 90 ) sin? ? ?° ? = ? ° = 0 0 0 0sin(90 ) sin( 90 ) ( cos ) cos? ? ? ?° ? = ? ? ° = ? ? = + Thus 0 0 0 02 0 0 0 0 1 2 cos 2 sin ( sin ) ( cos ) . v v R v v R g g ? ?? ?? ? ? ?= = =? ? ? ?? ? ? ? (b) 2 0 0sin 2vR g ?= so 2 0 2 2 0 (0.25 m)(9.80 m/s ) sin 2 . (2.2 m/s) Rg v ? = = This gives 15? = ° or 75 .° EVALUATE: 20 0( sin 2 ) / ,R v g?= so the result in part (a) requires that 2 20 0sin (2 ) sin (180 2 ),? ?= ° ? which is true. (Try some values of 0? and see!) 3.62. IDENTIFY: Mary Belle moves in projectile motion. SET UP: Let y+ be upward. 0xa = , ya g= ? . EXECUTE: (a) Eq.(3.27) with 8.2 mx = , 6.1 my = and 0 53? = ° gives 0 13 8 m/sv .= . (b) When she reached Joe Bob, 0 8.2 m 0.9874 s cos53 t v = =° . 0 8.31 m/sx xv v= = and 0 1.34 m/sy y yv v a t= + = + . 8 4 m/sv .= , at an angle of 9.16° . (c) The graph of ( )xv t is a horizontal line. The other graphs are sketched in Figure 3.62. (d) Use Eq. (3.27), which becomes 1 2(1.327) (0.071115 m )y x x?= ? . Setting 8.6 my = ? gives 23.8 mx = as the positive solution. Figure 3.62 3.63. (a) IDENTIFY: Projectile motion. Take the origin of coordinates at the top of the ramp and take y+ to be upward. The problem specifies that the object is displaced 40.0 m to the right when it is 15.0 m below the origin. Figure 3.63 We don?t know t, the time in the air, and we don?t know 0.v Write down the equations for the horizontal and vertical displacements. Combine these two equations to eliminate one unknown. SET UP: y-component: 0 15.0 m,y y? = ? 29.80 m/s ,ya = ? 0 0 sin53.0yv v= ° 21 0 0 2y yy y v t a t? = + EXECUTE: 2 2015.0 m ( sin53.0 ) (4.90 m/s )v t t? = ° ? 3-26 Chapter 3 SET UP: x-component: 0 40.0 m,x x? = 0,xa = 0 0 cos53.0xv v= ° 21 0 0 2x xx x v t a t? = + EXECUTE: 040.0 m ( )cos53.0v t= ° The second equation says 0 40.0 m 66.47 m. cos53.0 v t = =° Use this to replace 0v t in the first equation: 2 215.0 m (66.47 m)sin53 (4.90 m/s )t? = ° ? 2 2 (66.46 m)sin53 15.0 m 68.08 m 3.727 s. 4.90 m/s 4.90 m/s t ° += = = Now that we have t we can use the x-component equation to solve for 0:v 0 40.0 m 40.0 m 17.8 m/s. cos53.0 (3.727 s)cos53.0 v t = = =° ° EVALUATE: Using these values of 0v and t in the 10 0 2y yy y v a t 2= = + equation verifies that 0 15.0 m.y y? = ? (b) IDENTIFY: 0 (17.8 m/s) / 2 8.9 m/sv = = This is less than the speed required to make it to the other side, so he lands in the river. Use the vertical motion to find the time it takes him to reach the water: SET UP: 0 100 m;y y? = ? 0 0 sin53.0 7.11 m/s;yv v= + ° = 29.80 m/sya = ? 21 0 0 2y yy y v t a t? = + gives 100 7.11 4.90t t 2? = ? EXECUTE: 24.90 7.11 100 0t t? ? = and ( )219.80 7.11 (7.11) 4(4.90)( 100)t = ± ? ? 0.726 s 4.57 st = ± so 5.30 s.t = The horizontal distance he travels in this time is 0 0 0( cos53.0 ) (5.36 m/s)(5.30 s) 28.4 m.xx x v t v t? = = ° = = He lands in the river a horizontal distance of 28.4 m from his launch point. EVALUATE: He has half the minimum speed and makes it only about halfway across. 3.64. IDENTIFY: The rock moves in projectile motion. SET UP: Let y+ be upward. 0xa = , ya g= ? . Eqs.(3.22) and (3.23) give xv and yv . EXECUTE: Combining equations 3.25, 3.22 and 3.23 gives 2 2 2 2 2 2 2 2 0 0 0 0 0 0 0 0 0cos ( sin ) (sin cos ) 2 sin ( )v v v gt v v gt gt? ? ? ? ?= + ? = + ? + . 2 2 2 2 0 0 0 0 1 2 ( sin ) 2 2 v v g v t gt v gy?= ? ? = ? , where Eq.(3.21) has been used to eliminate t in favor of y. For the case of a rock thrown from the roof of a building of height h, the speed at the ground is found by substituting y h= ? into the above expression, yielding 20 2v v gh= + , which is independent of 0? . EVALUATE: This result, as will be seen in the chapter dealing with conservation of energy (Chapter 7), is valid for any y, positive, negative or zero, as long as 20 2 0v gy? > . 3.65. IDENTIFY and SET UP: Take y+ to be upward. The rocket moves with projectile motion, with 0 40.0 m/syv = + and 0 30.0 m/sxv = relative to the ground. The vertical motion of the rocket is unaffected by its horizontal velocity. EXECUTE: (a) 0yv = (at maximum height), 0 40.0 m/s,yv = + 29.80 m/s ,ya = ? 0 ?y y? = 2 2 0 02 ( )y y yv v a y y= + ? gives 0 81.6 my y? = (b) Both the cart and the rocket have the same constant horizontal velocity, so both travel the same horizontal distance while the rocket is in the air and the rocket lands in the cart. (c) Use the vertical motion of the rocket to find the time it is in the air. 0 40 m/s,yv = 29.80 m/s ,ya = ? 40 m/s,yv = ? ?t = 0y y yv v a t= + gives 8.164 st = Then 0 0 (30.0 m/s)(8.164 s) 245 m.xx x v t? = = = Motion in Two or Three Dimensions 3-27 (d) Relative to the ground the rocket has initial velocity components 0 30.0 m/sxv = and 0 40.0 m/s,yv = so it is traveling at 53.1° above the horizontal. (e) (i) Figure 3.65a Relative to the cart, the rocket travels straight up and then straight down (ii) Figure 3.65b Relative to the ground the rocket travels in a parabola. EVALUATE: Both the cart and rocket have the same constant horizontal velocity. The rocket lands in the cart. 3.66. IDENTIFY: The ball moves in projectile motion. SET UP: The woman and ball travel for the same time and must travel the same horizontal distance, so for the ball 0 6.00 m/sxv = . EXECUTE: (a) 0 0 0cosxv v ?= . 00 0 6.00 m/s cos 20.0 m/s xv v ? = = and 0 72.5? = ° . (b) Relative to the ground the ball moves in a parabola. The ball and the runner have the same horizontal component of velocity, so relative to the runner the ball has only vertical motion. The trajectories as seen by each observer are sketched in Figure 3.66. EVALUATE: The ball could be thrown with a different speed, so long as the angle at which it was thrown was adjusted to keep 0 6.00 m/sxv = . Figure 3.66 3.67. IDENTIFY: The boulder moves in projectile motion. SET UP: Take y+ downward. 0 0xv v= , 0 0yv = . 0xa = , 29.80 m/sya = + . EXECUTE: (a) Use the vertical motion to find the time for the boulder to reach the level of the lake: 21 0 0 2y yy y v t a t? = + with 0 20 my y? = + gives 0 22( ) 2(20 m) 2.02 s9.80 m/sy y y t a ?= = = . The rock must travel horizontally 100 m during this time. 210 0 2x xx x v t a t? = + gives 00 0 100 m 49.5 m/s2.02 sx x x v v t ?= = = = (b) In going from the edge of the cliff to the plain, the boulder travels downward a distance of 0 45 my y? = . 0 2 2( ) 2(45 m) 3.03 s 9.80 m/sy y y t a ?= = = and 0 0 (49.5 m/s)(3.03 s) 150 mxx x v t? = = = . The rock lands 150 m 100 m 50 m? = beyond the foot of the dam. EVALUATE: The boulder passes over the dam 2.02 s after it leaves the cliff and then travels an additional 1.01 s before landing on the plain. If the boulder has an initial speed that is less than 49 m/s, then it lands in the lake. 3.68. IDENTIFY: The bagels move in projectile motion. Find Henrietta?s location when the bagels reach the ground, and require the bagels to have this horizontal range. 3-28 Chapter 3 SET UP: Let y+ be downward and let 0 0 0x y= = . 0xa = , ya g= + . When the bagels reach the ground, 43.9 my = . EXECUTE: (a) When she catches the bagels, Henrietta has been jogging for 9.00 s plus the time for the bagels to fall 43.9 m from rest. Get the time to fall: 2 1 2 y gt= , 2 2143.9 m (9.80 m/s ) 2 t= and 2.99 st = . So, she has been jogging for 9.00 s 2.99 s 12.0 s+ = . During this time she has gone (3.05 m/s)(12.0 s) 36.6 mx vt= = = . Bruce must throw the bagels so they travel 36.6 m horizontally in 2.99 s. This gives x vt= . 36.6 m (2 99 s)v .= and 12.2 m/sv = . (b) 36.6 m from the building. EVALUATE: If 12.2 m/sv > the bagels land in front of her and if 12.2 m/sv < they land behind her. There is a range of velocities greater than 12.2 m/s for which she would catch the bagels in the air, at some height above the sidewalk. 3.69. IDENTIFY: The shell moves in projectile motion. To find the horizontal distance between the tanks we must find the horizontal velocity of one tank relative to the other. Take y+ to be upward. (a) SET UP: The vertical motion of the shell is unaffected by the horizontal motion of the tank. Use the vertical motion of the shell to find the time the shell is in the air: 0 0 sin 43.4 m/s,yv v ?= = 29.80 m/s ,ya = ? 0 0y y? = (returns to initial height), ?t = EXECUTE: 210 0 2y yy y v t a t? = + gives 8.86 st = SET UP: Consider the motion of one tank relative to the other. EXECUTE: Relative to tank #1 the shell has a constant horizontal velocity 0 cos 246.2 m/s.v ? = Relative to the ground the horizontal velocity component is 246.2 m/s 15.0 m/s 261.2 m/s.+ = Relative to tank #2 the shell has horizontal velocity component 261.2 m/s 35.0 m/s 226.2 m/s.? = The distance between the tanks when the shell was fired is the (226.2 m/s)(8.86 s) 2000 m= that the shell travels relative to tank #2 during the 8.86 s that the shell is in the air. (b) The tanks are initially 2000 m apart. In 8.86 s tank #1 travels 133 m and tank #2 travels 310 m, in the same direction. Therefore, their separation increases by 310 m 133 m 177 m.? = So, the separation becomes 2180 m (rounding to 3 significant figures). EVALUATE: The retreating tank has greater speed than the approaching tank, so they move farther apart while the shell is in the air. We can also calculate the separation in part (b) as the relative speed of the tanks times the time the shell is in the air: (35.0 m/s 15.0 m/s)(8.86 s) 177 m.? = 3.70. IDENTIFY: The object moves with constant acceleration in both the horizontal and vertical directions. SET UP: Let y+ be downward and let x+ be the direction in which the firecracker is thrown. EXECUTE: The firecracker?s falling time can be found from the vertical motion: 2ht g = . The firecracker?s horizontal position at any time t (taking the student?s position as 0x = ) is 212x vt at= ? . 0x = when cracker hits the ground, so 2 /t v a= . Combining this with the expression for the falling time gives 2 2v h a g = and 2 2 2v g h a = . EVALUATE: When h is smaller, the time in the air is smaller and either v must be smaller or a must be larger. 3.71. IDENTIFY: The velocity T/Gv! of the tank relative to the ground is related to the velocity R/Gv! of the rocket relative to the ground and the velocity T/Rv ! of the tank relative to the rocket by T/G T/R R/Gv = v + v ! ! ! . SET UP: Let y+ be upward and take 0y = at the ground. Let x+ be in the direction of the horizontal component of the tank's motion. Once the tank is released it has 0xa = , 29.80 m/sya = ? , relative to the ground. EXECUTE: (a) For the rocket 20 (1.75m/s )(22.0 s) 38.5 m/sy y yv v a t= + = = and 0xv = . The rocket has speed 38.5 m/s at the instant when the fuel tank is released. (b) (i) The rocket's path is vertical, so relative to the crew member T/R- 25.0 m/sxv = + and T/R- 0yv = . (ii) R/Gv! is vertical and T/Rv ! is horizontal, so T/G- 25.0 m/sxv = + and T/G 38.5 m/syv ? = + . (c) (i) The tank initially moves horizontally, at an angle of zero. (ii) T/G-0 T/G- 38.5 m/s tan 25.0 m/s y x v v ? = = and 0 57.0? = ° . Motion in Two or Three Dimensions 3-29 (d) Consider the motion of the tank, in the reference frame of the technician on the ground. At the instant the tank is released the rocket at a height 2 2 21 10 0 2 2 (1.75 m/s )(22.0 s) 423.5 my yy y v t a t? = + = = . So, for the tank 0 423.5 my = , 0 38.5 m/syv = and 29.80 m/sya = ? . 0yv = at the maximum height. 2 20 02 ( )y y yv v a y y= + ? gives 2 2 2 0 0 2 0 (38.5 m/s) 75.6 m 2 2(9.80 m/s ) y y y v v y y a ? ?? = = = . 423.5 m 75.6 m 499 my = + = . The tank reaches a height of 499 m above the launch pad. EVALUATE: Relative to the crew member in the rocket the jettisoned tank has an acceleration of 2 2 21.75 m/s 9.80 m/s 11.5 m/s+ = , downward. Relative to the rocket the tank follows a parabolic path, but with zero initial vertical velocity and with a downward acceleration that has magnitude greater than g . 3.72. IDENTIFY: The velocity R/Gv! of the rocket relative to the ground is related to the velocity S/Gv! of the secondary rocket relative to the ground and the velocity S/Rv ! of the secondary rocket relative to the rocket by S/G S/R R/Gv = v + v ! ! ! . SET UP: Let y+ be upward and let 0y = at the ground. Let x+ be in the direction of the horizontal component of the secondary rocket's motion. After it is launched the secondary rocket has 0xa = and 29.80 m/sya = ? , relative to the ground. EXECUTE: (a) (i) S/R- (12.0 m/s)cos53.0 7.22 m/sxv = =° and S/R-y (12.0 m/s)sin53.0 9.58 m/sv = =° . (ii) R/G- 0xv = and R/G- 8.50 m/syv = . S/G- S/R- R/G- 7.22 m/sx x xv v v= + = and S/G- S/R- R/G-y y yv v v= + = 9.58 m/s 8.50 m/s 18.1 m/s+ = . (b) 2 2S/G S/G- S/G-( ) ( ) 19.5 m/sx yv v v= + = . S/G-0 S/G- 18.1 m/s tan 7.22 m/s y x v v ? = = and 0 68.3? = ° . (c) Relative to the ground the secondary rocket has 0 145 my = , 0 18.1 m/syv = + , 29.80 m/sya = ? and 0yv = (at the maximum height). 2 20 02 ( )y y yv v a y y= + ? gives 2 2 2 0 0 2 0 (18.1 m/s) 16.7 m 2 2( 9.80 m/s ) y y y v v y y a ? ?? = = =? . 145 m 16.7 m 162 my = + = . EVALUATE: The secondary rocket reaches its maximum height in time 0 218.1 m/s 1.85 s9.80 m/s y y y v v t a ? ?= = =? after it is launched. At this time the primary rocket has height 145 m (8.50 m/s)(1.85 s) 161 m+ = , so is at nearly the same height as the secondary rocket. The secondary rocket first moves upward from the primary rocket but then loses vertical velocity due to the acceleration of gravity. 3.73. IDENTIFY: The original firecracker moves as a projectile. At its maximum height it's velocity is horizontal. The velocity A/Gv ! of fragment A relative to the ground is related to the velocity F/Gv ! of the original firecracker relative to the ground and the velocity A/Fv ! of the fragment relative to the original firecracker by A/G A/F F/Gv = v + v ! ! ! . Fragment B obeys a similar equation. SET UP: Let x+ be along the direction of the horizontal motion of the firecracker before it explodes and let y+ be upward. Fragment A moves at 53.0° above the x+ direction and fragment B moves at 53.0° below the x+ direction. Before it explodes the firecracker has 0xa = and 29.80 m/sya = ? EXECUTE: The horizontal component of the firecracker's velocity relative to the ground is constant (since 0xa = ), so F/G- (25.0 m/s)cos30.0 21.65 m/sxv = =° . At the time of the explosion, F/G- 0yv = . For fragment A, A/F- (20.0 m/s)cos53.0 12.0 m/sxv = =° and A/F- (20.0 m/s)sin53.0 16.0 m/syv = =° . A/G- A/F- F/G- 12.0 m/s 21.65 m/s 33.7 m/sx x xv v v= + = + = . A/G- A/F- F/G- 16.0 m/sy y yv v v= + = . A/G- 0 A/G- 16.0 m/s tan 33.7 m/s y x v v ? = = and 0 25.4? = ° . The calculation for fragment B is the same, except A/F- 16.0 m/syv = ? . The fragments move at 25.4° above and 25.4° below the horizontal. EVALUATE: As the initial velocity of the firecracker increases the angle with the horizontal for the fragments, as measured from the ground, decreases. 3.74. IDENTIFY: The grenade moves in projectile motion. 110 km/h 30.6 m/s= . The horizontal range R of the grenade must be 15.8 m plus the distance d that the enemy's car travels while the grenade is in the air. 3-30 Chapter 3 SET UP: For the grenade take y+ upward, so 0xa = , ya g= ? . Let 0v be the magnitude of the velocity of the grenade relative to the hero. 0 0 cos45xv v= ° , 0 0 sin 45yv v= ° . 90 km/h 25 m/s= ; The enemy?s car is traveling away from the hero?s car with a relative velocity of rel 30.6 m/s 25 m/s 5.6 m/sv = ? = . EXECUTE: 210 0 2y yy y v t a t? = + with 0 0y y? = gives 0 02 2 sin 45y y v v t a g = ? = ° . 0 relrel 2v vd v t g= = . 2 2 0 0 0 0 2 sin 45 cos45 (cos45x v v R v t v t g g = = = =° °°) . 15.8 mR d= + gives that 2 0 rel 0 2 15.8 m v v v g g = + . 2 0 rel 02 (15.8 m) 0v v v g? ? = . 20 07.92 154.8 0v v? ? = . The quadratic formula gives 0 17.0 m/s 61.2 km/hv = = . The grenade has velocity of magnitude 61.2 km/h relative to the hero. Relative to the hero the velocity of the grenade has components 0 0 cos45 43.3 km/hxv v= =° and 0 0 sin 45 43.3 km/hyv v= =° . Relative to the earth the velocity of the grenade has components E 43.3 km/h 90 km/h = 133.3 km/hxv = + and E 43.3 km/hyv = . The magnitude of the velocity relative to the earth is 2 2E E E 140 km/hx yv v v= + = . EVALUATE: The time the grenade is in the air is 0 22 sin 45 2(17.0 m/s)sin 45 2.45 s9.80 m/s v t g = = =° ° . During this time the grenade travels a horizontal distance 0 (133.3 km/h)(2.45 s)(1 h /3600 s) 90.7 mx x? = = , relative to the earth, and the enemy?s car travels a horizontal distance 0 (110 km/h)(2.45 s)(1 h /3600 s) 74.9 mx x? = = , relative to the earth. The grenade has traveled 15.8 m farther. 3.75. IDENTIFY and SET UP: Use Eqs. (3.4) and (3.12) to get the velocity and acceleration components from the position components. EXECUTE: cos ,x R t?= siny R t?= (a) 2 2 2 2 2 2 2 2 2 2cos sin (sin cos ) ,r x y R t R t R t t R R? ? ? ?= + = + = + = = since 2 2sin cos 1.t t? ?+ = (b) sin ,x dx v R t dt ? ?= = ? cosy dyv R tdt ? ?= = ( sin )( cos ) ( cos )( sin )x yv x v y R t R t R t R t? ? ? ? ? ?? = + = ? +v r! ! 2 ( sin cos sin cos ) 0,R t t t t? ? ? ? ?? = ? + =v r! ! so v! is perpendicular to .r! (c) 2 2cosxx dv a R t x dt ? ? ?= = ? = ? 2 2sinyy dv a R t y dt ? ? ?= = ? = ? 2 2 4 2 4 2 2 2 2 2.x ya a a x y x y R? ? ? ?= + = + = + = 2 2? ? ? ?( ) .x ya a x y? ?+ ? ?a = i j = i + j = r! ! Since 2? is positive this means that the direction of a! is opposite to the direction of .r! (d) 2 2 2 2 2 2 2 2 2 2 2 2sin cos (sin cos ).x yv v v R t R t R t t? ? ? ? ? ? ?= + = + = + 2 2 .v R R? ?= = (e) 2 ,a R?= / ,v R? = so 2 2 2( / ) / .a R v R v R= = EVALUATE: The rock moves in uniform circular motion. The position vector is radial, the velocity is tangential, and the acceleration is radially inward. 3.76. IDENTIFY: All velocities are constant, so the distance traveled is B/Ed v t= , where B/Ev is the magnitude of the velocity of the boat relative to the earth. The relative velocities B/Ev ! , S/Wv ! (boat relative to the water) and W/Ev ! (water relative to the earth) are related by B/E B/W W/Ev = v + v ! ! ! . SET UP: Let x+ be east and let y+ be north. W/E- 30.0 m/minxv = + and W/E- 0yv = . B/W 100.0 m/minv = . The direction of B/Wv ! is the direction in which the boat is pointed or aimed. EXECUTE: (a) B/W- 100.0 m/minyv = + and B/W- 0xv = . B/E- B/W- W/E- 30.0 m/minx x xv v v= + = and B/E- B/W- W/E- 100.0 m/miny y yv v v= + = . The time to cross the river is 0 B/E- 400.0 m 4.00 min 100.0 m/miny y y t v ?= = = . Motion in Two or Three Dimensions 3-31 0 (30.0 m/min)(4.00 min) 120.0 mx x? = = . You will land 120.0 m east of point B, which is 45.0 m east of point C. The distance you will have traveled is 2 2(400.0 m) (120.0 m) 418 m+ = . (b) B/Wv ! is directed at angle ? east of north, where 75.0 mtan 400.0 m ? = and 10.6? = ° . B/W- (100.0 m/min)sin10.6 18.4 m/minxv = =° and B/W- (100.0 m/min)cos10.6 98.3 m/minyv = =° . B/E- B/W- W/E- 18.4 m/min 30.0 m/min 48.4 m/minx x xv v v= + = + = . B/E- B/W- W/E- 98.3 m/miny y yv v v= + = . 0 B/E- 400.0 m 4.07 min 98.3 m/miny y y t v ?= = = . 0 (48.4 m/min)(4.07 min) 197 mx x? = = . You will land 197 m downstream from B, so 122 m downstream from C. (c) (i) If you reach point C, then B/Ev ! is directed at 10.6° east of north, which is 79.4° north of east. We don't know the magnitude of B/Ev ! and the direction of B/Wv ! . In part (a) we found that if we aim the boat due north we will land east of C, so to land at C we must aim the boat west of north. Let B/Wv ! be at an angle ? of north of west. The relative velocity addition diagram is sketched in Figure 3.76. The law of sines says W/E B/W sin sin 79.4 v v ? = ° . 30.0 m/min sin sin 79.4 100.0 m/min ? ? ?= ? ?? ? ° and 17.15? = ° . Then 180 79.4 17.15 83.5? = ? ? =° ° ° ° . The boat will head 83.5° north of west, so 6.5° west of north. B/E- B/W- W/E- (100.0 m/min)cos83.5 30.0 m/min 18.7 m/minx x xv v v= + = ? + =° . B/E- B/W- W/E- (100.0 m/min)sin83.5 99.4 m/miny y yv v v= + = ? =° . Note that these two components do give the direction of B/Ev ! to be 79.4° north of east, as required. (ii) The time to cross the river is 0 B/E- 400.0 m 4.02 min 99.4 m/miny y y t v ?= = = . (iii) You travel from A to C, a distance of 2 2(400.0 m) (75.0 m) 407 m+ = . (iv) 2 2B/E B/E- B/E-( ) ( ) 101 m/minx yv v v= + = . Note that B/E 406 mv t = , the distance traveled (apart from a small difference due to rounding). EVALUATE: You cross the river in the shortest time when you head toward point B, as in part (a), even though you travel farther than in part (c). Figure 3.76 3.77. IDENTIFY: /xv dx dt= , /yv dy dt= , /x xa dv dt= and /y ya dv dt= . SET UP: (sin ) cos( )d t t dt ? ? ?= and (cos ) sin( )d t t dt ? ? ?= ? . EXECUTE: (a) The path is sketched in Figure 3.77. (b) To find the velocity components, take the derivative of x and y with respect to time: (1 cos ),xv R ?t?= ? and sin .yv R ?t?= To find the acceleration components, take the derivative of xv and yv with respect to time: 2 sinxa R t,? ?= and 2 cosya R t.? ?= (c) The particle is at rest ( 0)y xv v= = every period, namely at 0 2 / 4 / ....t , ? ?, ? ?,= At that time, 0 2 4 ...;x , ?R, ?R,= and 0.y = The acceleration is 2a R?= in the -y+ direction. 3-32 Chapter 3 (d) No, since ( ) ( ) 1/ 22 22 2 2sin cos .a R t R t R? ? ? ? ?? ?= + =? ?? ? The magnitude of the acceleration is the same as for uniform circular motion. EVALUATE: The velocity is tangent to the path. 0xv is always positive; yv changes sign during the motion. Figure 3.77 3.78. IDENTIFY: At the highest point in the trajectory the velocity of the projectile relative to the earth is horizontal. The velocity P/Ev ! of the projectile relative to the earth, the velocity F/Pv ! of a fragment relative to the projectile, and the velocity F/Ev ! of a fragment relative to the earth are related by F/E F/P P/Ev = v + v ! ! ! . SET UP: Let x+ be along the horizontal component of the projectile motion. Let the speed of each fragment relative to the projectile be v. Call the fragments 1 and 2, where fragment 1 travels in the x+ direction and fragment 2 is in the -directionx? , and let the speeds just after the explosion of the two fragments relative to the earth be 1v and 2v . Let pv be the speed of the projectile just before the explosion. EXECUTE: F/E- F/P- P/E-x x xv v v= + gives 1 pv v v= + and 2 pv v v? = ? . Both fragments start from the same height with zero vertical component of velocity relative to the earth, so they both fall for the same time t, and this is also the same time as it took for the projectile to travel a horizontal distance D, so pv t D= . Since fragment 2 lands at A it travels a horizontal distance D as it falls and 2v t D= . 2 pv v v? = + ? gives p 2v v v= + and p 2 2vt v t v t D= + = . Then 1 p 3v t v t vt D= + = . This fragment lands a horizontal distance 3D from the point of explosion and hence 4D from A. EVALUATE: Fragment 1, that is ejected in the direction of the motion of the projectile travels with greater speed relative to the earth than the fragment that travels in the opposite direction. 3.79. IDENTIFY: 2 2 rad 2 4v R a R T ?= = . All points on the centrifuge have the same period T . SET UP: The period T in seconds is related to n , the number of revolutions per minute, by 60 s/minn T = . EXECUTE: (a) 2 rad 2 4a R T ?= , which is constant. rad,1 rad,2 1 2 a a R R = . Let 1R R= , so rad,1 5.00a g= and let 2 / 2R R= . 2 rad,2 rad,1 1 (5.00 )(1/ 2) 2.50 R a a g g R ? ?= = =? ?? ? . (b) 60 s/min T n ? ?= ? ?? ? and 2 rad 2 4 R a T ?= gives 2 2 2rad 4 /(60 s/min)a Rn?= . 2 rad 2 2 4 (60 s/min) a R n ?= , which is constant. rad,1 rad,2 2 2 1 2 a a n n = . Let rad,1 5.00a g= , so 1n n= and rad,2 Mercury5 5(0.378)a g g= = . Then rad,2 2 1 rad,1 5(0.378) 0.615 5.00 a g n n n n a g = = = . EVALUATE: The radial acceleration is less for points closer to the rotation axis. Since Mercuryg g< , a smaller rotation rate is required to produce Mercury5 g than to produce 5 g . 3.80. IDENTIFY: Use the relation that relates the relative velocities. SET UP: The relative velocities are the raindrop relative to the earth, R/Ev! , the raindrop relative to the train, R/Tv! , and the train relative to the earth, T/Ev ! . R/E R/T T/E= +v v v! ! ! . T/Ev! is due east and has magnitude 12.0 m/s. R/Tv! is 30.0° west of vertical. R/Ev! is vertical. The relative velocity addition diagram is given in Figure 3.80. EXECUTE: (a) R/Ev! is vertical and has zero horizontal component. The horizontal component of R/Tv! is T/E?v! , so is 12.0 m/s westward. (b) T/ER/E 12.0 m/s 20.8 m/s tan30.0 tan30.0 v v = = =° ° . T/E R/T 12.0 m/s 24.0 m/s sin30.0 sin30.0 v v = = =° ° . Motion in Two or Three Dimensions 3-33 EVALUATE: The speed of the raindrop relative to the train is greater than its speed relative to the earth, because of the motion of the train. Figure 3.80 3.81. IDENTIFY: Relative velocity problem. The plane?s motion relative to the earth is determined by its velocity relative to the earth. SET UP: Select a coordinate system where y+ is north and x+ is east. The velocity vectors in the problem are: P/E ,v ! the velocity of the plane relative to the earth. P/A ,v ! the velocity of the plane relative to the air (the magnitude P/Av is the air speed of the plane and the direction of P/Av ! is the compass course set by the pilot). A/E ,v ! the velocity of the air relative to the earth (the wind velocity). The rule for combining relative velocities gives P/E P/A A/E.v = v + v ! ! ! (a) We are given the following information about the relative velocities: P/Av ! has magnitude 220 km/h and its direction is west. In our coordinates is has components P/A( ) 220 km/hxv = ? and P/A( ) 0.yv = From the displacement of the plane relative to the earth after 0.500 h, we find that P/Ev ! has components in our coordinate system of P/E 120 km ( ) 240 km/h 0.500 hx v = ? = ? (west) P/E 20 km ( ) 40 km/h 0.500 hy v = ? = ? (south) With this information the diagram corresponding to the velocity addition equation is shown in Figure 3.81a. Figure 3.81a We are asked to find A/E ,v ! so solve for this vector: P/E P/A A/Ev = v + v ! ! ! gives A/E P/E P/A.v = v v ! ! !? EXECUTE: The x-component of this equation gives A/E P/E P/A( ) ( ) ( ) 240 km/h ( 220 km/h) 20 km/h.x x xv v v= ? = ? ? ? = ? The y-component of this equation gives A/E P/E P/A( ) ( ) ( ) 40 km/h.y y yv v v= ? = ? 3-34 Chapter 3 Now that we have the components of A/Ev ! we can find its magnitude and direction. 2 2 A/E A/E A/E( ) ( )x yv v v= + 2 2 A/E ( 20 km/h) ( 40 km/h) 44.7 km/hv = ? + ? = 40 km/h tan 2.00; 20 km/h ? = = 63.4? = ° The direction of the wind velocity is 63.4 S° of W, or 26.6 W° of S. Figure 3.81b EVALUATE: The plane heads west. It goes farther west than it would without wind and also travels south, so the wind velocity has components west and south. (b) SET UP: The rule for combining the relative velocities is still P/E P/A A/E ,v = v + v! ! ! but some of these velocities have different values than in part (a). P/Av ! has magnitude 220 km/h but its direction is to be found. A/Ev ! has magnitude 40 km/h and its direction is due south. The direction of P/Ev ! is west; its magnitude is not given. The vector diagram for P/E P/A A/Ev = v + v ! ! ! and the specified directions for the vectors is shown in Figure 3.81c. Figure 3.81c The vector addition diagram forms a right triangle. EXECUTE: A/E P/A 40 km/h sin 0.1818; 220 km/h v v ? = = = 10.5 .? = ° The pilot should set her course 10.5° north of west. EVALUATE: The velocity of the plane relative to the air must have a northward component to counteract the wind and a westward component in order to travel west. 3.82. IDENTIFY: Both the bolt and the elevator move vertically with constant acceleration. SET UP: Let y+ be upward and let 0y = at the initial position of the floor of the elevator, so 0y for the bolt is 3.00 m. EXECUTE: (a) The position of the bolt is 2 23.00 m (2.50 m/s) (1/ 2)(9.80 m/s )t t+ ? and the position of the floor is (2.50 m/s)t. Equating the two, 2 23.00 m (4.90 m/s )t= . Therefore, 0.782 st = . (b) The velocity of the bolt is 22.50 m/s (9.80 m/s )(0.782 s) 5.17 m/s? = ? relative to Earth, therefore, relative to an observer in the elevator 5.17 m/s 2.50 m/s 7.67 m/s.v = ? ? = ? (c) As calculated in part (b), the speed relative to Earth is 5.17 m/s. (d) Relative to Earth, the distance the bolt traveled is 2 2 2 2(2.50 m/s) (1/ 2)(9.80 m/s ) (2.50 m/s)(0.782 s) (4.90 m/s )(0.782 s) 1.04 mt t? = ? = ? . EVALUATE: As viewed by an observer in the elevator, the bolt has 0 0yv = and 29.80 m/sya = ? , so in 0.782 s it falls 2 212 (9.80 m/s )(0.782 s) 3.00 m? = ? . 3.83. IDENTIFY: In an earth frame the elevator accelerates upward at 24.00 m/s and the bolt accelerates downward at 29.80 m/s . Relative to the elevator the bolt has a downward acceleration of 2 2 24.00 m/s 9.80 m/s 13.80 m/s+ = . In either frame, that of the earth or that of the elevator, the bolt has constant acceleration and the constant acceleration equations can be used. SET UP: Let y+ be upward. The bolt travels 3.00 m downward relative to the elevator. EXECUTE: (a) In the frame of the elevator, 0 0yv = , 0 3.00 my y? = ? , 213.8 m/sya = ? . 21 0 0 2y yy y v t a t? = + gives 0 22( ) 2( 3.00 m) 0.659 s13.8 m/sy y y t a ? ?= = =? . Motion in Two or Three Dimensions 3-35 (b) 0y y yv v a t= + . 0 0yv = and 0.659 st = . (i) 213.8 m/sya = ? and 9.09 m/syv = ? . The bolt has speed 9.09 m/s when it reaches the floor of the elevator. (ii) 29.80 m/sya = ? and 6.46 m/syv = ? . In this frame the bolt has speed 6.46 m/s when it reaches the floor of the elevator. (c) 210 0 2y yy y v t a t? = + . 0 0yv = and 0.659 st = . (i) 213.8 m/sya = ? and 2 21 0 2 ( 13.8 m/s )(0.659 s) 3.00 my y? = ? = ? . The bolt falls 3.00 m, which is correctly the distance between the floor and roof of the elevator. (ii) 29.80 m/sya = ? and 2 210 2 ( 9.80 m/s )(0.659 s) 2.13 my y? = ? = ? . The bolt falls 2.13 m. EVALUATE: In the earth's frame the bolt falls 2.13 m and the elevator rises 2 21 2 (4.00 m/s )(0.659 s) 0.87 m= during the time that the bolt travels from the ceiling to the floor of the elevator. 3.84. IDENTIFY: The velocity P/Ev! of the plane relative to the earth is related to the velocity P/Av! of the plane relative to the air and the velocity A/Ev ! of the air relative to the earth (the wind velocity) by P/E P/A A/Ev = v + v ! ! ! . SET UP: Let x+ be to the east. With no wind P/A P/E 5550 km 840.9 km/h6.60 hv v= = = . A/E- 225 km/hxv = + . The distance between A and B is 2775 km. EXECUTE: P/E- P/A- A/E-x x xv v v= + . For the trip A to B, P/A- 840.9 km/hxv = + and P/E- 840.9 km/h 225 km/h 1065.9 km/hxv = + = and the travel time is 2775 km 2.60 h1065.9 km/hABt = = . For the trip B to A, P/A- 840.9 km/hxv = ? and P/E- 840.9 km/h 225 km/h 615.9 km/hxv = ? + = ? and the travel time is 2775 km 4.51 h 615.9 km/hBA t ?= =? . The total time for the round trip will be 7.11 hAB BAt t t= + = . EVALUATE: The round trip takes longer when the wind blows, even though the plane travels with the wind for one leg of the trip. The arithmetic average of the speeds for each leg is 1065.9 km/h 615.9 km/h 840.9 km/h 2 + = , the same speed when there is no wind. But the plane spends more time traveling at the slower speed relative to the ground and the average speed is less than the arithmetic average of the speeds for each half of the trip. 3.85. IDENTIFY: Relative velocity problem. SET UP: The three relative velocities are: J/G ,v ! Juan relative to the ground. This velocity is due north and has magnitude J/G 8.00 m/s.v = B/G ,v ! the ball relative to the ground. This vector is 37.0° east of north and has magnitude B/G 12.00 m/s.v = B/J ,v ! the ball relative to Juan. We are asked to find the magnitude and direction of this vector. The relative velocity addition equation is B/G B/J J/G ,v = v + v ! ! ! so B/J B/G J/G .?v = v v! ! ! The relative velocity addition diagram does not form a right triangle so we must do the vector addition using components. Take y+ to be north and x+ to be east. EXECUTE: B/J B/G sin37.0 7.222 m/sxv v= + ° = B/J B/G J/Gcos37.0 1.584 m/syv v v= + °? = These two components give B/J 7.39 m/sv = at 12.4° north of east. EVALUATE: Since Juan is running due north, the ball?s eastward component of velocity relative to him is the same as its eastward component relative to the earth. The northward component of velocity for Juan and the ball are in the same direction, so the component for the ball relative to Juan is the difference in their components of velocity relative to the ground. 3.86. IDENTIFY: (a) The ball moves in projectile motion. When it is moving horizontally, 0yv = . SET UP: Let x+ be to the right and let y+ be upward. 0xa = , ya g= ? . EXECUTE: (a) 20 2 2(9.80 m/s )(4.90 m) 9.80 m/s.yv gh= = = (b) 0 / 1.00 syv g = . (c) The horizontal component of the velocity of the ball relative to the man is 2 2(10.8 m/s) (9.80 m/s) 4.54 m/s? = , the horizontal component of the velocity relative to the hoop is 4.54 m/s 9.10 m/s 13.6 m/s+ = , and the man must be 13.6 m in front of the hoop at release. 3-36 Chapter 3 (d) Relative to the flat car, the ball is projected at an angle 1 9.80 m/s tan 65 . 4.54 m/s ? ? ? ?= = °? ?? ? Relative to the ground the angle is 1 9.80 m/s tan 35.7 4.54 m/s 9.10 m/s ? ? ? ?= = °? ?+? ? . EVALUATE: In both frames of reference the ball moves in a parabolic path with 0xa = and ya g= ? . The only difference between the description of the motion in the two frames is the horizontal component of the ball?s velocity. 3.87. IDENTIFY: The pellets move in projectile motion. The vertical motion determines their time in the air. SET UP: 0 0 cos1.0xv v= ° , 0 0 sin1.0yv v= ° . EXECUTE: (a) 02 yvt g = . 0 0xx x v t? = gives 00 0 sin1.0( cos1.0 80 mvx x v g ? ?? = =? ?? ? 2 °°) . (b) The probability is 1000 times the ratio of the area of the top of the person?s head to the area of the circle in which the pellets land. 2 2 3 2 (10 10 m) (1000) 1.6 10 . (80 m) ? ? ? ?? ?× = ×? ?? ? (c) The slower rise will tend to reduce the time in the air and hence reduce the radius. The slower horizontal velocity will also reduce the radius. The lower speed would tend to increase the time of descent, hence increasing the radius. As the bullets fall, the friction effect is smaller than when they were rising, and the overall effect is to decrease the radius. EVALUATE: The small angle of deviation from the vertical still causes the pellets to spread over a large area because their time in the air is large. 3.88. IDENTIFY: Write an expression for the square of the distance 2( )D from the origin to the particle, expressed as a function of time. Then take the derivative of 2D with respect to t, and solve for the value of t when this derivative is zero. If the discriminant is zero or negative, the distance D will never decrease. SET UP: 2 2 2D x y= + , with ( )x t and ( )y t given by Eqs.(3.20) and (3.21). EXECUTE: Following this process, 1sin 8/9 70.5 .? = ° EVALUATE: We know that if the object is thrown straight up it moves away from P and then returns, so we are not surprised that the projectile angle must be less than some maximum value for the distance to always increase with time. 3.89. IDENTIFY: The baseball moves in projectile motion. SET UP: Use coordinates where the x-axis is horizontal and the y-axis is vertical. EXECUTE: (a) The trajectory of the projectile is given by Eq. (3.27), with 0 ? ?,? = + and the equation describing the incline is tan .y x ?= Setting these equal and factoring out the 0x = root (where the projectile is on the incline) gives a value for 0;x the range measured along the incline is 2 2 02 cos ( )/ cos [tan( ) tan ] . cos v ? ? x ? ? ? g ? ?? ? ? ?+= + ?? ? ? ?? ? ? ? (b) Of the many ways to approach this problem, a convenient way is to use the same sort of substitution, involving double angles, as was used to derive the expression for the range along a horizontal incline. Specifically, write the above in terms of ? ?,? = + as 2 20 2 2 [sin cos cos cos sin ] . cos v R g ? ? ? ? ?? ? ?= ?? ?? ? The dependence on ? and hence ? is in the second term. Using the identities 2sin cos (1/ 2)sin 2 and cos (1/ 2)(1 cos2 ),? ? ? ? ?= = + this term becomes (1/ 2)[cos sin 2 sin cos2 sin ] (1/ 2)[sin(2 ) sin ] .? ? ? ? ?? ? ?? ? = ? ? This will be a maximum when sin(2 )?? ? is a maximum, at 2 2 90? ? ? ,? ? = + = ° or 45 / 2.? ?= ° ? EVALUATE: Note that the result reduces to the expected forms when 0? = (a flat incline, 45? = ° and when 90? = ? ° (a vertical cliff), when a horizontal launch gives the greatest distance). 3.90. IDENTIFY: The arrow moves in projectile motion. SET UP: Use coordinates that for which the axes are horizontal and vertical. Let ? be the angle of the slope and let? be the angle of projection relative to the sloping ground. Motion in Two or Three Dimensions 3-37 EXECUTE: The horizontal distance x in terms of the angles is 2 2 0 1 tan tan( ) . 2 cos ( ) gx v ? ? ? ? ? ? ?= + ? ? ? +? ? Denote the dimensionless quantity 20/ 2g x v by ;? in this case 2 2 (9.80 m/s )(60.0 m)cos30.0 0.2486. 2(32.0 m/s) ? °= = The above relation can then be written, on multiplying both sides by the product cos cos( ),? ? ?+ cos sin cos( ) sin( )cos , cos( ) ? ?? ? ? ? ? ? ? ?+ = + ? + and so cos sin( )cos cos( )sin . cos( ) ? ?? ? ? ? ? ? ? ?+ ? + = + The term on the left is sin(( ) ) sin ,? ? ? ?+ ? = so the result of this combination is sin cos( ) cos .? ? ? ? ?+ = Although this can be done numerically (by iteration, trial-and-error, or other methods), the expansion 1 2sin cos (sin( ) sin( ))a b a b a b= + + ? allows the angle ? to be isolated; specifically, then 1 (sin(2 ) sin( )) cos , 2 ? ? ? ? ?+ + ? = with the net result that sin(2 ) 2 cos sin .? ? ? ? ?+ = + (a) For 30 ,? = ° and ? as found above, 19.3? = ° and the angle above the horizontal is 49.3 .? ?+ = ° For level ground, using 0.2871,? = gives 17.5 .? = ° (b) For 30 ,? = ? ° the same ? as with 30? = ° may be used (cos30 cos( 30 )),° = ? ° giving 13.0? = ° and 17.0 .? ?+ = ? ° EVALUATE: For 0? = the result becomes 20sin(2 ) 2 /g x v? ?= = . This is equivalent to the expression 2 0 0sin(2 )vR g ?= derived in Example 3.8. 3.91. IDENTIFY: Find ?v! and use this to calculate the magnitude and direction of the average acceleration. SET UP: In a time t,? the velocity vector has moved through an angle (in radians) v t R ? ?? = (see Figure 3.28 in the textbook). By considering the isosceles triangle formed by the two velocity vectors, the magnitude ? v! is seen to be 2 sin( / 2)v ? . EXECUTE: av 10 m/s2 sin sin([1.0 / s] t)2 t v v v t t t R ? ?? ?= = = ?? ?? ? ?? ?a !# Using the given values gives magnitudes of 2 29.59 m/s 9.98 m/s, and 210.0 m/s . The changes in direction of the velocity vectors are given by v t R ? ?? = and are, respectively, 1.0 rad, 0.2 rad, and 0.1 rad. Therefore, the angle of the average acceleration vector with the original velocity vector is / 2 1/ 2 rad(or 118.6 ), 2 ? ? ?+ ? = + ° / 2 0.1 rad(or 95.7 ),? + ° and / 2 0.05 rad(or 92.9 ).? + ° EVALUATE: The instantaneous acceleration magnitude, 2 2 2/ (5.00 m/s) /(2.50 m 10 0 m/sv R ) .= = is indeed approached in the limit at 0.t? ? Also, the direction of ava # approaches the radially inward direction as 0t? ? . 3.92. IDENTIFY: The rocket has two periods of constant acceleration motion. SET UP: Let y+ be upward. During the free-fall phase, 0xa = and ya g= ? . After the engines turn on, (3.00 )cos30.0xa g= ° and (3.00 )sin30.0ya g= ° . Let t be the total time since the rocket was dropped and let T be the time the rocket falls before the engine starts. EXECUTE: (i) The diagram is given in Figure 3.92a. (ii) The x-position of the plane is (236 m/s)t and the x-position of the rocket is 2 2(236 m/s) (1/ 2)(3.00)(9.80 m/s )cos30 ( ) .t t T+ ° ? The graphs of these two equations are sketched in Figure 3.92b. 3-38 Chapter 3 (iii) If we take 0y = to be the altitude of the airliner, then 2 2 2( ) 1/ 2 ( ) 1/ 2(3.00)(9.80 m/s )(sin30 )( )y t gT gT t T t T= ? ? ? + ° ? for the rocket. The airliner has constant y. The graphs are sketched in Figure 3.92b. In each of the Figures 3.92a-c, the rocket is dropped at 0t = and the time T when the motor is turned on is indicated. By setting 0y = for the rocket, we can solve for t in terms of T : 2 2 2 2 20 (4.90 m/s ) (9.80 m/s ) ( ) (7.35 m/s )( )T T t T t T= ? ? ? + ? . Using the quadratic formula for the variable x t T= ? we find 2 2 2 2 2 2 (9.80 m/s ) (9.80 m/s ) (4)(7.35 m/s )(4.9) 2(7.35 m/s ) T T T x t T + += ? = , or 2.72 .t T= Now, using the condition that rocket plane 1000 mx x ,? = we find 2 2(236 m/s) (12.7 m/s ( ) (236 m/s) 1000 m,t ) t T t+ ? ? = or 2 2(1.72 ) 78.6 s .T = Therefore 5.15 s.T = EVALUATE: During the free-fall phase the rocket and airliner have the same x coordinate but the rocket moves downward from the airliner. After the engines fire, the rocket starts to move upward and its horizontal component of velocity starts to exceed that of the airliner. Figure 3.92 3.93. IDENTIFY: Apply the relative velocity relation. SET UP: Let C/Wv be the speed of the canoe relative to water and W/Gv be the speed of the water relative to the ground. EXECUTE: (a) Taking all units to be in km and h, we have three equations. We know that heading upstream C/W W/G 2v v? = . We know that heading downstream for a time C/W W/G, ( ) 5.t v v t+ = We also know that for the bottle W/G ( 1) 3.v t + = Solving these three equations for W/G C/W, 2v x v x,= = + therefore (2 ) 5x x t+ + = or (2 2 ) 5.x t+ = Also 3/ 1t x ,= ? so 3(2 2 ) 1 5x x ? ?+ ? =? ?? ? or 22 6 0.x x+ ? = The positive solution is W/G 1.5 km/h.x v= = (b) C/W W/G2 km/h 3.5 km/h.v v= + = EVALUATE: When they head upstream, their speed relative to the ground is 3.5 km/h 1.5 km/h 2.0 km/h? = . When they head downstream, their speed relative to the ground is 3.5 km/h 1.5 km/h 5.0 km/h+ = . The bottle is moving downstream at 1.5 km/s relative to the earth, so they are able to overtake it. NEWTON ? S LAWS OF M OTION 4 4.1. IDENTIFY: Consider the vector sum in each case. SET UP: Call the two forces 1F ! and 2F ! . Let 1F ! be to the right. In each case select the direction of 2F ! such that 1= 2F F + F ! ! ! has the desired magnitude. EXECUTE: (a) For the magnitude of the sum to be the sum of the magnitudes, the forces must be parallel, and the angle between them is zero. The two vectors and their sum are sketched in Figure 4.1a. (b) The forces form the sides of a right isosceles triangle, and the angle between them is 90 . The two vectors and their sum are sketched in Figure 4.1b. ° (c) For the sum to have zero magnitude, the forces must be antiparallel, and the angle between them is 180 . The two vectors are sketched in Figure 4.1c. ° EVALUATE: The maximum magnitude of the sum of the two vectors is 2F, as in part (a). Figure 4.1 4.2. IDENTIFY: Add the three forces by adding their components. SET UP: In the new coordinates, the 120-N force acts at an angle of 53 from the ° x? -axis, or from the 233° x+ -axis, and the 50-N force acts at an angle of 32 from the 3° x+ -axis. EXECUTE: (a) The components of the net force are (120 N)cos233 (50 N)cos323 32 NxR = °+ ° = ? (250 N) (120 N)sin 233 (50 N)sin323 124 N.yR = + °+ ° = (b) 2 2 128 N,x yR R R= + = 124arctan 10432 ? ? = °? ??? ? . The results have the same magnitude as in Example 4.1, and the angle has been changed by the amount (37 that the coordinates have been rotated. )° EVALUATE: We can use any set of coordinate axes that we wish to and can therefore select axes for which the analysis of the problem is the simplest. 4.3. IDENTIFY: Use right-triangle trigonometry to find the components of the force. SET UP: Let x+ be to the right and let y+ be downward. EXECUTE: The horizontal component of the force is to the right and the vertical component is down. (10 N)cos45 7.1 N° = (10 N)sin 45 7.1 N° = EVALUATE: In our coordinates each component is positive; the signs of the components indicate the directions of the component vectors. 4.4. IDENTIFY: cosxF F ?= , sinyF F ?= . SET UP: Let x+ be parallel to the ramp and directed up the ramp. Let y+ be perpendicular to the ramp and directed away from it. Then . 30.0? = ° 4-1 4-2 Chapter 4 EXECUTE: (a) 60.0N 69.3 N. cos cos30 xFF ?= = =° (b) sin tan 34.6 N.y xF F F? ?= = = EVALUATE: We can verify that . The signs of 2 2x yF F F+ = 2 xF and yF show their direction. 4.5. IDENTIFY: Vector addition. SET UP: Use a coordinate system where the -axisx+ is in the direction of ,AF ! the force applied by dog A. The forces are sketched in Figure 4.5. EXECUTE: 270 N,AxF = + 0AyF = cos60.0 (300 N)cos60.0 150 NBx BF F= ° = ° = + sin60.0 (300 N)sin 60.0 260 NBy BF F= ° = ° = + Figure 4.5a A B= +R F F ! ! ! 270 N 150 N 420 Nx Ax BxR F F= + = + + = + 0 260 N 260 Ny Ay ByR F F= + = + = + 2 2 x yR R R= + 2 2(420 N) (260 N) 494 NR = + = tan 0.619y x R R ? = = 31.8? = ° Figure 4.5b EVALUATE: The forces must be added as vectors. The magnitude of the resultant force is less than the sum of the magnitudes of the two forces and depends on the angle between the two forces. 4.6. IDENTIFY: Add the two forces using components. SET UP: cosxF F ?= , sinyF F ?= , where ? is the angle F ! makes with the x+ axis. EXECUTE: (a) 1 2 (9.00 N)cos120 (6.00 N)cos(233.1 ) 8.10 Nx xF F+ = ° + ° = ? 1 2 (9.00 N)sin120 (6.00 N)sin(233.1 ) 3.00 N.y yF F+ = ° + = +° (b) 2 2 2 2(8.10 N) (3.00 N) 8.64 N.x yR R R= + = + = EVALUATE: Since and , 0xF < 0yF > F ! is in the second quadrant. 4.7. IDENTIFY: Apply m?F = a! ! . SET UP: Let x+ be in the direction of the force. EXECUTE: . 2/ (132 N) (60 kg) 2.2 m /sx xa F m /= = = EVALUATE: The acceleration is in the direction of the force. 4.8. IDENTIFY: Apply m?F = a! ! . SET UP: Let x+ be in the direction of the acceleration. EXECUTE: 2(135 kg)(1.40 m/s ) 189 N.x xF ma= = = EVALUATE: The net force must be in the direction of the acceleration. 4.9. IDENTIFY: Apply m=?F a! ! to the box. SET UP: Let x+ be the direction of the force and acceleration. 48.0 NxF =? . Newton?s Laws of Motion 4-3 EXECUTE: x xF ma=? gives 248.0 N 16.0 kg3.00 m/sxx F m a = = =? . EVALUATE: The vertical forces sum to zero and there is no motion in that direction. 4.10. IDENTIFY: Use the information about the motion to find the acceleration and then use x xF ma=? to calculate m. SET UP: Let x+ be the direction of the force. 80.0 NxF =? . EXECUTE: (a) , , 0 11.0 mx x? = 5.00 st = 0 0xv = . 210 0 2x xx x v t a t? = + gives 20 2 2 2( ) 2(11.0 m) 0.880 m/s . (5.00 s)x x x a t ?= = = 280.0 N 90.9 kg0.880 m/s x x F m a = = =? . (b) and 0xa = xv is constant. After the first 5.0 s, . 20 (0.880 m/s )(5.00 s) 4.40 m/sx x xv v a t= + = = 21 0 0 2 (4.40 m/s)(5.00 s) 22.0 mx xx x v t a t? = + = = . EVALUATE: The mass determines the amount of acceleration produced by a given force. The block moves farther in the second 5.00 s than in the first 5.00 s. 4.11. IDENTIFY and SET UP: Use Newton?s second law in component form (Eq.4.8) to calculate the acceleration produced by the force. Use constant acceleration equations to calculate the effect of the acceleration on the motion. EXECUTE: (a) During this time interval the acceleration is constant and equal to 20.250 N 1.562 m/s 0.160 kg x x F a m = = = We can use the constant acceleration kinematic equations from Chapter 2. 2 21 1 0 0 2 20 (1.562 m/s )(2.00 s) ,x xx x v t a t? = + = + 2 so the puck is at 3.12 m.x = 2 0 0 (1.562 m/s )(2.00 s) 3.13 m/s.x x xv v a t= + = + = (b) In the time interval from to 5.00 s the force has been removed so the acceleration is zero. The speed stays constant at The distance the puck travels is At the end of the interval it is at 2.00 st = 3.12 m/s.xv = 0 0 (3.12 m/s)(5.00 s 2.00 s) 9.36 m.xx x v t? = = ? = 0 9.36 m 12.5 m.x x= + = In the time interval from to 7.00 s the acceleration is again At the start of this interval and 5.00 st = 21.562 m/s .xa = 0 3.12 m/sxv = 0 12.5 m.x = 2 21 1 0 0 2 2(3.12 m/s)(2.00 s) (1.562 m/s )(2.00 s) .x xx x v t a t? = + = + 2 0 6.24 m 3.12 m 9.36 m.x x? = + = Therefore, at the puck is at 7.00 st = 0 9.36 m 12.5 m 9.36 m 21.9 m.x x= + = + = 2 0 3.12 m/s (1.562 m/s )(2.00 s) 6.24 m/sx x xv v a t= + = + = EVALUATE: The acceleration says the puck gains 1.56 m/s of velocity for every second the force acts. The force acts a total of 4.00 s so the final velocity is (1.56 m/s)(4.0 s) 6.24 m/s.= 4.12. IDENTIFY: Apply m?F = a! ! . Then use a constant acceleration equation to relate the kinematic quantities. SET UP: Let x+ be in the direction of the force. EXECUTE: (a) 2/ (140 N) /(32.5 kg) 4.31 m/s .x xa F m= = = (b) 210 0 2x xx x v t a t? = + . With 210 20, 215 mxv x at= = = . (c) 0x x xv v a t= + . With . 0 0, 2 / 43.0 m/sx x xv v a t x t= = = = EVALUATE: The acceleration connects the motion to the forces. 4.13. IDENTIFY: The force and acceleration are related by Newton?s second law. SET UP: x xF ma=? , where xF? is the net force. 4.50 kgm = . EXECUTE: (a) The maximum net force occurs when the acceleration has its maximum value. . This maximum force occurs between 2.0 s and 4.0 s. 2(4.50 kg)(10.0 m/s ) 45.0 Nx xF ma= = =? (b) The net force is constant when the acceleration is constant. This is between 2.0 s and 4.0 s. (c) The net force is zero when the acceleration is zero. This is the case at 0t = and 6.0 st = . EVALUATE: A graph of xF? versus t would have the same shape as the graph of xa versus t. 4-4 Chapter 4 4.14. IDENTIFY: The force and acceleration are related by Newton?s second law. xx dva dt= , so xa is the slope of the graph of xv versus t. SET UP: The graph of xv versus t consists of straight-line segments. For 0t = to 2.00 st = , . For to 6.00 s, . For to 10.0 s, . 24.00 m/sxa = 2.00 st = 0xa = 6.00 st = 21.00 m/sxa = x xF ma=? , with . 2.75 kgm = xF? is the net force. EXECUTE: (a) The maximum net force occurs when the acceleration has its maximum value. . This maximum occurs in the interval to . 2(2.75 kg)(4.00 m/s ) 11.0 Nx xF ma= = =? 0t = 2.00 st = (b) The net force is zero when the acceleration is zero. This is between 2.00 s and 6.00 s. (c) Between 6.00 s and 10.0 s, , so 21.00 m/sxa = 2(2.75 kg)(1.00 m/s ) 2.75 N xF = =? . EVALUATE: The net force is largest when the velocity is changing most rapidly. 4.15. IDENTIFY: The net force and the acceleration are related by Newton?s second law. When the rocket is near the surface of the earth the forces on it are the upward force F ! exerted on it because of the burning fuel and the downward force gravF ! of gravity. . gravF m= g SET UP: Let be upward. The weight of the rocket is . y+ 2grav (8.00 kg)(9.80 m/s ) 78.4 NF = = EXECUTE: (a) At , . At 0t = 100.0 NF A= = 2.00 st = , and 2(4.00 s ) 150.0 NF A B= + = 2 2 150.0 N 100.0 N 12.5 N/s 4.00 s B ?= = . (b) (i) At , . The net force is 0t = 100.0 NF A= = grav 100.0 N 78.4 N 21.6 NyF F F= ? = ? =? . 221.6 N 2.70 m/s 8.00 kg y y F a m = = =? . (ii) At 3.00 s,t = 2(3.00 s) 212.5 NF A B= + = . . 212.5 N 78.4 N 134.1 NyF = ? =? 2134.1 N 16.8 m/s8.00 kgyy F a m = = =? . (c) Now and . grav 0F = 212.5 NyF F= =? 2212.5 N 26.6 m/s8.00 kgy = =a . EVALUATE: The acceleration increases as F increases. 4.16. IDENTIFY: Use constant acceleration equations to calculate xa and t. Then use m=?F a! ! to calculate the net force. SET UP: Let x+ be in the direction of motion of the electron. EXECUTE: (a) , , . gives 0 0xv = 20( ) 1.80 10 mx x ?? = × 63.00 10 m/sxv = × 2 20 02 ( )x x xv v a x x= + ? 2 2 6 2 14 20 2 0 (3.00 10 m/s) 0 2.50 10 m/s 2( ) 2(1.80 10 m) x x x v v a x x ? ? × ?= = = ×? × (b) 0x x xv v a t= + gives 6 80 14 2 3.00 10 m/s 0 1.2 10 s 2.50 10 m/s x x x v v t a ?? × ?= = = ×× (c) . 31 14 2 16(9.11 10 kg)(2.50 10 m/s ) 2.28 10 Nx xF ma ? ?= = × × = ×? EVALUATE: The acceleration is in the direction of motion since the speed is increasing, and the net force is in the direction of the acceleration. 4.17. IDENTIFY and SET UP: We must use .F ma= w mg= to find the mass of the boulder. EXECUTE: 22400 N 244.9 kg9.80 m/s w m g = = = Then 2(244.9 kg)(12.0 m/s ) 2940 N.F ma= = = EVALUATE: We must use mass in Newton?s second law. Mass and weight are proportional. 4.18. IDENTIFY: Apply m?F = a! ! . SET UP: . 2/ (71.2 N) /(9.80 m/s ) 7.27 kgm w g= = = EXECUTE: 2160 N 22.0 m/s 7.27 kg x x F a m = = = EVALUATE: The weight of the ball is a vertical force and doesn?t affect the horizontal acceleration. However, the weight is used to calculate the mass. Newton?s Laws of Motion 4-5 4.19. IDENTIFY and SET UP: . The mass of the watermelon is constant, independent of its location. Its weight differs on earth and Jupiter?s moon. Use the information about the watermelon?s weight on earth to calculate its mass: w mg= EXECUTE: gives that w mg= 244.0 N 4.49 kg.9.80 m/s w m g = = = On Jupiter?s moon, the same as on earth. Thus the weight on Jupiter?s moon is 4.49 kg,m = 2(4.49 kg)(1.81 m/s ) 8.13 N.w mg= = = EVALUATE: The weight of the watermelon is less on Io, since g is smaller there. 4.20. IDENTIFY: Weight and mass are related by w mg= . The mass is constant but g and w depend on location. SET UP: On earth, . 29.80 m/sg = EXECUTE: (a) w m g = , which is constant, so E A E A w w g g = . , , and . E 17.5 Nw = 2E 9.80 m/sg = A 3.24 Nw = 2 2A A E E 3.24 N (9.80 m/s ) 1.81 m/s 17.5 N w g g w ? ? ? ?= = =? ? ? ?? ?? ? . (b) E 2 E 17.5 N 1.79 kg 9.80 m/s w m g = = = . EVALUATE: The weight at a location and the acceleration due to gravity at that location are directly proportional. 4.21. IDENTIFY: Apply x xF ma=? to find the resultant horizontal force. SET UP: Let the acceleration be in the x+ direction. EXECUTE: . The force is exerted by the blocks. The blocks push on the sprinter because the sprinter pushes on the blocks. 2(55 kg)(15 m/s ) 825 Nx xF ma= = =? EVALUATE: The force the blocks exert on the sprinter has the same magnitude as the force the sprinter exerts on the blocks. The harder the sprinter pushes, the greater the force on him. 4.22. IDENTIFY: m=?F a! ! refers to forces that all act on one object. The third law refers to forces that a pair of objects exert on each other. SET UP: An object is in equilibrium if the vector sum of all the forces on it is zero. A third law pair of forces have the same magnitude regardless of the motion of either object. EXECUTE: (a) the earth (gravity) (b) 4 N; the book (c) no, these two forces are exerted on the same object (d) 4 N; the earth; the book; upward (e) 4 N, the hand; the book; downward (f) second (The two forces are exerted on the same object and this object has zero acceleration.) (g) third (The forces are between a pair of objects.) (h) No. There is a net upward force on the book equal to 1 N. (i) No. The force exerted on the book by your hand is 5 N, upward. The force exerted on the book by the earth is 4 N, downward. (j) Yes. These forces form a third-law pair and are equal in magnitude and opposite in direction. (k) Yes. These forces form a third-law pair and are equal in magnitude and opposite in direction. (l) One, only the gravity force. (m) No. There is a net downward force of 5 N exerted on the book. EVALUATE: Newton?s second and third laws give complementary information about the forces that act. 4.23. IDENTIFY: Identify the forces on the bottle. SET UP: Classify forces as contact or noncontact forces. The noncontact force is gravity and the contact forces come from things that touch the object. Gravity is always directed downward toward the center of the earth. Air resistance is always directed opposite to the velocity of the object relative to the air. EXECUTE: (a) The free-body diagram for the bottle is sketched in Figure 4.23a The only forces on the bottle are gravity (downward) and air resistance (upward). Figure 4.23a (b) 4-6 Chapter 4 w is the force of gravity that the earth exerts on the bottle. The reaction to this force is ,w? force that the bottle exerts on the earth Figure 4.23b Note that these two equal and opposite forces produce very different accelerations because the bottle and the earth have very different masses. airF is the force that the air exerts on the bottle and is upward. The reaction to this force is a downward force airF ? that the bottle exerts on the air. These two forces have equal magnitudes and opposite directions. EVALUATE: The only thing in contact with the bottle while it is falling is the air. Newton?s third law always deals with forces on two different objects. 4.24. IDENTIFY: The reaction forces in Newton?s third law are always between a pair of objects. In Newton?s second law all the forces act on a single object. SET UP: Let be downward. . y+ /m w g= EXECUTE: The reaction to the upward normal force on the passenger is the downward normal force, also of magnitude 620 N, that the passenger exerts on the floor. The reaction to the passenger?s weight is the gravitational force that the passenger exerts on the earth, upward and also of magnitude 650 N. y y F a m =? gives 2 2 650 N 620 N 0.452 m/s (650 N)/(9.80 m/s )y a ?= = . The passenger?s acceleration is , downward. 20.452 m /s EVALUATE: There is a net downward force on the passenger and the passenger has a downward acceleration. 4.25. IDENTIFY: Apply Newton?s second law to the earth. SET UP: The force of gravity that the earth exerts on her is her weight, By Newton?s 3rd law, she exerts an equal and opposite force on the earth. 2(45 kg)(9.8 m/s ) 441 N.w mg= = = Apply m=?F a! ! to the earth, with 441 N,w= =?F! but must use the mass of the earth for m. EXECUTE: 23 224441 N 7.4 10 m/s .6.0 10 kg w a m ?= = = ×× EVALUATE: This is much smaller than her acceleration of The force she exerts on the earth equals in magnitude the force the earth exerts on her, but the acceleration the force produces depends on the mass of the object and her mass is much less than the mass of the earth. 29.8 m/s . 4.26. IDENTIFY and SET UP: The only force on the ball is the gravity force, gravF ! . This force is , downward and is independent of the motion of the object. mg EXECUTE: The free-body diagram is sketched in Figure 4.26. The free-body diagram is the same in all cases. EVALUATE: Some forces, such as friction, depend on the motion of the object but the gravity force does not. Figure 4.26 4.27. IDENTIFY: Identify the forces on each object. SET UP: In each case the forces are the noncontact force of gravity (the weight) and the forces applied by objects that are in contact with each crate. Each crate touches the floor and the other crate, and some object applies F ! to crate A. EXECUTE: (a) The free-body diagrams for each crate are given in Figure 4.27. ABF (the force on due toAm Bm ) and BAF (the force on Bm due to ) form an action-reaction pair. Am (b) Since there is no horizontal force opposing F, any value of F, no matter how small, will cause the crates to accelerate to the right. The weight of the two crates acts at a right angle to the horizontal, and is in any case balanced by the upward force of the surface on them. Newton?s Laws of Motion 4-7 EVALUATE: Crate B is accelerated by BAF and crate A is accelerated by the net force . The greater the total weight of the two crates, the greater their total mass and the smaller will be their acceleration. ABF F? Figure 4.27 4.28. IDENTIFY: The surface of block B can exert both a friction force and a normal force on block A. The friction force is directed so as to oppose relative motion between blocks B and A. Gravity exerts a downward force w on block A. SET UP: The pull is a force on B not on A. EXECUTE: (a) If the table is frictionless there is a net horizontal force on the combined object of the two blocks, and block B accelerates in the direction of the pull. The friction force that B exerts on A is to the right, to try to prevent A from slipping relative to B as B accelerates to the right. The free-body diagram is sketched in Figure 4.28a. f is the friction force that B exerts on A and n is the normal force that B exerts on A. (b) The pull and the friction force exerted on B by the table cancel and the net force on the system of two blocks is zero. The blocks move with the same constant speed and B exerts no friction force on A. The free-body diagram is sketched in Figure 4.28b. EVALUATE: If in part (b) the pull force is decreased, block B will slow down, with an acceleration directed to the left. In this case the friction force on A would be to the left, to prevent relative motion between the two blocks by giving A an acceleration equal to that of B. Figure 4.28 4.29. IDENTIFY: Since the observer in the train sees the ball hang motionless, the ball must have the same acceleration as the train car. By Newton?s second law, there must be a net force on the ball in the same direction as its acceleration. SET UP: The forces on the ball are gravity, which is w, downward, and the tension T in the string, which is directed along the string. ! EXECUTE: (a) The acceleration of the train is zero, so the acceleration of the ball is zero. There is no net horizontal force on the ball and the string must hang vertically. The free-body diagram is sketched in Figure 4.29a. (b) The train has a constant acceleration directed east so the ball must have a constant eastward acceleration. There must be a net horizontal force on the ball, directed to the east. This net force must come from an eastward component of T ! and the ball hangs with the string displaced west of vertical. The free-body diagram is sketched in Figure 4.29b. EVALUATE: When the motion of an object is described in an inertial frame, there must be a net force in the direction of the acceleration. Figure 4.29 4.30. IDENTIFY: Identify the forces for each object. Action-reaction pairs of forces act between two objects. SET UP: Friction is parallel to the surfaces and is directly to oppose relative motion between the surfaces. EXECUTE: The free-body diagram for the box is given in Figure 4.30a. The free body diagram for the truck is given in Figure 4.30b. The box?s friction force on the truck bed and the truck bed?s friction force on the box form an action-reaction pair. There would also be some small air-resistance force action to the left, presumably negligible at this speed. 4-8 Chapter 4 EVALUATE: The friction force on the box, exerted by the bed of the truck, is in the direction of the truck's acceleration. This friction force can't be large enough to give the box the same acceleration that the truck has and the truck acquires a greater speed than the box. Figure 4.30 4.31. IDENTIFY: Identify the forces on the chair. The floor exerts a normal force and a friction force. SET UP: Let be upward and let y+ x+ be in the direction of the motion of the chair. EXECUTE: (a) The free-body diagram for the chair is given in Figure 4.31. (b) For the chair, so 0ya = y yF ma=? gives sin37 0n mg F? ? ° = and 142 Nn = . EVALUATE: n is larger than the weight because F! has a downward component. Figure 4.31 4.32. IDENTIFY: Identify the forces on the skier and apply m=?F a! ! . Constant speed means . 0a = SET UP: Use coordinates that are parallel and perpendicular to the slope. EXECUTE: (a) The free-body diagram for the skier is given in Figure 4.32. (b) x xF ma=? with gives . 0xa = 2sin (65.0 kg)(9.80 m/s )sin 26.0 279 NT mg ?= = ° = EVALUATE: T is less than the weight of the skier. It is equal to the component of the weight that is parallel to the incline. Figure 4.32 4.33. IDENTIFY: m?F = a! ! must be satisfied for each object. Newton?s third law says that the force that the car exerts on the truck is equal in magnitude and opposite in direction to the force C on TF ! T on CF ! that the truck exerts on the car. SET UP: The only horizontal force on the car is the force T on CF ! exerted by the truck. The car exerts a force on the truck. There is also a horizontal friction force C on TF ! f ! that the highway surface exerts on the truck. Assume the system is accelerating to the right in the free-body diagrams. EXECUTE: (a) The free-body diagram for the car is sketched in Figure 4.33a (b) The free-body diagram for the truck is sketched in Figure 4.33b. Newton?s Laws of Motion 4-9 (c) The friction force f ! accelerates the system forward. The tires of the truck push backwards on the highway surface as they rotate, so by Newton?s third law the roadway pushes forward on the tires. EVALUATE: and each equal the tension T in the rope. Both objects have the same acceleration T on CF C on TF a! . and CT m a= Tf T m a? = , so C T( )f m m a= + . The acceleration of the two objects is proportional to f. Figure 4.33 4.34. IDENTIFY: Use a constant acceleration equation to find the stopping time and acceleration. Then use m=?F a! ! to calculate the force. SET UP: Let x+ be in the direction the bullet is traveling. F! is the force the wood exerts on the bullet. EXECUTE: (a) , and . 0 350 m/sxv = 0xv = 0( ) 0.130 mx x? = 00 2 x xv v( )x t +? ?? = ? ?? ? x gives 40 0 2( ) 2(0.130 m) 7.43 10 s 350 m/sx x x x t v v ??= = = ×+ . (b) gives 2 20 02 ( )x x xv v a x x= + ? 2 2 2 5 20 0 0 (350 m/s) 4.71 10 m/s 2( ) 2(0.130 m) x x x v v a x x ? ?= = = ? ×? x xF ma=? gives xF ma? = and . 3 5 2(1.80 10 kg)( 4.71 10 m/s ) 848 NxF ma ?= ? = ? × ? × = EVALUATE: The acceleration and net force are opposite to the direction of motion of the bullet. 4.35. IDENTIFY: Vector addition problem. Write the vector addition equation in component form. We know one vector and its resultant and are asked to solve for the other vector. SET UP: Use coordinates with the along -axisx+ 1F ! and the -axisy+ along ;R! as shown in Figure 4.35a. 1 1300 N,xF = + 1 0yF = 0,xR = 1300 NyR = + Figure 4.35a 1 2 ,F F R ! ! !+ = so 2 1F R F ! ! != ? EXECUTE: 2 1 0 1300 N 1300 Nx x xF R F= ? = ? = ? 2 1 1300 N 0 1300 Ny y yF R F= ? = + ? = + The components of 2F ! are sketched in Figure 4.35b. 2 2 2 2 2 2 ( 1300 N) (1300 N)x yF F F= + = ? + 2 1840 NF = 2 2 1300 N tan 1.00 1300 N y x F F ? += = = ?? 135? = ° Figure 4.35b The magnitude of 2F ! is 1840 N and its direction is counterclockwise from the direction of 135° 1.F ! EVALUATE: 2F ! has a negative x-component to cancel 1F ! and a y-component to equal .R ! 4.36. IDENTIFY: Use the motion of the ball to calculate g , the acceleration of gravity on the planet. Then w mg= . SET UP: Let be downward and take y+ 0 0y = . 0 0yv = since the ball is released from rest. 4-10 Chapter 4 EXECUTE: Get g on X: 21 2 y gt= gives 2110.0 m (2.2 s) 2 g= . and then . 24.13 m /sg = 2 X X (0.100 kg)(4.03 m /s ) 0.41 Nw mg= = = EVALUATE: g on Planet X is smaller than on earth and the object weighs less than it would on earth. 4.37. IDENTIFY: If the box moves in the -directionx+ it must have 0,ya = so 0.yF =? The smallest force the child can exert and still produce such motion is a force that makes the y-components of all three forces sum to zero, but that doesn?t have any x-component. Figure 4.37 SET UP: 1F ! and 2F ! are sketched in Figure 4.37. Let 3F ! be the force exerted by the child. y yF ma=? implies so 1 2 3 0,y y yF F F+ + = 3 1 2( ).y y yF F F= ? + EXECUTE: 1 1 sin60 (100 N)sin 60 86.6 NyF F= + ° = ° = 2 2 2sin( 30 ) sin30 (140 N)sin30 70.0 NyF F F= + ? ° = ? ° = ? ° = ? Then 3 1 2( ) (86.6 N 70.0 N) 16.6 N;y y yF F F= ? + = ? ? = ? 3 0xF = The smallest force the child can exert has magnitude 17 N and is directed at clockwise from the shown in the figure. 90° -axisx+ (b) IDENTIFY and SET UP: Apply .x xF ma=? We know the forces and xa so can solve for m. The force exerted by the child is in the and has no x-component. -directiony? EXECUTE: 1 1 cos60 50 NxF F= ° = 2 2 cos30 121.2 NxF F= ° = 1 2 50 N 121.2 N 171.2 Nx x xF F F= + = + =? 2 171.2 N 85.6 kg 2.00 m/s x x F m a = = =? Then 840 N.w mg= = EVALUATE: In part (b) we don?t need to consider the y-component of Newton?s second law. so the mass doesn?t appear in the 0ya = y yF ma=? equation. 4.38. IDENTIFY: Use m=?F a! ! to calculate the acceleration of the tanker and then use constant acceleration kinematic equations. SET UP: Let x+ be the direction the tanker is moving initially. Then . /xa F= ? m 0EXECUTE: says that if the reef weren't there the ship would stop in a distance of 2 20 2 ( )x x xv v a x x= + ? 2 2 2 7 2 0 0 0 0 4 (3.6 10 kg)(1.5 m /s) 506 m, 2 2( / ) 2 2(8.0 10 N) x x v v mv x x a F m F ×? = ? = = = =× so the ship would hit the reef. The speed when the tanker hits the reef is found from , so it is 2 20 02 ( )x x xv v a x x= + ? 4 2 2 0 7 2(8.0 10 N)(500 m) (2 / ) (1.5 m/s) 0.17 m/s, (3.6 10 kg) v v Fx m ×= ? = ? =× and the oil should be safe. EVALUATE: The force and acceleration are directed opposite to the initial motion of the tanker and the speed decreases. 4.39. IDENTIFY: We can apply constant acceleration equations to relate the kinematic variables and we can use Newton?s second law to relate the forces and acceleration. (a) SET UP: First use the information given about the height of the jump to calculate the speed he has at the instant his feet leave the ground. Use a coordinate system with the -axisy+ upward and the origin at the position when his feet leave the ground. Newton?s Laws of Motion 4-11 0yv = (at the maximum height), 0 ?,yv = 29.80 m/s ,ya = ? 0 1.2 my y? = + 2 2 0 02 ( )y y yv v a y y= + ? EXECUTE: 20 02 ( ) 2( 9.80 m/s )(1.2 m) 4.85 m/sy yv a y y= ? ? = ? ? = (b) SET UP: Now consider the acceleration phase, from when he starts to jump until when his feet leave the ground. Use a coordinate system where the -axisy+ is upward and the origin is at his position when he starts his jump. EXECUTE: Calculate the average acceleration: 0 2 av 4.89 m/s 0 ( ) 16.2 m/s 0.300 s y y y v v a t ? ?= = = (c) SET UP: Finally, find the average upward force that the ground must exert on him to produce this average upward acceleration. (Don?t forget about the downward force of gravity.) The forces are sketched in Figure 4.39. EXECUTE: 2 890 N / 90.8 kg 9.80 m/s m w g= = = y yF ma=? av av( ) yF mg m a? = av av( ( ) )yF m g a= + 2 2 av 90.8 kg(9.80 m/s 16.2 m/s )F = + av 2360 NF = Figure 4.39 This is the average force exerted on him by the ground. But by Newton?s 3rd law, the average force he exerts on the ground is equal and opposite, so is 2360 N, downward. EVALUATE: In order for him to accelerate upward, the ground must exert an upward force greater than his weight. 4.40. IDENTIFY: Use constant acceleration equations to calculate the acceleration xa that would be required. Then use x xF ma=? to find the necessary force. SET UP: Let x+ be the direction of the initial motion of the auto. EXECUTE: with 2 20 02 ( )x x xv v a x x= + ? 0xv = gives 2 0 02( ) x x v a x x = ? ? . The force F is directed opposite to the motion and x F a m = ? . Equating these two expressions for xa gives 2 2 60 2 0 (12.5 m /s) (850 kg) 3.7 10 N. 2( ) 2(1.8 10 m) xvF m x x ? = = = ×? × EVALUATE: A very large force is required to stop such a massive object in such a short distance. 4.41. IDENTIFY: Apply Newton?s second law to calculate a. (a) SET UP: The free-body diagram for the bucket is sketched in Figure 4.41. The net force on the bucket is ,T mg? upward. Figure 4.41 4-12 Chapter 4 (b) EXECUTE: y yF ma=? gives T m g ma? = 2 275.0 N (4.80 kg)(9.80 m/s ) 75.0 N 47.04 N 5.82 m/s . 4.80 kg 4.80 kg T mg a m ? ? ?= = = = EVALUATE: The weight of the bucket is 47.0 N. The upward force exerted by the cord is larger than this, so the bucket accelerates upward. 4.42. IDENTIFY: Apply m=?F a! ! to the parachutist. SET UP: Let be upward. y+ airF ! is the force of air resistance. EXECUTE: (a) 2(55.0 kg)(9.80 m/s ) 539 Nw mg= = = (b) The free-body diagram is given in Fig. 4.42. air 620 N 539 N 81 NyF F w= ? = ? =? . The net force is upward. (c) 2 81 N 1.5 m/s 55.0 kg y y F a m = = =? , upward. EVALUATE: Both the net force and the acceleration are upward. Since her velocity is downward and her acceleration is upward, her speed decreases. Figure 4.42 4.43. IDENTIFY: Use Newton?s 2nd law to relate the acceleration and forces for each crate. (a) SET UP: Since the crates are connected by a rope, they both have the same acceleration, 22.50 m/s . (b) The forces on the 4.00 kg crate are shown in Figure 4.43a. EXECUTE: x xF ma=? 2 1 (4.00 kg)(2.50 m/s ) 10.0 N.T m a= = = Figure 4.43a (c) SET UP: Forces on the 6.00 kg crate are shown in Figure 4.43b The crate accelerates to the right, so the net force is to the right. F must be larger than T. Figure 4.43b (d) EXECUTE: x xF ma=? gives 2F T m a? = 2 2 10.0 N (6.00 kg)(2.50 m/s ) 10.0 N 15.0 N 25.0 NF T m a= + = + = + = Newton?s Laws of Motion 4-13 EVALUATE: We can also consider the two crates and the rope connecting them as a single object of mass The free-body diagram is sketched in Figure 4.43c. 1 2 10.0 kg.m m m= + = x xF ma=? 2(10.0 kg)(2.50 m/s ) 25.0 NF ma= = = This agrees with our answer in part (d). Figure 4.43c 4.44. IDENTIFY: Apply Newton's second and third laws. SET UP: Action-reaction forces act between a pair of objects. In the second law all the forces act on the same object. EXECUTE: (a) The force the astronaut exerts on the cable and the force that the cable exerts on the astronaut are an action-reaction pair, so the cable exerts a force of 80.0 N on the astronaut. (b) The cable is under tension. (c) 2 80.0 N 0.762 m /s 105.0 kg F a m = = = . (d) There is no net force on the massless cable, so the force that the shuttle exerts on the cable must be 80.0 N (this is not an action-reaction pair). Thus, the force that the cable exerts on the shuttle must be 80.0 N. (e) 4 24 80.0 N 8.84 10 m /s 9.05 10 kg F a m ?= = = ×× . EVALUATE: Since the cable is massless the net force on it is zero and the tension is the same at each end. 4.45. IDENTIFY and SET UP: Take derivatives of ( )x t to find xv and .xa Use Newton?s second law to relate the acceleration to the net force on the object. EXECUTE: (a) 3 2 2 4 3(9.0 10 m/s ) (8.0 10 m/s )x t t3= × ? × 0x = at 0t = When 0.025 s,t = 3 2 2 4 3 3(9.0 10 m/s )(0.025 s) (8.0 10 m/s )(0.025 s) 4.4 m.x = × ? × = The length of the barrel must be 4.4 m. (b) 3 2 4 3(18.0 10 m/s ) (24.0 10 m/s )x dx v t dt = = × ? × 2t At 0 (object starts from rest). 0,t = xv = At when the object reaches the end of the barrel, 0.025 s,t = 2 4 3 2(18.0 10 m/s )(0.025 s) (24.0 10 m/s )(0.025 s) 300 m/sxv 3= × ? × = (c) ,x xF ma=? so must find .xa 3 2 318.0 10 m/s (48.0 10 m/s )xx dv a t dt 4= = × ? × (i) At and 0,t = 318.0 10 m/sxa = × 2 3 2 4(1.50 kg)(18.0 10 m/s ) 2.7 10 N.xF = × = ×? (ii) At and 0.025 s,t = 3 2 4 318 10 m/s (48.0 10 m/s )(0.025 s) 6.0 10 m/sxa 3= × ? × = × 2 2 3(1.50 kg)(6.0 10 m/s ) 9.0 10 N.xF 3= × = ×? EVALUATE: The acceleration and net force decrease as the object moves along the barrel. 4.46. IDENTIFY: Apply m=?F a! ! and solve for the mass m of the spacecraft. SET UP: . Let be upward. w mg= y+ EXECUTE: (a) The velocity of the spacecraft is downward. When it is slowing down, the acceleration is upward. When it is speeding up, the acceleration is downward. (b) In each case the net force is in the direction of the acceleration. Speeding up: and the net force is downward. Slowing down: and the net force is upward. w F> w F< 4-14 Chapter 4 (c) Denote the y-component of the acceleration when the thrust is by and the y-component of the acceleration when the thrust is by and . The forces and accelerations are then related by Dividing the first of these by the second to eliminate the mass gives 1F 1a 2F 1.a 21.20 m/sya = + 22 0.80 m/sa = ? 1 1 2,F w ma F w ma? = ? = 2. 1 2 2 , F w a F w a ? =? 1 and solving for the weight w gives 1 2 2 1 1 2 . a F a F w a a ?= ? Substituting the given numbers, with y+ upward, gives 2 3 2 3 3 2 2 (1.20 m /s )(10.0 10 N) ( 0.80 m /s )(25.0 10 N) 16.0 10 N. 1.20 m /s ( 0.80 m /s ) w × ? ? ×= =? ? × EVALUATE: The acceleration due to gravity at the surface of Mercury did not need to be found. 4.47. IDENTIFY: The ship and instrument have the same acceleration. The forces and acceleration are related by Newton?s second law. We can use a constant acceleration equation to calculate the acceleration from the information given about the motion. SET UP: Let be upward. The forces on the instrument are the upward tension Ty+ ! exerted by the wire and the downward force of gravity. w ! 2(6.50 kg)(9.80 m/s ) 63.7 Nw mg= = = EXECUTE: (a) The free-body diagram is sketched in Figure 4.47. The acceleration is upward, so T w . , , . > 0 276 my y? = 15.0 st = 0 0yv = 210 0 2y yy y v t a t? = + gives 202 22( ) 2(276 m) 2.45 m/s(15.0 s)y y y a t ?= = = . y yF ma=? gives and . T w ma? = 263.7 N (6.50 kg)(2.45 m/s ) 79.6 NT w ma= + = + = EVALUATE: There must be a net force in the direction of the acceleration. Figure 4.47 4.48. If the rocket is moving downward and its speed is decreasing, its acceleration is upward, just as in Problem 4.47. The solution is identical to that of Problem 4.47. 4.49. IDENTIFY: Apply m=?F a! ! to the gymnast. SET UP: The upward force on the gymnast gives the tension in the rope. The free-body diagram for the gymnast is given in Figure 4.49. EXECUTE: (a) If the gymnast climbs at a constant rate, there is no net force on the gymnast, so the tension must equal the weight; T . mg= (b) No motion is no acceleration, so the tension is again the gymnast?s weight. (c) T w T mg ma m? = ? = = a! (the acceleration is upward, the same direction as the tension), so ( )T m g= + a! . (d) T w T mg ma m? = ? = = ? a! (the acceleration is downward, the opposite direction to the tension), so ( )T m g= ? a! . EVALUATE: When she accelerates upward the tension is greater than her weight and when she accelerates downward the tension is less than her weight. Figure 4.49 Newton?s Laws of Motion 4-15 4.50. IDENTIFY: Apply m=?F a! ! to the elevator to relate the forces on it to the acceleration. (a) SET UP: The free-body diagram for the elevator is sketched in Figure 4.50. The net force is T mg? (upward). Figure 4.50 Take the to be upward since that is the direction of the acceleration. The maximum upward acceleration is obtained from the maximum possible tension in the cables. -directiony+ EXECUTE: y yF ma=? gives T m g ma? = 2 228,000 N (2200 kg)(9.80 m/s ) 2.93 m/s . 2200 kg T mg a m ? ?= = = (b) What changes is the weight mg of the elevator. 2 228,000 N (2200 kg)(1.62 m/s ) 11.1 m/s . 2200 kg T mg a m ? ?= = = EVALUATE: The cables can give the elevator a greater acceleration on the moon since the downward force of gravity is less there and the same T then gives a greater net force. 4.51. IDENTIFY: He is in free-fall until he contacts the ground. Use the constant acceleration equations and apply m=?F a! ! . SET UP: Take downward. While he is in the air, before he touches the ground, his acceleration is . y+ 29.80 m/sya = EXECUTE: (a) , , and . gives 0 0yv = 0 3.10 my y? = 29.80 m/sya = 2 20 02 ( )y y yv v a y y= + ? 2 02 ( ) 2(9.80 m/s )(3.10 m) 7.79 m/sy yv a y y= ? = = (b) , , . gives 0 7.79 m/syv = 0yv = 0 0.60 my y? = 2 20 02 ( )y y yv v a y y= + ? 2 2 2 0 2 0 0 (7.79 m/s) 50.6 m/s 2( ) 2(0.60 m) y y y v v a y y ? ?= = = ?? . The acceleration is upward. (c) The free-body diagram is given in Fig. 4.51. F ! is the force the ground exerts on him. y yF ma=? gives . , upward. mg F ma? = ? 2 2( ) (75.0 kg)(9.80 m/s 50.6 m/s ) 4.53 10 NF m g a= + = + = × 3 3 2 4.53 10 N 6.16 (75.0 kg)(9.80 m/s ) F w ×= = , so 6.16F w= . By Newton's third law, the force his feet exert on the ground is ?F! . EVALUATE: The force the ground exerts on him is about six times his weight. Figure 4.51 4.52. IDENTIFY: Apply m=?F a! ! to the hammer head. Use a constant acceleration equation to relate the motion to the acceleration. SET UP: Let be upward. y+ EXECUTE: (a) The free-body diagram for the hammer head is sketched in Figure 4.52. (b) The acceleration of the hammer head is given by with 2 20 02 ( )y y yv v a y y= + ? 0yv = , and . . The mass of the hammer 2 0 3.2 m/syv = ? 0 0.0045 my y? = ? 2 20 0/ 2( ) (3.2 m /s) / 2(0.0045 cm) 1.138 10 m /sy ya v y y= ? = = × 3 2 4-16 Chapter 4 head is its weight divided by , and so the net force on the hammer head is This is the sum of the forces on the hammer head: the upward force that the nail exerts, the downward weight and the downward 15-N force. The force that the nail exerts is then 590 N, and this must be the magnitude of the force that the hammer head exerts on the nail. 2, (4.9 N) /(9.80 m /s ) 0.50 kgg = 3 2(0.50 kg)(1.138 10 m /s ) 570 N.× = (c) The distance the nail moves is 0.12 m, so the acceleration will be , and the net force on the hammer head will be 2133 N. The magnitude of the force that the nail exerts on the hammer head, and hence the magnitude of the force that the hammer head exerts on the nail, is 2153 N, or about 2200 N. 24267 m /s EVALUATE: For the shorter stopping distance the acceleration has a larger magnitude and the force between the nail and hammer head is larger. Figure 4.52 4.53. IDENTIFY: Apply m=?F a! ! to some portion of the cable. SET UP: The free-body diagrams for the whole cable, the top half of the cable and the bottom half are sketched in Figure 4.53. The cable is at rest, so in each diagram the net force is zero. EXECUTE: (a) The net force on a point of the cable at the top is zero; the tension in the cable must be equal to the weight w. (b) The net force on the cable must be zero; the difference between the tensions at the top and bottom must be equal to the weight w, and with the result of part (a), there is no tension at the bottom. (c) The net force on the bottom half of the cable must be zero, and so the tension in the cable at the middle must be half the weight, . Equivalently, the net force on the upper half of the cable must be zero. From part (a) the tension at the top is w, the weight of the top half is and so the tension in the cable at the middle must be . / 2w / 2w / 2 / 2w w w? = (d) A graph of T vs. distance will be a negatively sloped line. EVALUATE: The tension decreases linearly from a value of w at the top to zero at the bottom of the cable. Figure 4.53 4.54. IDENTIFY: Note that in this problem the mass of the rope is given, and that it is not negligible compared to the other masses. Apply m=?F a! ! to each object to relate the forces to the acceleration. (a) SET UP: The free-body diagrams for each block and for the rope are given in Figure 4.54a. Figure 4.54a Newton?s Laws of Motion 4-17 tT is the tension at the top of the rope and bT is the tension at the bottom of the rope. EXECUTE: (b) Treat the rope and the two blocks together as a single object, with mass Take 6.00 kg 4.00 kg 5.00 kg 15.0 kg.m = + + = y+ upward, since the acceleration is upward. The free-body diagram is given in Figure 4.54b. y yF ma=? F mg ma? = F mg a m ?= 2200 N (15.0 kg)(9.80 m/s ) 3.53 m/s 15.0 kg a 2 ?= = Figure 4.54b (c) Consider the forces on the top block ( 6.00 kg),m = since the tension at the top of the rope will be one of these forces. t( )T y yF ma=? tF mg T ma? ? = t ( )T F m g a= ? + 2 2200 N (6.00 kg)(9.80 m/s 3.53 m/s ) 120 NT = ? + = Figure 4.54c Alternatively, can consider the forces on the combined object rope plus bottom block ( 9.00 kg):m = y yF ma=? tT mg ma? = 2 2 t ( ) 9.00 kg(9.80 m/s 3.53 m/s ) 120 N,T m g a= + = + = which checks Figure 4.54d (d) One way to do this is to consider the forces on the top half of the rope ( 2.00 kg).m = Let be the tension at the midpoint of the rope. mT y yF ma=? t mT T mg ma? ? = 2 2 m t ( ) 120 N 2.00 kg(9.80 m/s 3.53 m/s ) 93.3 NT T m g a= ? + = ? + = Figure 4.54e To check this answer we can alternatively consider the forces on the bottom half of the rope plus the lower block taken together as a combined object ( 2.00 kg 5.00 kg 7.00 kg):m = + = y yF ma=? mT mg ma? = 2 2 m ( ) 7.00 kg(9.80 m/s 3.53 m/s ) 93.3 N,T m g a= + = + = which checks Figure 4.54f 4-18 Chapter 4 EVALUATE: The tension in the rope is not constant but increases from the bottom of the rope to the top. The tension at the top of the rope must accelerate the rope as well the 5.00-kg block. The tension at the top of the rope is less than F; there must be a net upward force on the 6.00-kg block. 4.55. IDENTIFY: Apply m=?F a! ! to the barbell and to the athlete. Use the motion of the barbell to calculate its acceleration. SET UP: Let be upward. y+ EXECUTE: (a) The free-body diagrams for the baseball and for the athlete are sketched in Figure 4.55. (b) The athlete?s weight is . The upward acceleration of the barbell is found from 2(90.0 kg)(9.80 m /s ) 882 Nmg = = 21 0 0 2y yy y v t a t? = + . 202 22( ) 2(0.600 m) 0.469 m/s(1.6 s)y y y a t ?= = = . The force needed to lift the barbell is given by lift barbell yF w ma? = . The barbell?s mass is , so . 2(490 N) (9 80 m s ) 50 0 kg/ . / = . 2 lift barbell 490 N (50.0 kg)(0.469 m /s ) 490 N 23 N 513 NF w ma= + = + = + = The athlete is not accelerating, so floor lift athlete 0F F w? ? = . . floor lift athlete 513 N 882 N 1395 NF F w= + = + = EVALUATE: Since the athlete pushes upward on the barbell with a force greater than its weight the barbell pushes down on him and the normal force on the athlete is greater than the total weight, 1362 N, of the athlete plus barbell. Figure 4.55 4.56. IDENTIFY: Apply m=?F a! ! to the balloon and its passengers and cargo, both before and after objects are dropped overboard. SET UP: When the acceleration is downward take y+ to be downward and when the acceleration is upward take to be upward. y+ EXECUTE: (a) The free-body diagram for the descending balloon is given in Figure 4.56. L is the lift force. (b) y yF ma? = gives and ( /3)Mg L M g? = 2 /3L Mg= . (c) Now is upward, so , where m is the mass remaining. y+ ( / 2)L mg m g? = 2 /3L Mg= , so . Mass 5 must be dropped overboard. 4 /9m M= /9M EVALUATE: In part (b) the lift force is greater than the total weight and in part (c) the lift force is less than the total weight. Figure 4.56 Newton?s Laws of Motion 4-19 4.57. IDENTIFY: Apply m=?F a! ! to the entire chain and to each link. SET UP: mass of one link. Let m = y+ be upward. EXECUTE: (a) The free-body diagrams are sketched in Figure 4.57. is the force the top and middle links exert on each other. is the force the middle and bottom links exert on each other. topF middleF (b) (i) The weight of each link is . Using the free-body diagram for the whole chain: 2(0.300 kg)(9.80 m /s ) 2.94 Nmg = = 2student 3 12 N 3(2.94 N) 3.18 N 3.53 m /s 3 0.900 kg 0.900 kg F mg a m ? ?= = = = (ii) The top link also accelerates at , so . . 23.53 m /s student topF F mg ma? ? = 2 2 top student ( ) 12 N (0.300 kg)(9.80 m/s 3.53 m/s ) 8.0 NF F m g a= ? + = ? + = EVALUATE: The force exerted by the middle link on the bottom link is given by and . We can verify that with our results middleF mg m? = a Nmiddle ( ) 4.0 F m g a= + = y yF ma=? is satisfied for the middle link. Figure 4.57 4.58. IDENTIFY: Calculate from . Then a! 2 /d dt=a r! ! 2 net m=F a ! ! . SET UP: w mg= EXECUTE: Differentiating twice, the acceleration of the helicopter as a function of time is 3 2? ?(0.120 m /s ) (0.12 m /s )t ?a = i k! and at 5 0st = . , the acceleration is 2 2? ?(0.60 m /s ) (0.12 m /s )?a = i k! . The force is then 5 2 2 4 2 (2.75 10 N) ? ? ?(0.60 m /s ) (0.12 m /s ) (1.7 10 N) (3.4 10 N) (9.80 m /s ) w m g × ? ?? × ?? ? 3 ?×F = a = a = i k = i k ! ! ! EVALUATE: The force and acceleration are in the same direction. They are both time dependent. 4.59. IDENTIFY: x xF ma= and 2 2x d x a dt = . SET UP: 1( )n nd t nt dt ?= EXECUTE: The velocity as a function of time is and the acceleration as a function of time is , and so the force as a function of time is . 2( ) 3xv t A Bt= ? ( ) 6xa t Bt= ? ( ) ( ) 6xF t ma t mBt= = ? EVALUATE: Since the acceleration is along the x-axis, the force is along the x-axis. 4.60. IDENTIFY: . . /ma = F!! 0 0 t dt?v = v + a ! ! ! SET UP: since the object is initially at rest. 0 0v = EXECUTE: 4210 1 1 ? ?( ) 4 t k t dt k t m m ? ?? ?? ??v = F = i + t j !! . EVALUATE: F! has both x and y components, so v! develops x and y components. 4.61. IDENTIFY: Follow the steps specified in the problem. SET UP: The chain rule for differentiating says dv dv dv dv v dt dx dt dx = = . 4-20 Chapter 4 EXECUTE: (a) The equation of motion, 2 dvCv m dt ? = cannot be integrated with respect to time, as the unknown function is part of the integrand. The equation must be separated before integration; that is, ( )v t 2 C d dt m v ? = v and 0 1 1 , Ct m v v ? = ? + where is the constant of integration that gives 0v 0v v= at 0t = . Note that this form shows that if , there is no motion. This expression may be rewritten as 0 0v = 1 0 1 , dx Ct v dt v m ?? ?= = +? ?? ? which may be integrated to obtain 00 ln 1 . m Ctv x x C m ? ?? = +? ?? ? To obtain x as a function of v, the time t must be eliminated in favor of v; from the expression obtained after the first integration, 0 0 1 Ctv v m v = ? , so 00 ln .m vx x C v ? ?? = ? ?? ? (b) Applying the chain rule, dv dv F m mv dt dx = =? . Using the given expression for the net force, 2 dvCv v m dx ? ?? = ? ?? ? . C d dx m v ? = v . Integrating gives 0 0 ( ) ln C v x x m v ? ?? ? = ? ?? ? and 00 ln . m v x x C v ? ?? = ? ?? ? EVALUATE: If C is positive, our expression for shows it decreases from its value of . As v decreases, so does the acceleration and therefore the rate of decrease of v. ( )v t 0v 4.62. IDENTIFY: 0 t xx v dt= ? and , and similar equations apply to the y-component. 0tx xv a d= ? t SET UP: In this situation, the x-component of force depends explicitly on the y-component of position. As the y- component of force is given as an explicit function of time, yv and y can be found as functions of time and used in the expression for . ( )xa t EXECUTE: , so and , where the initial conditions have been used. Then, the expressions for 3( / )ya k m= t 23( / 2 )yv k m t= 33( / 6 )y k m t= 0 00, 0yv y= = ,x xa v and x are obtained as functions of time: 31 2 3 26x k k k a t m m = + , 41 2 3 224x k k k v t m m = + t and 2 51 2 3 22 120 k k k x t t m m = + . In vector form, 2 51 2 3 32 ? ? 2 120 6 k k k k t t t m m m ? ? ? 3+? ? ?? ? ?i + ???r = j ! and 4 21 2 3 32 ? ? 24 2 k k k k t t t m m m ? ? ? ?+? ? ? ?? ? ? ?v = i + ! j . EVALUATE: xa depends on time because it depends on y, and y is a function of time. 5-1 APPLYING NEWTON ? S LAWS 5.1. IDENTIFY: 0a = for each object. Apply y yF ma=? to each weight and to the pulley. SET UP: Take y+ upward. The pulley has negligible mass. Let rT be the tension in the rope and let cT be the tension in the chain. EXECUTE: (a) The free-body diagram for each weight is the same and is given in Figure 5.1a. y yF ma=? gives r 25.0 NT w= = . (b) The free-body diagram for the pulley is given in Figure 5.1b. c r2 50.0 NT T= = . EVALUATE: The tension is the same at all points along the rope. Figure 5.1a, b 5.2. IDENTIFY: Apply m=?F a! ! to each weight. SET UP: Two forces act on each mass: w down and ( )T w= up. EXECUTE: In all cases, each string is supporting a weight w against gravity, and the tension in each string is w. EVALUATE: The tension is the same in all three cases. 5.3. IDENTIFY: Both objects are at rest and 0a = . Apply Newton?s first law to the appropriate object. The maximum tension maxT is at the top of the chain and the minimum tension is at the bottom of the chain. SET UP: Let y+ be upward. For the maximum tension take the object to be the chain plus the ball. For the minimum tension take the object to be the ball. For the tension T three-fourths of the way up from the bottom of the chain, take the chain below this point plus the ball to be the object. The free-body diagrams in each of these three cases are sketched in Figures 5.3a, 5.3b and 5.3c. b+c 75.0 kg 26.0 kg 101.0 kgm = + = . b 75.0 kgm = . m is the mass of three-fourths of the chain: 34 (26.0 kg) 19.5 kgm = = . EXECUTE: (a) From Figure 5.3a, 0yF =? gives max b+c 0T m g? = and 2max (101.0 kg)(9.80 m/s ) 990 NT = = . From Figure 5.3b, 0yF =? gives min b 0T m g? = and 2min (75.0 kg)(9.80 m/s ) 735 NT = = . (b) From Figure 5.3c, 0yF =? gives b( ) 0T m m g? + = and 2(19.5 kg 75.0 kg)(9.80 m/s ) 926 NT = + = . 5 5-2 Chapter 5 EVALUATE: The tension in the chain increases linearly from the bottom to the top of the chain. Figure 5.3a?c 5.4. IDENTIFY: Apply Newton?s 1st law to the person. Each half of the rope exerts a force on him, directed along the rope and equal to the tension T in the rope. SET UP: (a) The force diagram for the person is given in Figure 5.4 1T and 2T are the tensions in each half of the rope. Figure 5.4 EXECUTE: 0xF =? 2 1cos cos 0T T? ?? = This says that 1 2T T T= = (The tension is the same on both sides of the person.) 0yF =? 1 2sin sin 0T T mg? ?+ ? = But 1 2 ,T T T= = so 2 sinT mg? = 2(90.0 kg)(9.80 m/s ) 2540 N 2sin 2sin10.0 mg T ?= = =° (b) The relation 2 sinT mg? = still applies but now we are given that 42.50 10 NT = × (the breaking strength) and are asked to find .? 2 4 (90.0 kg)(9.80 m/s ) sin 0.01764, 2 2(2.50 10 N) mg T ? = = =× 1.01 .? = ° EVALUATE: /(2sin )T mg ?= says that / 2T mg= when 90? = ° (rope is vertical). T ? ? when 0? ? since the upward component of the tension becomes a smaller fraction of the tension. 5.5. IDENTIFY: Apply m=?F a! ! to the frame. SET UP: Let w be the weight of the frame. Since the two wires make the same angle with the vertical, the tension is the same in each wire. 0.75T w= . EXECUTE: The vertical component of the force due to the tension in each wire must be half of the weight, and this in turn is the tension multiplied by the cosine of the angle each wire makes with the vertical. 3 cos 2 4 w w ?= and 23arccos 48? = = ° . EVALUATE: If 0? = ° , / 2T w= and T ? ? as 90? ? ° . Therefore, there must be an angle where 3 / 4T w= . Applying Newton?s Laws 5-3 5.6. IDENTIFY: Apply Newton?s 1st law to the car. The forces are the same as in Example 5.5. SET UP: The free-body diagram is sketched in Figure 5.6. EXECUTE: x xF ma=? cos sin 0T n? ?? = cos sinT n? ?= y yF ma=? cos sin 0n T w? ?+ ? = cos sinn T w? ?+ = Figure 5.6 The first equation gives cos . sin n T ? ? ? ?= ? ?? ? Use this in the second equation to eliminate n : cos cos sin sin T T w ? ? ?? ? ? + =? ?? ? Multiply this equation by sin :? 2 2(cos sin ) sinT w? ? ?+ = sinT w ?= (since 2 2cos sin 1? ?+ = ). Then cos cos sin cos . sin sin n T w w ? ?? ?? ? ? ? ? ?= = =? ? ? ?? ? ? ? EVALUATE: These results are the same as obtained in Example 5.5. The choice of coordinate axes is up to us. Some choices may make the calculation easier, but the results are the same for any choice of axes. 5.7. IDENTIFY: Apply m=?F a! ! to the car. SET UP: Use coordinates with x+ parallel to the surface of the street. EXECUTE: 0xF =? gives sinT w ?= . 2 3sin (1390 kg)(9.80 m/s )sin17.5 4.10 10 NF mg ?= = ° = × . EVALUATE: The force required is less than the weight of the car by the factor sin? . 5.8. IDENTIFY: Apply Newton?s 1st law to the wrecking ball. Each cable exerts a force on the ball, directed along the cable. SET UP: The force diagram for the wrecking ball is sketched in Figure 5.8. Figure 5.8 EXECUTE: (a) y yF ma=? cos40 0BT mg° ? = 2 4(4090 kg)(9.80 m/s ) 5.23 10 N cos40 cos40B mg T = = = ×° ° (b) x xF ma=? sin 40 0B AT T° ? = 4sin 40 3.36 10 NA BT T= ° = × EVALUATE: If the angle 40° is replaces by 0° (cable B is vertical), then BT mg= and 0.AT = 5-4 Chapter 5 5.9. IDENTIFY: Apply m=?F a! ! to the object and to the knot where the cords are joined. SET UP: Let y+ be upward and x+ be to the right. EXECUTE: (a) , sin30 sin 45 , and cos30 cos45 0.C A B C A BT w T T T w T T= ° + ° = = ° ? ° = Since sin 45 cos45 ,° = ° adding the last two equations gives (cos30 sin30 ) ,AT w° + ° = and so 0.732 .1.366A w T w= = Then, cos30 0.897 . cos45B A T T w °= =° (b) Similar to part (a), , cos60 sin 45 ,C A BT w T T w= ? ° + ° = and sin 60 cos45 0.A BT T° ? ° = Adding these two equations, 2.73 , (sin60 cos60 )A w T w= =° ? ° and sin60 3.35 . cos45B A T T w °= =° EVALUATE: In part (a), A BT T w+ > since only the vertical components of AT and BT hold the object against gravity. In part (b), since AT has a downward component BT is greater than w. 5.10. IDENTIFY: Apply Newton?s first law to the car. SET UP: Use x and y coordinates that are parallel and perpendicular to the ramp. EXECUTE: (a) The free-body diagram for the car is given in Figure 5.10. The vertical weight w and the tension T in the cable have each been replaced by their x and y components. (b) 0xF =? gives cos31.0 sin 25.0 0T w? =° ° and 2sin 25.0 sin 25.0(1130 kg)(9.80 m/s ) 5460 Ncos31.0 cos31.0T w= = =° °° ° . (c) 0yF =? gives sin31.0 cos25.0 0n T w+ ? =° ° and 2cos25.0 sin31.0 kg 9.80 m/s )cos25.0 (5460 N)sin31.0 7220 Nn w T= ? =°? °=(1130 )( ° ° EVALUATE: We could also use coordinates that are horizontal and vertical and would obtain the same values of n and T . Figure 5.10 5.11. IDENTIFY: Since the velocity is constant, apply Newton?s first law to the piano. The push applied by the man must oppose the component of gravity down the incline. SET UP: The free-body diagrams for the two cases are shown in Figures 5.11a and b. F! is the force applied by the man. Use the coordinates shown in the figure. EXECUTE: (a) 0xF =? gives sin11.0 0F w? =° and 2(180 kg)(9.80 m/s )sin11.0 337 NF = °= . (b) 0yF =? gives cos11.0 0n w? =° and cos11.0wn = ° . 0xF =? gives sin11.0 0F n? =° and sin11.0 tan11.0 N cos11.0 w F w? ?= ? ?? ? °= °=343° . Applying Newton?s Laws 5-5 EVALUATE: A slightly greater force is required when the man pushes parallel to the floor. If the slope angle of the incline were larger, sin? and tan? would differ more and there would be more difference in the force needed in each case. Figure 5.11a, b 5.12. IDENTIFY: Apply Newton?s 1st law to the hanging weight and to each knot. The tension force at each end of a string is the same. (a) Let the tensions in the three strings be T , ,T ? and ,T ?? as shown in Figure 5.12a. Figure 5.12a SET UP: The free-body diagram for the block is given in Figure 5.12b. EXECUTE: 0yF =? 0T w? ? = 60.0 NT w? = = Figure 5.12b SET UP: The free-body diagram for the lower knot is given in Figure 5.12c. EXECUTE: 0yF =? sin 45 0T T ?° ? = 60.0 N 84.9 N sin 45 sin 45 T T ?= = =° ° Figure 5.12c 5-6 Chapter 5 (b) Apply 0xF =? to the force diagram for the lower knot: 0xF =? 2 cos45 (84.9 N)cos45 60.0 NF T= ° = ° = SET UP: The free-body diagram for the upper knot is given in Figure 5.12d. EXECUTE: 0xF =? 1cos45 0T F° ? = 1 (84.9 N)cos45F = ° 1 60.0 NF = Figure 5.12d Note that 1 2.F F= EVALUATE: Applying 0yF =? to the upper knot gives sin 45 60.0 N .T T w?? = ° = = If we treat the whole system as a single object, the force diagram is given in Figure 5.12e. 0xF =? gives 2 1,F F= which checks 0yF =? gives ,T w?? = which checks Figure 5.12e 5.13. IDENTIFY: Apply Newton?s first law to the ball. The force of the wall on the ball and the force of the ball on the wall are related by Newton?s third law. SET UP: The forces on the ball are its weight, the tension in the wire, and the normal force applied by the wall. To calculate the angle ? that the wire makes with the wall, use Figure 5.13a. 16.0 cmsin 46.0 cm ? = and 20.35? = ° EXECUTE: (a) The free-body diagram is shown in Figure 5.13b. Use the x and y coordinates shown in the figure. 0yF =? gives cos 0T w? ? = and 2(45.0 kg)(9.80 m/s ) 470 Ncos cos20.35wT ?= = =° (b) 0xF =? gives sin 0T n? ? = . (470 N)sin 20.35 163 Nn = =° . By Newton?s third law, the force the ball exerts on the wall is 163 N, directed to the right. EVALUATE: sin tan cos w n w? ?? ? ?= =? ?? ? . As the angle ? decreases (by increasing the length of the wire), T decreases and n decreases. Figure 5.13a, b 5.14. IDENTIFY: Apply m=?F a! ! to each block. 0a = . SET UP: Take y+ perpendicular to the incline and x+ parallel to the incline. Applying Newton?s Laws 5-7 EXECUTE: The free-body diagrams for each block, A and B, are given in Figure 5.14. (a) For B, x xF ma=? gives 1 sin 0T w ?? = and 1 sinT w ?= . (b) For block A, x xF ma=? gives 1 2 sin 0T T w ?? ? = and 2 2 sinT w ?= . (c) y yF ma=? for each block gives cosA Bn n w ?= = . (d) For 0? ? , 1 2 0T T= ? and A Bn n w= ? . For 90? ? ° , 1T w= , 2 2T w= and 0A Bn n= = . EVALUATE: The two tensions are different but the two normal forces are the same. Figure 5.14a, b 5.15. IDENTIFY: Apply Newton?s first law to the ball. Treat the ball as a particle. SET UP: The forces on the ball are gravity, the tension in the wire and the normal force exerted by the surface. The normal force is perpendicular to the surface of the ramp. Use x and y axes that are horizontal and vertical. EXECUTE: (a) The free-body diagram for the ball is given in Figure 5.15. The normal force has been replaced by its x and y components. (b) 0yF =? gives cos35.0 0n w? =° and 1.22cos35.0mgn mg= =° . (c) 0xF =? gives sin35.0 0T n? =° and (1.22 )sin35.0 0.700T mg mg= =° . EVALUATE: Note that the normal force is greater than the weight, and increases without limit as the angle of the ramp increases towards 90° . The tension in the wire is tanw ? , where ? is the angle of the ramp and T also increases without limit as 90? ? ° . Figure 5.15 5.16. IDENTIFY: Apply Newton?s second law to the rocket plus its contents and to the power supply. Both the rocket and the power supply have the same acceleration. SET UP: The free-body diagrams for the rocket and for the power supply are given in Figures 5.16a and b. Since the highest altitude of the rocket is 120 m, it is near to the surface of the earth and there is a downward gravity force on each object. Let y+ be upward, since that is the direction of the acceleration. The power supply has mass 2ps (15.5 N) /(9.80 m/s ) 1.58 kgm = = 5-8 Chapter 5 EXECUTE: (a) y yF ma=? applied to the rocket gives r rF m g m a? = . 2 2r r 1720 N (125 kg)(9.80 m/s ) 3.96 m/s 125 kg F m g a m ? ?= = = . (b) y yF ma=? applied to the power supply gives ps psn m g m a? = . 2 2 ps ( ) (1.58 kg)(9.80 m/s 3.96 m/s ) 21.7 Nn m g a= + = + = . EVALUATE: The acceleration is constant while the thrust is constant and the normal force is constant while the acceleration is constant. The altitude of 120 m is not used in the calculation. Figure 5.16a, b 5.17. IDENTIFY: Use the kinematic information to find the acceleration of the capsule and the stopping time. Use Newton?s second law to find the force F that the ground exerted on the capsule during the crash. SET UP: Let y+ be upward. 311 km/h 86.4 m/s= . The free-body diagram for the capsule is given in Figure 15.17. EXECUTE: 0 0.810 my y? = ? , 0 86.4 m/syv = ? , 0yv = . 2 20 02 ( )y y yv v a y y= + ? gives 2 2 2 0 2 0 0 ( 86.4 m/s) 4610 m/s 470 2( ) 2( 0.810) m y y y v v a g y y ? ? ?= = = =? ? . (b) y yF ma=? applied to the capsule gives F mg ma? = and 2 2 5( ) (210 kg)(9.80 m/s 4610 m/s ) 9.70 10 N 471 .F m g a w= + = + = × = (c) 00 2 y yv vy y t +? ?? = ? ?? ? gives 0 2 0 2( ) 2( 0.810 m) 0.0187 s 86.4 m/s 0y y y y t v v ? ?= = =+ ? + EVALUATE: The upward force exerted by the ground is much larger than the weight of the capsule and stops the capsule in a short amount of time. After the capsule has come to rest, the ground still exerts a force mg on the capsule, but the large 59.00 10 N× force is exerted only for 0.0187 s. Figure 5.17 5.18. IDENTIFY: Apply Newton?s second law to the three sleds taken together as a composite object and to each individual sled. All three sleds have the same horizontal acceleration a. SET UP: The free-body diagram for the three sleds taken as a composite object is given in Figure 5.18a and for each individual sled in Figure 5.18b-d. Let x+ be to the right, in the direction of the acceleration. tot 60.0 kgm = . EXECUTE: (a) x xF ma=? for the three sleds as a composite object gives totP m a= and 2 tot 125 N 2.08 m/s 60.0 kg P a m = = = . Applying Newton?s Laws 5-9 (b) x xF ma=? applied to the 10.0 kg sled gives 10AP T m a? = and 2 10 125 N (10.0 kg)(2.08 m/s ) 104 NAT P m a= ? = ? = . x xF ma=? applied to the 30.0 kg sled gives 2 30 (30.0 kg)(2.08 m/s ) 62.4 NBT m a= = = . EVALUATE: If we apply x xF ma=? to the 20.0 kg sled and calculate a from AT and BT found in part (b), we get 20A BT T m a? = . 2 20 104 N 62.4 N 2.08 m/s 20.0 kg A BT Ta m ? ?= = = , which agrees with the value we calculated in part (a). Figure 5.18a?d 5.19. IDENTIFY: Apply m=?F a! ! to the load of bricks and to the counterweight. The tension is the same at each end of the rope. The rope pulls up with the same force ( )T on the bricks and on the counterweight. The counterweight accelerates downward and the bricks accelerate upward; these accelerations have the same magnitude. (a) SET UP: The free-body diagrams for the bricks and counterweight are given in Figure 5.19. Figure 5.19 (b) EXECUTE: Apply y yF ma=? to each object. The acceleration magnitude is the same for the two objects. For the bricks take y+ to be upward since a! for the bricks is upward. For the counterweight take y+ to be downward since a ! is downward. bricks: y yF ma=? 1 1T m g m a? = counterweight: y yF ma=? 2 2m g T m a? = Add these two equations to eliminate T : 2 1 1 2( ) ( )m m g m m a? = + 2 22 1 1 2 28.0 kg 15.0 kg (9.80 m/s ) 2.96 m/s 15.0 kg 28.0 kg m m a g m m ? ? ? ?? ?= = =? ? ? ?+ +? ?? ? (c) 1 1T m g m a? = gives 2 21( ) (15.0 kg)(2.96 m/s 9.80 m/s ) 191 NT m a g= + = + = As a check, calculate T using the other equation. 2 2m g T m a? = gives 2 22 ( ) 28.0 kg(9.80 m/s 2.96 m/s ) 191 N,T m g a= ? = ? = which checks. 5-10 Chapter 5 EVALUATE: The tension is 1.30 times the weight of the bricks; this causes the bricks to accelerate upward. The tension is 0.696 times the weight of the counterweight; this causes the counterweight to accelerate downward. If 1 2 ,m m= 0a = and 1 2 .T m g m g= = In this special case the objects don?t move. If 1 0,m = a g= and 0;T = in this special case the counterweight is in free-fall. Our general result is correct in these two special cases. 5.20. IDENTIFY: In part (a) use the kinematic information and the constant acceleration equations to calculate the acceleration of the ice. Then apply m?F = a! ! . In part (b) use m?F = a! ! to find the acceleration and use this in the constant acceleration equations to find the final speed. SET UP: Figures 5.20a and b give the free-body diagrams for the ice both with and without friction. Let x+ be directed down the ramp, so y+ is perpendicular to the ramp surface. Let ? be the angle between the ramp and the horizontal. The gravity force has been replaced by its x and y components. EXECUTE: (a) 0 1.50 mx x? = , 0 0xv = , 2.50 m/sxv = . 2 20 02 ( )x x xv v a x x= + ? gives 2 2 2 20 0 (2.50 m/s) 0 2.08 m/s 2( ) 2(1.50 m) x x x v v a x x ? ?= = =? . x xF ma=? gives sinmg ma? = and 2 2 2.08 m/s sin 9.80 m/s a g ? = = . 12.3? = ° . (b) x xF ma=? gives sinmg f ma? ? = and 2 2sin (8.00 kg)(9.80 m/s )sin12.3 10.0 N 0.838 m/s 8.00 kg mg f a m ? ? ?= = =° . Then 0 1.50 mx x? = , 0 0xv = , 20.838 m/sxa = and 2 20 02 ( )x x xv v a x x= + ? gives 2 02 ( ) 2(0.838 m/s )(1.50 m) 1.59 m/sx xv a x x= ? = = EVALUATE: With friction present the speed at the bottom of the ramp is less. Figure 5.20a, b 5.21. IDENTIFY: Apply m?F = a! ! to each block. Each block has the same magnitude of acceleration a. SET UP: Assume the pulley is to the right of the 4.00 kg block. There is no friction force on the 4.00 kg block, the only force on it is the tension in the rope. The 4.00 kg block therefore accelerates to the right and the suspended block accelerates downward. Let x+ be to the right for the 4.00 kg block, so for it xa a= , and let y+ be downward for the suspended block, so for it ya a= . EXECUTE: (a) The free-body diagrams for each block are given in Figures 5.21a and b. (b) x xF ma=? applied to the 4.00 kg block gives (4.00 kg)T a= and 210.0 N 2.50 m/s4.00 kg 4.00 kgTa = = = . (c) y yF ma=? applied to the suspended block gives mg T ma? = and 2 2 10.0 N 1.37 kg 9.80 m/s 2.50 m/s T m g a = = =? ? . (d) The weight of the hanging block is 2(1.37 kg)(9.80 m/s ) 13.4 Nmg = = . This is greater than the tension in the rope; 0.75T mg= . Applying Newton?s Laws 5-11 EVALUATE: Since the hanging block accelerates downward, the net force on this block must be downward and the weight of the hanging block must be greater than the tension in the rope. Note that the blocks accelerate no matter how small m is. It is not necessary to have 4.00 kgm > , and in fact in this problem m is less than 4.00 kg. Figure 5.21a, b 5.22. IDENTIFY: (a) Consider both gliders together as a single object, apply m=?F a! ! , and solve for a. Use a in a constant acceleration equation to find the required runway length. (b) Apply m=?F a! ! to the second glider and solve for the tension gT in the towrope that connects the two gliders. SET UP: In part (a), set the tension tT in the towrope between the plane and the first glider equal to its maximum value, t 12,000 NT = . EXECUTE: (a) The free-body diagram for both gliders as a single object of mass 2 1400 kgm = is given in Figure 5.22a. x xF ma=? gives t 2 (2 )T f m a? = and 2t 2 12,000 N 5000 N 5.00 m/s2 1400 kgT fa m? ?= = = . Then 25.00 m/sxa = , 0 0xv = and 40 m/sxv = in 2 20 02 ( )x x xv v a x x= + ? gives 2 2 0 0( ) 160 m2 x x x v v x x a ?? = = . (b) The free-body diagram for the second glider is given in Figure 5.22b. x xF ma=? gives gT f ma? = and 22500 N + (700 kg)(5.00 m/s ) 6000 NT f ma= + = = . EVALUATE: We can verify that x xF ma=? is also satisfied for the first glider. Figure 5.22a, b 5.23. IDENTIFY: The maximum tension in the chain is at the top of the chain. Apply m?F = a! ! to the composite object of chain and boulder. Use the constant acceleration kinematic equations to relate the acceleration to the time. SET UP: Let y+ be upward. The free-body diagram for the composite object is given in Figure 5.23. chain2.50T w= . tot chain boulder 1325 kgm m m= + = . EXECUTE: (a) y yF ma=? gives tot totT m g m a? = . tot chain tot chain tot tot tot 2.50 2.50 1 T m g m g m g m a g m m m ? ?? ?= = = ?? ?? ? 2 22.50[575 kg] 1 (9.80 m/s ) 0.832 m/s 1325 kg a ? ?= ? =? ?? ? . 5-12 Chapter 5 (b) Assume the acceleration has its maximum value: 20.832 m/sya = , 0 125 my y? = and 0 0yv = . 21 0 0 2y yy y v t a t? = + gives 0 22( ) 2(125 m) 17.3 s0.832 m/sy y y t a ?= = = EVALUATE: The tension in the chain is 41.41 10 NT = × and the total weight is 41.30 10 N× . The upward force exceeds the downward force and the acceleration is upward. Figure 5.23 5.24. IDENTIFY: Apply m?F = a! ! to the composite object of elevator plus student ( tot 850 kgm = ) and also to the student ( 550 Nw = ). The elevator and the student have the same acceleration. SET UP: Let y+ be upward. The free-body diagrams for the composite object and for the student are given in Figure 5.24a and b. T is the tension in the cable and n is the scale reading, the normal force the scale exerts on the student. The mass of the student is / 56.1 kgm w g= = . EXECUTE: (a) y yF ma=? applied to the student gives yn mg ma? = . 2450 N 550 N 1.78 m/s 56.1 kgy n mg a m ? ?= = = ? . The elevator has a downward acceleration of 21.78 m/s . (b) 2 670 N 550 N 2.14 m/s 56.1 kgy a ?= = . (c) 0n = means ya g= ? . The student should worry; the elevator is in free-fall. (d) y yF ma=? applied to the composite object gives tot totT m g m a? = . tot ( )yT m a g= + . In part (a), 2 2(850 kg)( 1.78 m/s 9.80 m/s ) 6820 NT = ? + = . In part (c), ya g= ? and 0T = . EVALUATE: In part (b), 2 2(850 kg)(2.14 m/s 9.80 m/s ) 10,150 NT = + = . The weight of the composite object is 8330 N. When the acceleration is upward the tension is greater than the weight and when the acceleration is downward the tension is less than the weight. Figure 5.24a, b 5.25. IDENTIFY: Apply m=?F a! ! to the puck. Use the information about the motion to calculate the acceleration. The table must slope downward to the right. SET UP: Let ? be the angle between the table surface and the horizontal. Let the x+ -axis be to the right and parallel to the surface of the table. EXECUTE: x xF ma=? gives sin xmg ma? = . The time of travel for the puck is 0/L v , where 1.75 mL = and 0 3.80 m/sv = . 210 0 2x xx x v t a t? = + gives 2 0 2 2 2 2 x x xv a t L = = , where 0.0250 mx = . 2 0 2 2 sin x a xv g gL ? = = . ( ) 2 2 2 2 2(2.50 10 m)(3.80 m /s) arcsin 1.38 9.80 m /s (1.75 m) ? ?? ?×? ?= = °? ?? ? . Applying Newton?s Laws 5-13 EVALUATE: The table is level in the direction along its length, since the velocity in that direction is constant. The angle of slope to the right is small, so the acceleration and deflection in that direction are small. 5.26. IDENTIFY: Acceleration and velocity are related by yy dva dt= . Apply m?F = a ! ! to the rocket. SET UP: Let y+ be upward. The free-body diagram for the rocket is sketched in Figure 5.26. F! is the thrust force. EXECUTE: (a) 2yv At Bt= + . 2ya A Bt= + . At 0t = , 21.50 m/sya = so 21.50 m/sA = . Then 2.00 m/syv = at 1.00 st = gives 2 22.00 m/s (1.50 m/s )(1.00 s) (1.00 s)B= + and 30.50 m/sB = . (b) At 4.00 st = , 2 3 21.50 m/s 2(0.50 m/s )(4.00 s) 5.50 m/sya = + = . (c) y yF ma=? applied to the rocket gives T mg ma? = and 2 2 4( ) (2540 kg)(9.80 m/s 5.50 m/s ) 3.89 10 NT m a g= + = + = × . 1.56T w= . (d) When 21.50 m/sa = , 2 2 4(2540 kg)(9.80 m/s 1.50 m/s ) 2.87 10 NT = + = × EVALUATE: During the time interval when 2( )v t At Bt= + applies the magnitude of the acceleration is increasing, and the thrust is increasing. Figure 5.26 5.27. IDENTIFY: Consider the forces in each case. There is the force of gravity and the forces from objects that touch the object in question. SET UP: A surface exerts a normal force perpendicular to the surface, and a friction force, parallel to the surface. EXECUTE: The free-body diagrams are sketched in Figure 5.27a-c. EVALUATE: Friction opposes relative motion between the two surfaces. When one surface is stationary the friction force on the other surface is directed opposite to its motion. Figure 5.27a?c 5.28. IDENTIFY: s sf n?? and k kf n?= . The normal force n is determined by applying m?F = a! ! to the block. Normally, k s? ?? . sf is only as large as it needs to be to prevent relative motion between the two surfaces. SET UP: Since the table is horizontal, with only the block present 135 Nn = . With the brick on the block, 270 Nn = . EXECUTE: (a) The friction is static for 0P = to 75.0 NP = . The friction is kinetic for 75.0 NP > . (b) The maximum value of sf is s n? . From the graph the maximum sf is s 75.0 Nf = , so s s max 75.0 N 0.556 135 N f n ? = = = . k kf n?= . From the graph, k 50.0 Nf = and kk 50.0 N 0.370135 N f n ? = = = . (c) When the block is moving the friction is kinetic and has the constant value k kf n?= , independent of P . This is why the graph is horizontal for 75.0 NP > . When the block is at rest, sf P= since this prevents relative motion. This is why the graph for 75.0 NP < has slope 1.+ (d) smax f and kf would double. The values of f on the vertical axis would double but the shape of the graph would be unchanged. EVALUATE: The coefficients of friction are independent of the normal force. 5-14 Chapter 5 5.29. (a) IDENTIFY: Constant speed implies 0.a = Apply Newton?s 1st law to the box. The friction force is directed opposite to the motion of the box. SET UP: Consider the free-body diagram for the box, given in Figure 5.29a. Let F! be the horizontal force applied by the worker. The friction is kinetic friction since the box is sliding along the surface. EXECUTE: y yF ma=? 0n mg? = n mg= So k k kf n mg? ?= = Figure 5.29a x xF ma=? k 0F f? = 2 k k (0.20)(11.2 kg)(9.80 m/s ) 22 NF f mg?= = = = (b) IDENTIFY: Now the only horizontal force on the box is the kinetic friction force. Apply Newton?s 2nd law to the box to calculate its acceleration. Once we have the acceleration, we can find the distance using a constant acceleration equation. The friction force is k k ,f mg?= just as in part (a). SET UP: The free-body diagram is sketched in Figure 5.29b. EXECUTE: x xF ma=? k xf ma? = k xmg ma?? = 2 2 k (0.20)(9.80 m/s ) 1.96 m/sxa g?= ? = ? = ? Figure 5.29b Use the constant acceleration equations to find the distance the box travels: 0,xv = 0 3.50 m/s,xv = 21.96 m/s ,xa = ? 0 ?x x? = 2 2 0 02 ( )x x xv v a x x= + ? 2 2 2 0 0 2 0 (3.50 m/s) 3.1 m 2 2( 1.96 m/s ) x x x v v x x a ? ?? = = =? EVALUATE: The normal force is the component of force exerted by a surface perpendicular to the surface. Its magnitude is determined by .m=?F a! ! In this case n and mg are the only vertical forces and 0,ya = so .n mg= Also note that kf and n are proportional in magnitude but perpendicular in direction. 5.30. IDENTIFY: Apply m=?F a! ! to the box. SET UP: Since the only vertical forces are n and w, the normal force on the box equals its weight. Static friction is as large as it needs to be to prevent relative motion between the box and the surface, up to its maximum possible value of maxs sf n?= . If the box is sliding then the friction force is k kf n?= . EXECUTE: (a) If there is no applied force, no friction force is needed to keep the box at rest. (b) maxs s (0.40)(40.0 N) 16.0 Nf n?= = = . If a horizontal force of 6.0 N is applied to the box, then s 6.0 Nf = in the opposite direction. (c) The monkey must apply a force equal to maxsf , 16.0 N. (d) Once the box has started moving, a force equal to k k 8.0 Nf n?= = is required to keep it moving at constant velocity. EVALUATE: k s? ?< and less force must be applied to the box to maintain its motion than to start it moving. Applying Newton?s Laws 5-15 5.31. IDENTIFY: Apply m?F = a! ! to the crate. s sf n?? and k kf n?= . SET UP: Let y+ be upward and let x+ be in the direction of the push. Since the floor is horizontal and the push is horizontal, the normal force equals the weight of the crate: 441 Nn mg= = . The force it takes to start the crate moving equals smax f and the force required to keep it moving equals kf EXECUTE: smax 313 Nf = , so s 313 N 0.710441 N? = = . k 208 Nf = , so k 208 N 0.472 441 N ? = = . (b) The friction is kinetic. x xF ma=? gives kF f ma? = and 2k 208 (45.0 kg)(1.10 m/s ) 258 NF f ma= + = + = . (c) (i) The normal force now is 72.9 Nmg = . To cause it to move, s smax (0.710)(72.9 N) 51.8 NF f n?= = = = . (ii) kF f ma= + and 2k 258 N (0.472)(72.9 N) 4.97 m/s45.0 kg F f a m ? ?= = = EVALUATE: The kinetic friction force is independent of the speed of the object. On the moon, the mass of the crate is the same as on earth, but the weight and normal force are less. 5.32. IDENTIFY: Apply m=?F a! ! to the box and calculate the normal and friction forces. The coefficient of kinetic friction is the ratio k f n . SET UP: Let x+ be in the direction of motion. 20.90 m/sxa = ? . The box has mass 8.67 kg. EXECUTE: The normal force has magnitude 85 N 25 N 110 N.+ = The friction force, from H kF f ma? = is 2 k H 20 N (8.67 kg)( 0.90 m/s ) 28 Nf F ma= ? = ? ? = . k 28 N 0.25.110 N? = = EVALUATE: The normal force is greater than the weight of the box, because of the downward component of the push force. 5.33. IDENTIFY: Apply m?F = a! ! to the composite object consisting of the two boxes and to the top box. The friction the ramp exerts on the lower box is kinetic friction. The upper box doesn?t slip relative to the lower box, so the friction between the two boxes is static. Since the speed is constant the acceleration is zero. SET UP: Let x+ be up the incline. The free-body diagrams for the composite object and for the upper box are given in Figures 5.33a and b. The slope angle ? of the ramp is given by 2.50 mtan 4.75 m ? = , so 27.76? = ° . Since the boxes move down the ramp, the kinetic friction force exerted on the lower box by the ramp is directed up the incline. To prevent slipping relative to the lower box the static friction force on the upper box is directed up the incline. tot 32.0 kg 48.0 kg 80.0 kgm = + = . EXECUTE: (a) y yF ma=? applied to the composite object gives tot tot cosn m g ?= and k k tot cosf m g? ?= . x xF ma=? gives k tot sin 0f T m g ?+ ? = and 2 k tot(sin cos ) (sin 27.76 [0.444]cos27.76 )(80.0 kg)(9.80 m/s ) 57.1 NT m g? ? ?= ? = ? =° ° . The person must apply a force of 57.1 N, directed up the ramp. (b) x xF ma=? applied to the upper box gives 2s sin (32.0 kg)(9.80 m/s )sin 27.76 146 Nf mg ?= = =° , directed up the ramp. EVALUATE: For each object the net force is zero. Figure 5.33a, b 5-16 Chapter 5 5.34. IDENTIFY: Use m=?F a! ! to find the acceleration that can be given to the car by the kinetic friction force. Then use a constant acceleration equation. SET UP: Take x+ in the direction the car is moving. EXECUTE: (a) The free-body diagram for the car is shown in Figure 5.34. y yF ma=? gives n mg= . x xF ma=? gives k xn ma?? = . k xmg ma?? = and kxa g?= ? . Then 0xv = and 2 20 02 ( )x x xv v a x x= + ? gives 2 2 2 0 0 0 2 k (29.1 m/s) ( ) 54.0 m 2 2 2(0.80)(9.80 m/s ) x x x v v x x a g?? = ? = + = = . (b) 20 k 02 ( ) 2(0.25)(9.80 m/s )(54.0 m) 16.3 m/sxv g x x?= ? = = EVALUATE: For constant stopping distance 2 0 k xv ? is constant and 0xv is proportional to k? . The answer to part (b) can be calculated as (29.1 m/s) 0.25/ 0.80 16.3 m/s= . Figure 5.34 5.35. IDENTIFY: For a given initial speed, the distance traveled is inversely proportional to the coefficient of kinetic friction. SET UP: From Table 5.1 the coefficient of kinetic friction is 0.04 for Teflon on steel and 0.44 for brass on steel. EXECUTE: The ratio of the distances is 0.44 11 0.04 = . EVALUATE: The smaller the coefficient of kinetic friction the smaller the retarding force of friction, and the greater the stopping distance. 5.36. IDENTIFY: Constant speed means zero acceleration for each block. If the block is moving the friction force the tabletop exerts on it is kinetic friction. Apply m?F = a! ! to each block. SET UP: The free-body diagrams and choice of coordinates for each block are given by Figure 5.36. 4.59 kgAm = and 2.55 kgBm = . EXECUTE: (a) y yF ma=? with 0ya = applied to block B gives 0Bm g T? = and 25.0 NT = . x xF ma=? with 0xa = applied to block A gives k 0T f? = and k 25.0 Nf = . 45.0 NA An m g= = and kk 25.0 N 0.55645.0 NA f n ? = = = . (b) Now let A be block A plus the cat, so 9.18 kgAm = . 90.0 NAn = and k k (0.556)(90.0 N) 50.0 Nf n?= = = . x xF ma=? for A gives k A xT f m a? = . y yF ma=? for block B gives B B ym g T m a? = . xa for A equals ya for B, so adding the two equations gives k ( )B A B ym g f m m a? = + and 2k 25.0 N 50.0 N 2.13 m/s9.18 kg 2.55 kg B y A B m g f a m m ? ?= = = ?+ + . The acceleration is upward and block B slows down. Applying Newton?s Laws 5-17 EVALUATE: The equation k ( )B A B ym g f m m a? = + has a simple interpretation. If both blocks are considered together then there are two external forces: Bm g that acts to move the system one way and kf that acts oppositely. The net force of kBm g f? must accelerate a total mass of A Bm m+ . Figure 5.36 5.37. IDENTIFY: Apply m=?F a! ! to each crate. The rope exerts force T to the right on crate A and force T to the left on crate B. The target variables are the forces T and F. Constant v implies 0.a = SET UP: The free-body diagram for A is sketched in Figure 5.37a EXECUTE: y yF ma=? 0A An m g? = A An m g= k k kA A Af n m g? ?= = Figure 5.37a x xF ma=? k 0AT f? = k AT m g?= SET UP: The free-body diagram for B is sketched in Figure 5.37b. EXECUTE: y yF ma=? 0B Bn m g? = B Bn m g= k k kB B Bf n m g? ?= = Figure 5.37b x xF ma=? k 0BF T f? ? = k BF T m g?= + Use the first equation to replace T in the second: k k .A BF m g m g? ?= + (a) k ( )A BF m m g?= + (b) k AT m g?= EVALUATE: We can also consider both crates together as a single object of mass ( ).A Bm m+ x xF ma=? for this combined object gives k k ( ) ,A BF f m m g?= = + in agreement with our answer in part (a). 5-18 Chapter 5 5.38. IDENTIFY: rf n?= . Apply m=?F a! ! to the tire. SET UP: n mg= and f ma= . EXECUTE: 2 2 0 x v v a L ?= , where L is the distance covered before the wheel?s speed is reduced to half its original speed and 0 / 2v v= . 2 22 2 21 0 00 04 r 3 2 2 8 v va v v v g Lg Lg Lg ? ??= = = = . Low pressure, 18.1 mL = and 2 2 3 (3.50 m /s) 0.0259 8 (18.1 m)(9.80 m /s ) = . High pressure, 92.9 mL = and 2 2 3 (3.50 m /s) 0.00505 8 (3.50 m /s) = . EVALUATE: r? is inversely proportional to the distance L, so r1 2 r2 1 L L ? ? = . 5.39. IDENTIFY: Apply m=?F a! ! to the box. Use the information about sliding to calculate the mass of the box. SET UP: k kf n?= , r rf n?= and n mg= . EXECUTE: Without the dolly: n mg= and k 0F n?? = ( 0xa = since speed is constant). 2 k 160 N 34.74 kg (0.47) (9.80 m s ) F m g?= = = With the dolly: the total mass is 34.7 kg 5.3 kg 40.04 kg+ = and friction now is rolling friction, r r .f mg?= rF mg ma?? = . 2r 3.82 m sF mga m ??= = . EVALUATE: k k 160 Nf mg?= = and r r 4.36 Nf mg?= = , or, r r k k f f ? ?= . The rolling friction force is much less than the kinetic friction force. 5.40. IDENTIFY: Apply m=?F a! ! to the truck. For constant speed, 0a = and horiz rF f= . SET UP: r r rf n mg? ?= = . Let 2 11.42m m= and r2 r10.81? ?= . EXECUTE: Since the speed is constant and we are neglecting air resistance, we can ignore the 2.4 m/s, and netF in the horizontal direction must be zero. Therefore r r horiz 200 Nf n F?= = = before the weight and pressure changes are made. After the changes, r horiz(0.81 ) (1.42 ) ,n F? = because the speed is still constant and net 0F = . We can simply divide the two equations: r horiz r (0.81 )(1.42 ) 200 N n F ? n ? = and horiz(0.81) (1.42) (200 N) 230 NF= = . EVALUATE: The increase in weight increases the normal force and hence the friction force, whereas the decrease in r? reduces it. The percentage increase in the weight is larger, so the net effect is an increase in the friction force. 5.41. IDENTIFY: Apply m=?F a! ! to each block. The target variables are the tension T in the cord and the acceleration a of the blocks. Then a can be used in a constant acceleration equation to find the speed of each block. The magnitude of the acceleration is the same for both blocks. SET UP: The system is sketched in Figure 5.41a. For each block take a positive coordinate direction to be the direction of the block?s acceleration. Figure 5.41a Applying Newton?s Laws 5-19 block on the table: The free-body is sketched in Figure 5.41b. EXECUTE: y yF ma=? 0An m g? = An m g= k k k Af n m g? ?= = Figure 5.41b x xF ma=? k AT f m a? = k A AT m g m a?? = SET UP: hanging block: The free-body is sketched in Figure 5.41c. EXECUTE: y yF ma=? B Bm g T m a? = B BT m g m a= ? Figure 5.41c (a) Use the second equation in the first kB B A Am g m a m g m a?? ? = k( ) ( )A B B Am m a m m g?+ = ? 2 2k( ) (1.30 kg (0.45)(2.25 kg))(9.80 m/s ) 0.7937 m/s 2.25 kg 1.30 kg B A A B m m g a m m ?? ?= = =+ + SET UP: Now use the constant acceleration equations to find the final speed. Note that the blocks have the same speeds. 0 0.0300 m,x x? = 20.7937 m/s ,xa = 0 0,xv = ?xv = 2 2 0 02 ( )x x xv v a x x= + ? EXECUTE: 202 ( ) 2(0.7937 m/s )(0.0300 m) 0.218 m/s 21.8 cm/s.x xv a x x= ? = = = (b) 2 2( ) 1.30 kg(9.80 m/s 0.7937 m/s ) 11.7 NB B BT m g m a m g a= ? = ? = ? = Or, to check, k A AT m g m a?? = 2 2 k( ) 2.25 kg(0.7937 m/s (0.45)(9.80 m/s )) 11.7 N,AT m a g?= + = + = which checks. EVALUATE: The force T exerted by the cord has the same value for each block. BT m g< since the hanging block accelerates downward. Also, k k 9.92 N.Af m g?= = kT f> and the block on the table accelerates in the direction of T . 5.42. IDENTIFY: Apply m?F = a! ! to the box. When the box is ready to slip the static friction force has its maximum possible value, s sf n?= . SET UP: Use coordinates parallel and perpendicular to the ramp. EXECUTE: (a) The normal force will be cos w ? and the component of the gravitational force along the ramp is sin w ? . The box begins to slip when ssin cos ,w ? w ??> or stan 0.35,? ?> = so slipping occurs at arctan(0.35) 19.3? = = ° . (b) When moving, the friction force along the ramp is k cosw ?? , the component of the gravitational force along the ramp is sinw ? , so the acceleration is 2 k k( sin cos ) (sin cos ) 0.92 m s .w ? w ? m g ? ?? ?? = ? = (c) Since 0 0xv = , 22ax v= , so 1 2(2 )v ax= , or 2 1 2[(2)(0.92m s )(5 m)] 3 m/sv = = . EVALUATE: When the box starts to move, friction changes from static to kinetic and the friction force becomes smaller. 5-20 Chapter 5 5.43. (a) IDENTIFY: Apply m=?F a! ! to the crate. Constant v implies 0.a = Crate moving says that the friction is kinetic friction. The target variable is the magnitude of the force applied by the woman. SET UP: The free-body diagram for the crate is sketched in Figure 5.43. EXECUTE: y yF ma=? sin 0n mg F ?? ? = sinn mg F ?= + k k k k sinf n mg F? ? ? ?= = + Figure 5.43 x xF ma=? kcos 0F f? ? = k kcos sin 0F mg F? ? ? ?? ? = k k(cos sin )F mg? ? ? ?? = k kcos sin mg F ? ? ? ?= ? (b) IDENTIFY and SET UP: ?start the crate moving? means the same force diagram as in part (a), except that k? is replaced by s.? Thus s s . cos sin mg F ? ? ? ?= ? EXECUTE: F ? ? if scos sin 0.? ? ?? = This gives s cos 1 .sin tan ?? ? ?= = EVALUATE: F! has a downward component so .n mg> If 0? = (woman pushes horizontally), n mg= and k k .F f mg?= = 5.44. IDENTIFY: Apply m?F = a! ! to the box. SET UP: Let y+ be upward and x+ be horizontal, in the direction of the acceleration. Constant speed means 0a = . EXECUTE: (a) There is no net force in the vertical direction, so sin 0,n F w?+ ? = or sin sin .n w F ? mg F ?= ? = ? The friction force is k k k ( sin ).f n mg F ?? ?= = ? The net horizontal force is k kcos cos ( sin )F ? f F ? mg F ??? = ? ? , and so at constant speed, k kcos sin mg F ? ? ? ?= + (b) Using the given values, 2(0.35)(90 kg)(9.80m s ) 290 N (cos25 (0.35)sin 25 ) F = =° + ° . EVALUATE: If 0? = ° , kF mg?= . 5.45. IDENTIFY: Apply m?F = a! ! to each block. SET UP: For block B use coordinates parallel and perpendicular to the incline. Since they are connected by ropes, blocks A and B also move with constant speed. EXECUTE: (a) The free-body diagrams are sketched in Figure 5.45. (b) The blocks move with constant speed, so there is no net force on block A; the tension in the rope connecting A and B must be equal to the frictional force on block A, k (0.35) (25.0 N) 9 N.? = = (c) The weight of block C will be the tension in the rope connecting B and C; this is found by considering the forces on block B. The components of force along the ramp are the tension in the first rope (9 N, from part (a)), the component of the weight along the ramp, the friction on block B and the tension in the second rope. Thus, the weight of block C is k9 N (sin36.9 cos36.9 ) 9 N (25.0 N)(sin 36.9 (0.35)cos 36.9 ) 31.0 NC Bw w ?= + ° + ° = + ° + ° = The intermediate calculation of the first tension may be avoided to obtain the answer in terms of the common weight w of blocks A and B, k k( (sin cos )),Cw w ? ? ??= + + giving the same result. (d) Applying Newton?s Second Law to the remaining masses (B and C) gives: ( ) 2k( cos sin ) 1.54m s .C B B B Ca g w w ? w w w? ?= ? ? + = Applying Newton?s Laws 5-21 EVALUATE: Before the rope between A and B is cut the net external force on the system is zero. When the rope is cut the friction force on A is removed from the system and there is a net force on the system of blocks B and C. Figure 5.45 5.46. IDENTIFY and SET UP: The derivative of yv gives ya as a function of time, and the integral of yv gives y as a function of time. EXECUTE: Differentiating Eq. (5.10) with respect to time gives the acceleration ( ) ( ) t , k m t k m tka v e ge m ? ?? ?= =? ?? ? where Eq. (5.9), tv mg k= , has been used. Integrating Eq. (5.10) with respect to time with 0 0y = gives ( )( ) ( ) ( )t t t t0 [1 ] 1t k m t k m t k m tm m my v e dt v t e v v t ek k k? ? ?? ?? ? ? ? ? ?= ? = + ? = ? ?? ? ? ?? ? ? ?? ? ? ? ? ?? ?? . EVALUATE: We can verify that / ydy dt v= . 5.47. IDENTIFY and SET UP: Apply Eq.(5.13). EXECUTE: (a) Solving for D in terms of tv , 2 2 2 t (80 kg) (9.80 m s ) 0.44 kg m. (42 m s) mg D v = = = (b) 2 t (45 kg)(9.80 m s ) 42 m s. (0.25 kg m) mg v D = = = EVALUATE: tv is less for the daughter since her mass is less. 5.48. IDENTIFY: Apply m?F = a! ! to the ball. At the terminal speed, f mg= . SET UP: The fluid resistance is directed opposite to the velocity of the object. At half the terminal speed, the magnitude of the frictional force is one-fourth the weight. EXECUTE: (a) If the ball is moving up, the frictional force is down, so the magnitude of the net force is (5/4)w and the acceleration is (5/4)g , down. (b) While moving down, the frictional force is up, and the magnitude of the net force is (3/4)w and the acceleration is (3/4)g , down. EVALUATE: The frictional force is less than mg in each case and in each case the net force is downward and the acceleration is downward. 5.49. IDENTIFY: Apply m?F = a! ! to one of the masses. The mass moves in a circular path, so has acceleration 2 rad v a R = , directed toward the center of the path. SET UP: In each case, 0.200 mR = . In part (a), let x+ be toward the center of the circle, so radxa a= . In part (b) let y+ be toward the center of the circle, so radya a= . y+ is downward when the mass is at the top of the circle and y+ is upward when the mass is at the bottom of the circle. Since rada has its greatest possible value, F ! is in the direction of rada ! at both positions. EXECUTE: (a) x xF ma=? gives 2rad vF ma m R= = . 75.0 NF = and (75.0 N)(0.200 m) 3.61 m/s1.15 kgFRv m= = = . (b) The free-body diagrams for a mass at the top of the path and at the bottom of the path are given in figure 5.49. At the top, y yF ma=? gives radF ma mg= ? and at the bottom it gives radF mg ma= + . For a given rotation rate and hence value of rada , the value of F required is larger at the bottom of the path. (c) radF mg ma= + so 2v F g R m = ? and 275.0 N(0.200 m) 9.80 m/s 3.33 m/s 1.15 kg F v R g m ? ?? ?= ? = ? =? ?? ?? ? ? ? 5-22 Chapter 5 EVALUATE: The maximum speed is less for the vertical circle. At the bottom of the vertical path F! and the weight are in opposite directions so F must exceed radma by an amount equal to mg . At the top of the vertical path F and mg are in the same direction and together provide the required net force, so F must be larger at the bottom. Figure 5.49 5.50. IDENTIFY: Since the car travels in an arc of a circle, it has acceleration 2rad /a v R= , directed toward the center of the arc. The only horizontal force on the car is the static friction force exerted by the roadway. To calculate the minimum coefficient of friction that is required, set the static friction force equal to its maximum value, s sf n?= . Friction is static friction because the car is not sliding in the radial direction. SET UP: The free-body diagram for the car is given in Figure 5.50. The diagram assumes the center of the curve is to the left of the car. EXECUTE: (a) y yF ma=? gives n mg= . x xF ma=? gives 2s vn m R? = . 2 s v mg m R ? = and 2 2 s 2 (25.0 m/s) 0.290 (9.80 m/s )(220 m) v gR ? = = = (b) 2 s constant v Rg? = = , so 2 2 1 2 s1 s2 v v ? ?= . s2 s1 2 1 s1 s1 /3 (25.0 m/s) 14.4 m/sv v ? ? ? ?= = = . EVALUATE: A smaller coefficient of friction means a smaller maximum friction force, a smaller possible acceleration and therefore a smaller speed. Figure 5.50 5.51. IDENTIFY: We can use the analysis done in Example 5.23. As in that example, we assume friction is negligible. SET UP: From Example 5.23, the banking angle ? is given by 2 tan v g R ? = . Also, / cosn mg ?= . 65.0 mi/h 29.1 m/s= . EXECUTE: (a) 2 2 (29.1 m/s) tan (9.80 m/s )(225 m) ? = and 21.0? = ° . The expression for tan? does not involve the mass of the vehicle, so the truck and car should travel at the same speed. (b) For the car, 2 4 car (1125 kg)(9.80 m/s ) 1.18 10 N cos21.0 n = = ×° and 4 truck car2 2.36 10 Nn n= = × , since truck car2m m= . EVALUATE: The vertical component of the normal force must equal the weight of the vehicle, so the normal force is proportional to m. 5.52. IDENTIFY: The acceleration of the person is 2rad /a v R= , directed horizontally to the left in the figure in the problem. The time for one revolution is the period 2 R T v ?= . Apply m=?F a! ! to the person. Applying Newton?s Laws 5-23 SET UP: The person moves in a circle of radius 3.00 m (5.00 m)sin30.0 5.50 mR = + =° . The free-body diagram is given in Figure 5.52. F ! is the force applied to the seat by the rod. EXECUTE: (a) y yF ma=? gives cos30.0F mg=° and cos30.0mgF = ° . x xF ma=? gives 2 sin30.0 v F m R =° . Combining these two equations gives 2tan (5.50 m)(9.80 m/s ) tan30.0 5.58 m/sv Rg ?= = =° . Then the period is 2 2 (5.50 m) 6.19 s 5.58 m/s R T v ? ?= = = . (b) The net force is proportional to m so in m=?F a! ! the mass divides out and the angle for a given rate of rotation is independent of the mass of the passengers. EVALUATE: The person moves in a horizontal circle so the acceleration is horizontal. The net inward force required for circular motion is produced by a component of the force exerted on the seat by the rod. Figure 5.52 5.53. IDENTIFY: Apply m?F = a! ! to the composite object of the person plus seat. This object moves in a horizontal circle and has acceleration rada , directed toward the center of the circle. SET UP: The free-body diagram for the composite object is given in Figure 5.53. Let x+ be to the right, in the direction of rada ! . Let y+ be upward. The radius of the circular path is 7.50 mR = . The total mass is 2(255 N 825 N) /(9.80 m/s ) 110.2 kg+ = . Since the rotation rate is 32.0 rev/min 0.5333 rev/s= , the period T is 1 1.875 s 0.5333 rev/s = . EXECUTE: y yF ma=? gives cos40.0 0AT mg? =° and 255 N 825 N 1410 Ncos40.0 cos40.0A mgT += = =° ° . x xF ma=? gives radsin 40.0A BT T ma+ =° and 2 2 2 2 4 4 (7.50 m) sin 40.0 (110.2 kg) (1410 N)sin 40.0 8370 N (1.875 s)B A R T m T T ? ?= ? = ? =° ° . The tension in the horizontal cable is 8370 N and the tension in the other cable is 1410 N. EVALUATE: The weight of the composite object is 1080 N. The tension in cable A is larger than this since its vertical component must equal the weight. rad 9280 Nma = . The tension in cable B is less than this because part of the required inward force comes from a component of the tension in cable A. Figure 5.53 5-24 Chapter 5 5.54. IDENTIFY: Apply m?F = a! ! to the button. The button moves in a circle, so it has acceleration rada . SET UP: The situation is equivalent to that of Example 5.22. EXECUTE: (a) 2 s v Rg ? = . Expressing v in terms of the period T , 2 Rv T ?= so 2s 24 RT g ?? = . A platform speed of 40.0 rev/min corresponds to a period of 1.50 s, so 2 s 2 2 4 (0.150 m) 0.269. (1.50 s) (9.80 m s ) ? ?= = (b) For the same coefficient of static friction, the maximum radius is proportional to the square of the period (longer periods mean slower speeds, so the button may be moved further out) and so is inversely proportional to the square of the speed. Thus, at the higher speed, the maximum radius is (0.150 m) 240.0 0 067 m 60.0 ? ? = .? ?? ? . EVALUATE: 2 rad 2 4 R a T ?= . The maximum radial acceleration that friction can give is smg? . At the faster rotation rate T is smaller so R must be smaller to keep rada the same. 5.55. IDENTIFY: The acceleration due to circular motion is 2 rad 2 4 R a T ?= . SET UP: 800 mR = . 1/ T is the number of revolutions per second. EXECUTE: (a) Setting rada g= and solving for the period T gives 2 400 m 2 2 40.1 s, 9.80 m s R T ? ? g = = = so the number of revolutions per minute is (60 s min) (40.1 s) 1.5 rev min= . (b) The lower acceleration corresponds to a longer period, and hence a lower rotation rate, by a factor of the square root of the ratio of the accelerations, (1.5 rev min) 3.70 9.8 0.92 rev min.T ? = × = EVALUATE: In part (a) the tangential speed of a point at the rim is given by 2 rad v a R = , so rad 62.6 m/sv Ra Rg= = = ; the space station is rotating rapidly. 5.56. IDENTIFY: 2 RT v ?= . The apparent weight of a person is the normal force exerted on him by the seat he is sitting on. His acceleration is 2rad /a v R= , directed toward the center of the circle. SET UP: The period is 60.0 s.T = The passenger has mass / 90.0 kgm w g= = . EXECUTE: (a) 2 2 (50.0 m) 5.24 m/s 60.0 s R v T ? ?= = = . Note that 2 2 2 rad (5.24 m/s) 0.549 m/s 50.0 m v a R = = = . (b) The free-body diagram for the person at the top of his path is given in Figure 5.56a. The acceleration is downward, so take y+ downward. y yF ma=? gives radmg n ma? = . 2 2 rad( ) (90.0 kg)(9.80 m/s 0.549 m/s ) 833 Nn m g a= ? = ? = . The free-body diagram for the person at the bottom of his path is given in Figure 5.56b. The acceleration is upward, so take y+ upward. y yF ma=? gives radn mg ma? = and rad( ) 931 Nn m g a= + = . (c) Apparent weight 0= means 0n = and radmg ma= . 2v g R = and 22.1 m/sv gR= = . The time for one revolution would be 2 2 (50.0 m) 14.2 s 22.1 m/s R T v ? ?= = = . Note that rada g= . (d) rad( ) 2 2(882 N) 1760 Nn m g a mg= + = = = , twice his true weight. Applying Newton?s Laws 5-25 EVALUATE: At the top of his path his apparent weight is less than his true weight and at the bottom of his path his apparent weight is greater than his true weight. Figure 5.56a, b 5.57. IDENTIFY: Apply m=?F a! ! to the motion of the pilot. The pilot moves in a vertical circle. The apparent weight is the normal force exerted on him. At each point rada ! is directed toward the center of the circular path. (a) SET UP: ?the pilot feels weightless? means that the vertical normal force n exerted on the pilot by the chair on which the pilot sits is zero. The force diagram for the pilot at the top of the path is given in Figure 5.57a. EXECUTE: y yF ma=? radmg ma= 2v g R = Figure 5.57a Thus 2(9.80 m/s )(150 m) 38.34 m/sv gR= = = 3 1 km 3600 s (38.34 m/s) 138 km/h 10 m 1 h v ? ?? ?= =? ?? ?? ?? ? (b) SET UP: The force diagram for the pilot at the bottom of the path is given in Figure 5.57b. Note that the vertical normal force exerted on the pilot by the chair on which the pilot sits is now upward. EXECUTE: y yF ma=? 2v n mg m R ? = 2v n mg m R = + This normal force is the pilot?s apparent weight. Figure 5.57b 700 N,w = so 71.43 kgwm g = = 31 h 10 m (280 km/h) 77.78 m/s 3600 s 1 km v ? ?? ?= =? ?? ?? ?? ? Thus 2(77.78 m/s) 700 N 71.43 kg 3580 N. 150 m n = + = EVALUATE: In part (b), n mg> since the acceleration is upward. The pilot feels he is much heavier than when at rest. The speed is not constant, but it is still true that 2rad /a v R= at each point of the motion. 5.58. IDENTIFY: 2rad /a v R= , directed toward the center of the circular path. At the bottom of the dive, rada! is upward. The apparent weight of the pilot is the normal force exerted on her by the seat on which she is sitting. SET UP: The free-body diagram for the pilot is given in Figure 5.58. 5-26 Chapter 5 EXECUTE: (a) 2 rad v a R = gives 2 2 2 rad (95.0 m/s) 230 m 4.00(9.80 m/s ) v R a = = = . (b) y yF ma=? gives radn mg ma? = . 2 rad( ) ( 4.00 ) 5.00 (5.00)(50.0 kg)(9.80 m/s ) 2450 Nn m g a m g g mg= + = + = = = EVALUATE: Her apparent weight is five times her true weight, the force of gravity the earth exerts on her. Figure 5.58 5.59. IDENTIFY: Apply m=?F a! ! to the water. The water moves in a vertical circle. The target variable is the speed v; we will calculate rada and then get v from 2 rad /a v R= SET UP: Consider the free-body diagram for the water when the pail is at the top of its circular path, as shown in Figures 5.59a and b. The radial acceleration is in toward the center of the circle so at this point is downward. n is the downward normal force exerted on the water by the bottom of the pail. Figure 5.59a EXECUTE: y yF ma=? 2v n mg m R + = Figure 5.59b At the minimum speed the water is just ready to lose contact with the bottom of the pail, so at this speed, 0.n ? (Note that the force n cannot be upward.) With 0n ? the equation becomes 2 . v mg m R = 2(9.80 m/s )(0.600 m) 2.42 m/s.v gR= = = EVALUATE: At the minimum speed rad .a g= If v is less than this minimum speed, gravity pulls the water (and bucket) out of the circular path. 5.60. IDENTIFY: The ball has acceleration 2rad /a v R= , directed toward the center of the circular path. When the ball is at the bottom of the swing, its acceleration is upward. SET UP: Take y+ upward, in the direction of the acceleration. The bowling ball has mass / 7.27 kgm w g= = . EXECUTE: (a) 2 2 rad (4.20 m/s) 4.64 m/s 3.80 m v a R = = = , upward. (b) The free-body diagram is given in Figure 5.60. y yF ma=? gives radT mg ma? = . 2 2 rad( ) (7.27 kg)(9.80 m/s 4.64 m/s ) 105 NT m g a= + = + = Applying Newton?s Laws 5-27 EVALUATE: The acceleration is upward, so the net force is upward and the tension is greater than the weight. Figure 5.60 5.61. IDENTIFY: Apply m?F = a! ! to the knot. SET UP: 0a = . Use coordinates with axes that are horizontal and vertical. EXECUTE: (a) The free-body diagram for the knot is sketched in Figure 5.61. 1T is more vertical so supports more of the weight and is larger. You can also see this from :x xF ma? = 2 1cos40 cos60 0T T° ? ° = . 2 1cos40 cos60 0T T° ? ° = . (b) 1T is larger so set 1 5000 N.T = Then 2 1 1.532 3263.5 NT T= = . y yF ma? = gives 1 2sin 60 sin 40T T w° + ° = and 6400 Nw = . EVALUATE: The sum of the vertical components of the two tensions equals the weight of the suspended object. The sum of the tensions is greater than the weight. Figure 5.61 5.62. IDENTIFY: Apply m?F = a! ! to each object . Constant speed means 0a = . SET UP: The free-body diagrams are sketched in Figure 5.62. 1T is the tension in the lower chain, 2T is the tension in the upper chain and T F= is the tension in the rope. EXECUTE: The tension in the lower chain balances the weight and so is equal to w. The lower pulley must have no net force on it, so twice the tension in the rope must be equal to w and the tension in the rope, which equals F, is 2w . Then, the downward force on the upper pulley due to the rope is also w, and so the upper chain exerts a force w on the upper pulley, and the tension in the upper chain is also w. EVALUATE: The pulley combination allows the worker to lift a weight w by applying a force of only / 2w . Figure 5.62 5.63. IDENTIFY: Apply m?F = a! ! to the rope. SET UP: The hooks exert forces on the ends of the rope. At each hook, the force that the hook exerts and the force due to the tension in the rope are an action-reaction pair. EXECUTE: (a) The vertical forces that the hooks exert must balance the weight of the rope, so each hook exerts an upward vertical force of 2w on the rope. Therefore, the downward force that the rope exerts at each end is end sin 2T ? w= , so end (2sin ) (2sin ).T w ? Mg ?= = 5-28 Chapter 5 (b) Each half of the rope is itself in equilibrium, so the tension in the middle must balance the horizontal force that each hook exerts, which is the same as the horizontal component of the force due to the tension at the end; end middlecos ,T ? T= so middle cos (2sin ) (2tan ).T Mg ? ? Mg ?= = (c) Mathematically speaking, 0? ? because this would cause a division by zero in the equation for endT or middleT . Physically speaking, we would need an infinite tension to keep a non-massless rope perfectly straight. EVALUATE: The tension in the rope is not the same at all points along the rope. 5.64. IDENTIFY: Apply m=?F a! ! to the combined rope plus block to find a. Then apply m=?F a! ! to a section of the rope of length x. First note the limiting values of the tension. The system is sketched in Figure 5.64a. At the top of the rope T F= At the bottom of the rope ( )T M g a= + Figure 5.64a SET UP: Consider the rope and block as one combined object, in order to calculate the acceleration: The free- body diagram is sketched in Figure 5.64b. EXECUTE: y yF ma=? ( ) ( )F M m g M m a? + = + F a g M m = ?+ Figure 5.64b SET UP: Now consider the forces on a section of the rope that extends a distance x L< below the top. The tension at the bottom of this section is ( )T x and the mass of this section is ( / ).m x L The free-body diagram is sketched in Figure 5.64c. EXECUTE: y yF ma=? ( ) ( / ) ( / )F T x m x L g m x L a? ? = ( ) ( / ) ( / )T x F m x L g m x L a= ? ? Figure 5.64c Using our expression for a and simplifying gives ( ) 1 ( ) mx T x F L M m ? ?= ?? ?+? ? EVALUATE: Important to check this result for the limiting cases: 0 :x = The expression gives the correct value of .T F= :x L= The expression gives ( /( )).T F M M m= + This should equal ( ),T M g a= + and when we use the expression for a we see that it does. 5.65. IDENTIFY: Apply m?F = a! ! to each block. SET UP: Constant speed means 0a = . When the blocks are moving, the friction force is kf and when they are at rest, the friction force is sf . EXECUTE: (a) The tension in the cord must be 2m g in order that the hanging block move at constant speed. This tension must overcome friction and the component of the gravitational force along the incline, so ( )2 1 1sin coskm g m g ? m g? ?= + and 2 1(sin cos )km m ?? ?= + . (b) In this case, the friction force acts in the same direction as the tension on the block of mass 1m , so 2 1 k 1( sin cos )m g m g ? ? m g ?= ? , or 2 1 k(sin? cos )m m ? ?= ? . Applying Newton?s Laws 5-29 (c) Similar to the analysis of parts (a) and (b), the largest 2m could be is 1 s(sin cos )m ? ? ?+ and the smallest 2m could be is 1 s(sin cos )m ? ? ?? . EVALUATE: In parts (a) and (b) the friction force changes direction when the direction of the motion of 1m changes. In part (c), for the largest 2m the static friction force on 1m is directed down the incline and for the smallest 2m the static friction force on 1m is directed up the incline. 5.66. IDENTIFY: The system is in equilibrium. Apply Newton?s 1st law to block A, to the hanging weight and to the knot where the cords meet. Target variables are the two forces. (a) SET UP: The free-body diagram for the hanging block is given in Figure 5.66a. EXECUTE: y yF ma=? 3 0T w? = 3 12.0 NT = Figure 5.66a SET UP: The free-body diagram for the knot is given in Figure 5.66b. EXECUTE: y yF ma=? 2 3sin 45.0 0T T° ? = 3 2 12.0 N sin 45.0 sin 45.0 T T = =° ° 2 17.0 NT = Figure 5.66b x xF ma=? 2 1cos45.0 0T T° ? = 1 2 cos45.0 12.0 NT T= ° = SET UP: The free-body diagram for block A is given in Figure 5.66c. EXECUTE: x xF ma=? 1 s 0T f? = s 1 12.0 Nf T= = Figure 5.66c EVALUATE: Also can apply y yF ma=? to this block: 0An w? = 60.0 NAn w= = Then s (0.25)(60.0 N) 15.0 N;n? = = this is the maximum possible value for the static friction force. We see that s s ;f n?< for this value of w the static friction force can hold the blocks in place. (b) SET UP: We have all the same free-body diagrams and force equations as in part (a) but now the static friction force has its largest possible value, s s 15.0 N.f n?= = Then 1 s 15.0 N.T f= = EXECUTE: From the equations for the forces on the knot 2 1cos45.0 0T T° ? = implies 2 1 15.0 N/ cos45.0 21.2 Ncos45.0T T= ° = =° 2 3sin 45.0 0T T° ? = implies 3 2 sin 45.0 (21.2 N)sin 45.0 15.0 NT T= ° = ° = And finally 3 0T w? = implies 3 15.0 N.w T= = EVALUATE: Compared to part (a), the friction is larger in part (b) by a factor of (15.0/12.0) and w is larger by this same ratio. 5-30 Chapter 5 5.67. IDENTIFY: Apply m=?F a! ! to each block. Use Newton?s 3rd law to relate forces on A and on B. SET UP: Constant speed means 0a = . EXECUTE: (a) Treat A and B as a single object of weight 4.80 NA Bw w w= + = . The free-body diagram for this combined object is given in Figure 5.67a. y yF ma=? gives 4.80 Nn w= = . k k 1.44 Nf n?= = . x xF ma=? gives k 1.44 NF f= = (b) The free-body force diagrams for blocks A and B are given in Figure 5.67b. n and kf are the normal and friction forces applied to block B by the tabletop and are the same as in part (a). kBf is the friction force that A applies to B. It is to the right because the force from A opposes the motion of B. Bn is the downward force that A exerts on B. kAf is the friction force that B applies to A. It is to the left because block B wants A to move with it. An is the normal force that block B exerts on A. By Newton?s third law, k kB Af f= and these forces are in opposite directions. Also, A Bn n= and these forces are in opposite directions. y yF ma=? for block A gives 1.20 NA An w= = , so 1.20 NBn = . k k (0.300)(1.20 N) 0.36 NA Af n?= = = , and k 0.36 N.Bf = x xF ma=? for block A gives k 0.36 NAT f= = . x xF ma=? for block B gives k k 0.36 N 1.44 N 1.80 NBF f f= + = + = EVALUATE: In part (a) block A is at rest with respect to B and it has zero acceleration. There is no horizontal force on A besides friction, and the friction force on A is zero. A larger force F is needed in part (b), because of the friction force between the two blocks. Figure 5.67a?c 5.68. IDENTIFY: Apply m=?F a! ! to the brush. Constant speed means 0.a = Target variables are two of the forces on the brush. SET UP: Note that the normal force exerted by the wall is horizontal, since it is perpendicular to the wall. The kinetic friction force exerted by the wall is parallel to the wall and opposes the motion, so it is vertically downward. The free-body diagram is given in Figure 5.68. EXECUTE: x xF ma=? cos53.1 0n F? ° = cos53.1n F= ° k k k cos53.1f n F? ?= = ° Figure 5.68 y yF ma=? ksin53.1 0F w f° ? ? = ksin53.1 cos53.1 0F w F?° ? ? ° = k(sin53.1 cos53.1 )F w?° ? ° = ksin53.1 cos53.1 w F ?= ° ? ° Applying Newton?s Laws 5-31 (a) k 120 N 16.9 N sin53.1 cos53.1 sin53.1 (0.15)cos53.1 w F ?= = =° ? ° ° ? ° (b) cos53.1 (16.9 N)cos53.1 10.1 Nn F= ° = ° = EVALUATE: In the absence of friction sin53.1 ,w F= ° which agrees with our expression. 5.69. IDENTIFY: The net force at any time is netF ma= . SET UP: At 0t = , 62a g= . The maximum acceleration is 140g at 1.2 mst = . EXECUTE: (a) 9 2 4net 62 62(210 10 kg)(9.80 m/s ) 1.3 10 NF ma mg ? ?= = = × = × . This force is 62 times the flea?s weight. (b) 4net 140 2.9 10 NF mg ?= = × . (c) Since the initial speed is zero, the maximum speed is the area under the -xa t graph. This gives 1.2 m/s. EVALUATE: a is much larger than g and the net external force is much larger than the flea's weight. 5.70. IDENTIFY: Apply m?F = a! ! to the instrument and calculate the acceleration. Then use constant acceleration equations to describe the motion. SET UP: The free-body diagram for the instrument is given in Figure 5.70. The instrument has mass 1.531 kgm w g= = . EXECUTE: (a) For on the instrument, y yF ma? = gives T mg ma? = and 213.07 m sT mga m ?= = . 2 0 0, 330 m s, 13.07 m s , ?y y yv v a t= = = = Then 0y y yv v a t= + gives 25.3 s t = . Consider forces on the rocket; rocket has the same ya . Let F be the thrust of the rocket engines. F mg ma? = and 2 2 5( ) (25,000 kg) (9.80 m s 13.07 m s ) 5.72 10 NF m g a= + = + = × . (b) 210 0 02 gives 4170 m.y yy y v t a t y y? = + ? = EVALUATE: The rocket and instrument have the same acceleration. The tension in the wire is over twice the weight of the instrument and the upward acceleration is greater than g . Figure 5.70 5.71. IDENTIFY: /a dv dt= . Apply m?F = a! ! to yourself. SET UP: The reading of the scale is equal to the normal force the scale applies to you. EXECUTE: The elevator?s acceleration is 2 3 2 3( ) 3.0 m s 2(0.20 m s ) 3.0 m s (0.40 m s ) dv t a t t dt = = + = + At 2 3 24.0 s, 3.0 m s (0.40 m s )(4.0 s) 4.6 m st a= = + = . From Newton?s Second Law, the net force on you is net scaleF F w ma= ? = and 2 2 scale (72 kg)(9.8 m s ) (72 kg)(4.6 m s ) 1040 NF w ma= + = + = EVALUATE: a increases with time, so the scale reading is increasing. 5.72. IDENTIFY: Apply m?F = a! ! to the passenger to find the maximum allowed acceleration. Then use a constant acceleration equation to find the maximum speed. SET UP: The free-body diagram for the passenger is given in Figure 5.72. EXECUTE: y yF ma? = gives n mg ma? = . 1.6n mg= , so 20.60 5.88 m sa g= = . 2 0 03.0 m, 5.88 m s , 0y yy y a v? = = = so 2 20 02 ( )y y yv v a y y= + ? gives 5.0 m syv = . 5-32 Chapter 5 EVALUATE: A larger final speed would require a larger value of ya , which would mean a larger normal force on the person. Figure 5.72 5.73. IDENTIFY: Apply m?F = a! ! to the package. Calculate a and then use a constant acceleration equation to describe the motion. SET UP: Let x+ be directed up the ramp. EXECUTE: (a) net k ksin37 sin37 cos37F mg f mg mg ma?= ? ° ? = ? ° ? ° = and 2 2(9.8 m s )(0.602 (0.30)(0.799)) 8.25m sa = ? + = ? Since we know the length of the slope, we can use 2 20 02 ( )x x xv v a x x= + ? with 0 0x = and 0xv = at the top. 2 2 2 2 0 2 2( 8.25 m s )(8.0 m) 132 m sv ax= ? = ? ? = and 2 20 132 m s 11.5 m sv = = (b) For the trip back down the slope, gravity and the friction force operate in opposite directions to each other. net ksin37 cos37F mg ? mg ma= ? ° + ° = and 2 2( sin37 0.30 cos37 ) (9.8 m s )(( 0.602) (0.30)(0.799)) 3.55 m sa g= ? ° + ° = ? + = ? . Now we have 0 00, 8.0 m, 0v x x= = ? = and 2 2 2 2 20 02 ( ) 0 2( 3.55 m s )( 8.0 m) 56.8 m sv v a x x= + ? = + ? ? = , so 2 256.8 m s 7.54 m sv = = . EVALUATE: In both cases, moving up the incline and moving down the incline, the acceleration is directed down the incline. The magnitude of a is greater when the package is going up the incline, because sin37mg ° and kf are in the same direction whereas when the package is going down these two forces are in opposite directions. 5.74. IDENTIFY: Apply m?F = a! ! to the hammer. Since the hammer is at rest relative to the bus its acceleration equals that of the bus. SET UP: The free-body diagram for the hammer is given in Figure 5.74. EXECUTE: gives sin74 0 so sin 74 .y yF ma T mg T mg? = ° ? = ° = gives cos74 .x xF ma T ma? = ° = Divide the second equation by the first: 2 1 and 2.8 m s tan74 a a g = =° . EVALUATE: When the acceleration increases the angle between the rope and the ceiling of the bus decreases, and the angle the rope makes with the vertical increases. Figure 5.74 5.75. IDENTIFY: Apply m?F = a! ! to the washer and to the crate. Since the washer is at rest relative to the crate, these two objects have the same acceleration. SET UP: The free-body diagram for the washer is given in Figure 5.75. EXECUTE: It?s interesting to look at the string?s angle measured from the perpendicular to the top of the crate. This angle is string 90 angle measured from the top of the crate? = °? . The free-body diagram for the washer then leads to the following equations, using Newton?s Second Law and taking the upslope direction as positive: w slope string wsin sinm g ? T ? m a? + = and string w slopesin ( sin )T ? m a g ?= + w slope stringcos cos 0m g ? T ?? + = and string w slopecos cosT m g ?? = Applying Newton?s Laws 5-33 Dividing the two equations: slopestring slope sin tan cos a g ? ? g ? += For the crate, the component of the weight along the slope is c slopesinm g ?? and the normal force is c slopecos .m g ? Using Newton?s Second Law again: c slope k c slope csin cosm g ? m g ? m a?? + = . slopek slope sin cos a g ? g ? ? += . This leads to the interesting observation that the string will hang at an angle whose tangent is equal to the coefficient of kinetic friction: k stringtan tan(90 68 ) tan 22 0.40?? = = ° ? ° = ° = . EVALUATE: In the limit that k 0? ? , string 0? ? and the string is perpendicular to the top of the crate. As k? increases, string? increases. Figure 5.75 5.76. IDENTIFY: Apply m?F = a! ! to yourself and calculate a. Then use constant acceleration equations to describe the motion. SET UP: The free-body diagram is given in Figure 5.76. EXECUTE: (a) y yF ma? = gives cosn mg ?= . x xF ma? = gives ksinmg ? f ma? = . Combining these two equations, we have 2k(sin cos ) 3.094 m sa g ? ? ?= ? = ? . Find your stopping distance: 2 00, 3.094 m s , 20 m sx x xv a v= = ? = . 2 20 0 02 ( ) gives 64.6 m, x x xv v a x x x x= + ? ? = which is greater than 40 m. You don?t stop before you reach the hole, so you fall into it. (b) 2 03.094 m s , 40 m, 0x xa x x v= ? ? = = . 2 20 0 02 ( ) gives 16 m s.x x x xv v a x x v= + ? = EVALUATE: Your stopping distance is proportional to the square of your initial speed, so your initial speed is proportional to the square root of your stopping distance. To stop in 40 m instead of 64.6 m your initial speed must be 40 m (20 m/s) 16 m/s 64.6 m = . Figure 5.76 5.77. IDENTIFY: Apply m?F = a! ! to each block and to the rope. The key idea in solving this problem is to recognize that if the system is accelerating, the tension that block A exerts on the rope is different from the tension that block B exerts on the rope. (Otherwise the net force on the rope would be zero, and the rope couldn?t accelerate.) SET UP: Take a positive coordinate direction for each object to be in the direction of the acceleration of that object. All three objects have the same magnitude of acceleration. EXECUTE: The Second Law equations for the three different parts of the system are: Block A (The only horizontal forces on A are tension to the right, and friction to the left): k .A A Am g T m a?? + = Block B (The only vertical forces on B are gravity down, and tension up): .B B Bm g T m a? = Rope (The forces on the rope along the direction of its motion are the tensions at either end and the weight of the portion of the rope that hangs vertically): ( ) .R B A Rdm g T T m aL + ? = 5-34 Chapter 5 To solve for a and eliminate the tensions, add the left hand sides and right hand sides of the three equations: ( ) kk ( / )( ) , or .( )B R AA B R A B R A B Rm m d L mdm g m g m g m m m a a gL m m m?? + ?? + + = + + = + + (a) When k ( / ) 0, . ( ) B R A B R m m d L a g m m m ? += = + + As the system moves, d will increase, approaching L as a limit, and thus the acceleration will approach a maximum value of ( ) B R A B R m m a g m m m += + + . (b) For the blocks to just begin moving, 0,a > so solve s0 [ ( / ) ]B R Am m d L m?= + ? for d. Note that we must use static friction to find d for when the block will begin to move. Solving for d, s( )A B R L d m m m ?= ? or 1.0 m (0.25(2 kg) 0.4 kg) 0.63 m. 0.160 kg d = ? = (c) When 1.0 m0.04 kg, (0.25(2 kg) 0.4 kg) 2.50 m 0.04 kgR m d= = ? = . This is not a physically possible situation since .d L> The blocks won?t move, no matter what portion of the rope hangs over the edge. EVALUATE: For the blocks to move when released, the weight of B plus the weight of the rope that hangs vertically must be greater than the maximum static friction force on A, which is s 4.9 Nn? = . 5.78. IDENTIFY: Apply Newton?s 1st law to the rope. Let 1m be the mass of that part of the rope that is on the table, and let 2m be the mass of that part of the rope that is hanging over the edge. ( 1 2 ,m m m+ = the total mass of the rope). Since the mass of the rope is not being neglected, the tension in the rope varies along the length of the rope. Let T be the tension in the rope at that point that is at the edge of the table. SET UP: The free-body diagram for the hanging section of the rope is given in Figure 5.78a EXECUTE: y yF ma=? 2 0T m g? = 2T m g= Figure 5.78a SET UP: The free-body diagram for that part of the rope that is on the table is given in Figure 5.78b. EXECUTE: y yF ma=? 1 0n m g? = 1n m g= Figure 5.78b When the maximum amount of rope hangs over the edge the static friction has its maximum value: s s s 1f n m g? ?= = x xF ma=? s 0T f? = s 1T m g?= Use the first equation to replace T : 2 s 1m g m g?= 2 s 1m m?= The fraction that hangs over is 2 s 1 s 1 s 1 s . 1 m m m m m ? ? ? ?= =+ + EVALUATE: As s 0,? ? the fraction goes to zero and as s ,? ? ? the fraction goes to unity. 5.79. IDENTIFY: First calculate the maximum acceleration that the static friction force can give to the case. Apply m=?F a! ! to the case. Applying Newton?s Laws 5-35 (a) SET UP: The static friction force is to the right in Figure 5.79a (northward) since it tries to make the case move with the truck. The maximum value it can have is s s .f N?= EXECUTE: y yF ma=? 0n mg? = n mg= s s sf n mg? ?= = Figure 5.79a x xF ma=? sf ma= smg ma? = 2 2 s (0.30)(9.80 m/s ) 2.94 m/sa g?= = = The truck?s acceleration is less than this so the case doesn?t slip relative to the truck; the case?s acceleration is 22.20 m/sa = (northward). Then 2s (30.0 kg)(2.20 m/s ) 66 N,f ma= = = northward. (b) IDENTIFY: Now the acceleration of the truck is greater than the acceleration that static friction can give the case. Therefore, the case slips relative to the truck and the friction is kinetic friction. The friction force still tries to keep the case moving with the truck, so the acceleration of the case and the friction force are both southward. The free-body diagram is sketched in Figure 5.79b. SET UP: EXECUTE: y yF ma=? 0n mg? = n mg= 2 k k (0.20)(30.0 kg)(9.80 m/s )f mg?= = k 59 N,f = southward Figure 5.79b EVALUATE: kf ma= implies 2k 59 N 2.0 m/s .30.0 kg f a m = = = The magnitude of the acceleration of the case is less than that of the truck and the case slides toward the front of the truck. In both parts (a) and (b) the friction is in the direction of the motion and accelerates the case. Friction opposes relative motion between two surfaces in contact. 5.80. IDENTIFY: Apply m?F = a! ! to the car to calculate its acceleration. Then use a constant acceleration equation to find the initial speed. SET UP: Let x+ be in the direction of the car?s initial velocity. The friction force kf is then in the -directionx? . 192 ft 58.52 m= . EXECUTE: n mg= and k kf mg?= . x xF ma=? gives k xmg ma?? = and 2 2 k (0.750)(9.80 m/s ) 7.35 m/sxa g?= ? = ? = ? . 0xv = (stops), 0 58.52 mx x? = . 2 20 02 ( )x x xv v a x x= + ? gives 2 0 02 ( ) 2( 7.35 m/s )(58.52 m) 29.3 m/s 65.5 mi/hx xv a x x= ? ? = ? ? = = . He was guilty. EVALUATE: 2 2 2 0 0 0 2 2 x x x x x v v v x x a a ?? = = ? . If his initial speed had been 45 mi/h he would have stopped in 245 mi/h (192 ft) 91 ft 65.5 mi/h ? ? =? ?? ? . 5.81. IDENTIFY: Apply m?F = a! ! to the point where the three wires join and also to one of the balls. By symmetry the tension in each of the 35.0 cm wires is the same. 5-36 Chapter 5 SET UP: The geometry of the situation is sketched in Figure 5.81a. The angle ? that each wire makes with the vertical is given by 12.5 cm sin 47.5 cm ? = and 15.26? = ° . Let AT be the tension in the vertical wire and let BT be the tension in each of the other two wires. Neglect the weight of the wires. The free-body diagram for the left-hand ball is given in Figure 5.81b and for the point where the wires join in Figure 5.81c. n is the force one ball exerts on the other. EXECUTE: (a) y yF ma=? applied to the ball gives cos 0BT mg? ? = . 2(15.0 kg)(9.80 m/s ) 152 N cos cos15.26B mg T ?= = =° . Then y yF ma=? applied in Figure 5.81c gives 2 cos 0A BT T ?? = and 2(152 N)cos 294 NAT ?= = . (b) x xF ma=? applied to the ball gives sin 0Bn T ?? = and (152 N)sin15.26 40.0 Nn = =° . EVALUATE: AT equals the total weight of the two balls. Figure 5.81a?c 5.82. IDENTIFY: Apply m?F = a! ! to the box. Compare the acceleration of the box to the acceleration of the truck and use constant acceleration equations to describe the motion. SET UP: Both objects have acceleration in the same direction; take this to be the x+ -direction. EXECUTE: If the block were to remain at rest relative to the truck, the friction force would need to cause an acceleration of 22.20 m s ; however, the maximum acceleration possible due to static friction is 2 2(0.19)(9.80 m s ) 1.86 m s ,= and so the block will move relative to the truck; the acceleration of the box would be 2 2k (0.15)(9.80 m s ) 1.47 m s .g? = = The difference between the distance the truck moves and the distance the box moves ( i . e . , the distance the box moves relative to the truck) will be 1.80 m after a time 22 truck box 2 2(1.80 m) 2.221 s. (2.20 m s 1.47 m s ) x t a a ?= = =? ? In this time, the truck moves 2 2 21 1truck2 2 (2.20m s ) (2.221 s) 5.43 m.a t = = EVALUATE: To prevent the box from sliding off the truck the coefficient of static friction would have to be 2 s (2.20 m/s ) / 0.224g? = = . 5.83. IDENTIFY: Apply m=?F a! ! to each block. Forces between the blocks are related by Newton?s 3rd law. The target variable is the force F. Block B is pulled to the left at constant speed, so block A moves to the right at constant speed and 0a = for each block. SET UP: The free-body diagram for block A is given in Figure 5.83a. BAn is the normal force that B exerts on A. kBA BAf n?= is the kinetic friction force that B exerts on A. Block A moves to the right relative to B, and BAf opposes this motion, so BAf is to the left. Applying Newton?s Laws 5-37 Note also that F acts just on B, not on A. EXECUTE: y yF ma=? 0BA An w? = 1.40 NBAn = k (0.30)(1.40 N) 0.420 NBA BAf n?= = = Figure 5.83a x xF ma=? 0BAT f? = 0.420 NBAT f= = SET UP: The free-body diagram for block B is given in Figure 5.83b. Figure 5.83b EXECUTE: ABn is the normal force that block A exerts on block B. By Newton?s third law ABn and BAn are equal in magnitude and opposite in direction, so 1.40 N.ABn = ABf is the kinetic friction force that A exerts on B. Block B moves to the left relative to A and ABf opposes this motion, so ABf is to the right. k (0.30)(1.40 N) 0.420 N.AB ABf n?= = = n and kf are the normal and friction force exerted by the floor on block B; k k .f n?= Note that block B moves to the left relative to the floor and kf opposes this motion, so kf is to the right. y yF ma=? 0B ABn w n? ? = 4.20 N 1.40 N 5.60 NB ABn w n= + = + = Then k k (0.30)(5.60 N) 1.68 N.f n?= = = x xF ma=? k 0ABf T f F+ + ? = k 0.420 N 0.420 N 1.68 N 2.52 NABF T f f= + + = + + = EVALUATE: Note that ABf and BAf are a third law action-reaction pair, so they must be equal in magnitude and opposite in direction and this is indeed what our calculation gives. 5.84. IDENTIFY: Apply m?F = a! ! to the person to find the acceleration the PAPS unit produces. Apply constant acceleration equations to her free-fall motion and to her motion after the PAPS fires. SET UP: We take the upward direction as positive. EXECUTE: The explorer?s vertical acceleration is 23.7 m s? for the first 20 s. Thus at the end of that time her vertical velocity will be 2( 3.7 m s )(20 s) 74 m s.y yv a t= = ? = ? She will have fallen a distance av 74 m s (20 s) 740 m 2 d v t ?? ?= = = ?? ?? ? and will thus be 1200 m 740 m 460 m? = above the surface. Her vertical velocity must reach zero as she touches the ground; therefore, taking the ignition point of the PAPS as 5-38 Chapter 5 0 0,y = 2 20 02 ( )y y yv v a y y= + ? gives 2 2 2 0 2 0 0 ( 74 m s) 5.95 m s 2( ) 460 m y y y v v a y y ? ? ?= = =? ? , which is the vertical acceleration that must be provided by the PAPS. The time it takes to reach the ground is given by 0 2 0 ( 74 m s) 12.4 s 5.95 m s y y y v v t a ? ? ?= = = Using Newton?s Second Law for the vertical direction PAPSvF mg ma+ = . This gives 2 PAPSv ( ) (150 kg)(5.95 ( 3.7)) m s 1450 NF ma mg m a g= ? = + = ? ? = , which is the vertical component of the PAPS force. The vehicle must also be brought to a stop horizontally in 12.4 seconds; the acceleration needed to do this is 2 0 20 33 m s 2.66 m s 12.4 s y y y v v a t ? ?= = = and the force needed is 2PAPSh (150 kg)(2.66 m s ) 400 NF ma= = = , since there are no other horizontal forces. EVALUATE: The acceleration produced by the PAPS must bring to zero both her horizontal and vertical components of velocity. 5.85. IDENTIFY: Apply m=?F a! ! to each block. Parts (a) and (b) will be done together. Figure 5.85a Note that each block has the same magnitude of acceleration, but in different directions. For each block let the direction of a ! be a positive coordinate direction. SET UP: The free-body diagram for block A is given in Figure 5.85b. EXECUTE: y yF ma=? AB A AT m g m a? = ( )AB AT m a g= + 2 24.00 kg(2.00 m/s 9.80 m/s ) 47.2 NABT = + = Figure 5.85b SET UP: The free-body diagram for block B is given in Figure 5.85b. EXECUTE: y yF ma=? 0Bn m g? = Bn m g= Figure 5.85c 2 k k k (0.25)(12.0 kg)(9.80 m/s ) 29.4 NBf n m g? ?= = = = x xF ma=? kBC AB BT T f m a? ? = 2 k 47.2 N 29.4 N (12.0 kg)(2.00 m/s )BC AB BT T f m a= + + = + + 100.6 NBCT = Applying Newton?s Laws 5-39 SET UP: The free-body diagram for block C is sketched in Figure 5.85d. EXECUTE: y yF ma=? C BC Cm g T m a? = ( )C BCm g a T? = 2 2 100.6 N 12.9 kg 9.80 m/s 2.00 m/s BC C T m g a = = =? ? Figure 5.85d EVALUATE: If all three blocks are considered together as a single object and m=?F a! ! is applied to this combined object, k ( ) .C A B A B Cm g m g m g m m m a?? ? = + + Using the values for k ,? Am and Bm given in the problem and the mass Cm we calculated, this equation gives 22.00 m/s ,a = which checks. 5.86. IDENTIFY: Apply m=?F a! ! to each block. They have the same magnitude of acceleration, a. SET UP: Consider positive accelerations to be to the right (up and to the right for the left-hand block, down and to the right for the right-hand block). EXECUTE: (a) The forces along the inclines and the accelerations are related by (100 kg) sin30 (100 kg) and (50 kg) sin53 (50 kg) ,T g a g T a? ° = ° ? = where T is the tension in the cord and a the mutual magnitude of acceleration. Adding these relations, (50 kg sin 53 100 kg sin 30 ) (50 kg 100 kg) , or 0.067 .g a a g° ? ° = + = ? Since a comes out negative, the blocks will slide to the left; the 100-kg block will slide down. Of course, if coordinates had been chosen so that positive accelerations were to the left, a would be 0.067 .g+ (b) 220.067(9.80 m s ) 0.658 m s .a = = (c) Substituting the value of a (including the proper sign, depending on choice of coordinates) into either of the above relations involving T yields 424 N. EVALUATE: For part (a) we could have compared sinmg ? for each block to determine which direction the system would move. 5.87. IDENTIFY: Let the tensions in the ropes be 1T and 2.T Figure 5.87a Consider the forces on each block. In each case take a positive coordinate direction in the direction of the acceleration of that block. SET UP: The free-body diagram for 1m is given in Figure 5.87b. EXECUTE: x xF ma=? 1 1 1T m a= Figure 5.87b 5-40 Chapter 5 SET UP: The free-body diagram for 2m is given in Figure 5.87c. EXECUTE: y yF ma=? 2 2 2 2m g T m a? = Figure 5.87c This gives us two equations, but there are 4 unknowns ( 1,T 2 ,T 1,a and 2a ) so two more equations are required. SET UP: The free-body diagram for the moveable pulley (mass m) is given in Figure 5.87d. EXECUTE: y yF ma=? 2 12mg T T ma+ ? = Figure 5.87d But our pulleys have negligible mass, so 0mg ma= = and 2 12 .T T= Combine these three equations to eliminate 1T and 2 :T 2 2 2 2m g T m a? = gives 2 1 2 22 .m g T m a? = And then with 1 1 1T m a= we have 2 1 1 2 22 .m g m a m a? = SET UP: There are still two unknowns, 1a and 2.a But the accelerations 1a and 2a are related. In any time interval, if 1m moves to the right a distance d, then in the same time 2m moves downward a distance / 2.d One of the constant acceleration kinematic equations says 210 0 2 ,x xx x v t a t? = + so if 2m moves half the distance it must have half the acceleration of 1 :m 2 1 / 2,a a= or 1 22 .a a= EXECUTE: This is the additional equation we need. Use it in the previous equation and get 2 1 2 2 22 (2 ) .m g m a m a? = 2 1 2 2(4 )a m m m g+ = 2 2 1 24 m g a m m = + and 2 1 2 1 2 2 2 . 4 m g a a m m = = + EVALUATE: If 2 0m ? or 1 ,m ? ? 1 2 0.a a= = If 2 1,m m>> 2a g= and 1 2 .a g= 5.88. IDENTIFY: Apply m?F = a! ! to block B, to block A and B as a composite object and to block C. If A and B slide together all three blocks have the same magnitude of acceleration. SET UP: If A and B don?t slip the friction between them is static. The free-body diagrams for block B, for blocks A and B, and for C are given in Figures 5.88a-c. Block C accelerates downward and A and B accelerate to the right. In each case take a positive coordinate direction to be in the direction of the acceleration. Since block A moves to the right, the friction force sf on block B is to the right, to prevent relative motion between the two blocks. When C has its largest mass, sf has its largest value: s sf n?= . EXECUTE: x xF ma=? applied to the block B gives s Bf m a= . Bn m g= and s s Bf m g?= . s B Bm g m a? = and sa g?= . x xF ma=? applied to blocks A B+ gives sAB ABT m a m g?= = . y yF ma=? applied to block C gives C Cm g T m a? = . s sC AB Cm g m g m g? ?? = . s s 0.750 (5.00 kg 8.00 kg) 39.0 kg 1 1 0.750 AB C m m ? ? ? ?= = + =? ?? ?? ? . Applying Newton?s Laws 5-41 EVALUATE: With no friction from the tabletop, the system accelerates no matter how small the mass of C is. If Cm is less than 39.0 kg, the friction force that A exerts on B is less than s n? . If Cm is greater than 39.0 kg, blocks C and A have a larger acceleration than friction can give to block B and A accelerates out from under B. Figure 5.88 5.89. IDENTIFY: Apply the method of Exercise 5.19 to calculate the acceleration of each object. Then apply constant acceleration equations to the motion of the 2.00 kg object. SET UP: After the 5.00 kg object reaches the floor, the 2.00 kg object is in free-fall, with downward acceleration g . EXECUTE: The 2.00-kg object will accelerate upward at 5.00 kg 2.00 kg 3 7, 5.00 kg 2.00 kg g g? =+ and the 5.00-kg object will accelerate downward at 3 7.g Let the initial height above the ground be 0h . When the large object hits the ground, the small object will be at a height 02h , and moving upward with a speed given by 2 0 0 02 6 7.v ah gh= = The small object will continue to rise a distance 20 02 3 7,v g h= and so the maximum height reached will be 0 0 02 3 7 17 7 1.46 mh h h+ = = above the floor , which is 0.860 m above its initial height. EVALUATE: The small object is 1.20 m above the floor when the large object strikes the floor, and it rises an additional 0.26 m after that. 5.90. IDENTIFY: Apply m=?F a! ! to the box. SET UP: The box has an upward acceleration of 21.90 m/sa = . EXECUTE: The floor exerts an upward force n on the box, obtained from ,n mg ma? = or ( ).n m a g= + The friction force that needs to be balanced is 22 k k ( ) (0.32)(28.0 kg)(1.90 m s 9.80 m s ) 105 N.n m a g? ?= + = + = EVALUATE: If the elevator wasn't accelerating the normal force would be n mg= and the friction force that would have to be overcome would be 87.8 N. The upward acceleration increases the normal force and that increases the friction force. 5.91. IDENTIFY: Apply m=?F a! ! to the block. The cart and the block have the same acceleration. The normal force exerted by the cart on the block is perpendicular to the front of the cart, so is horizontal and to the right. The friction force on the block is directed so as to hold the block up against the downward pull of gravity. We want to calculate the minimum a required, so take static friction to have its maximum value, s s .f n?= SET UP: The free-body diagram for the block is given in Figure 5.91. EXECUTE: x xF ma=? n ma= s s sf n ma? ?= = Figure 5.91 y yF ma=? s 0f mg? = sma mg? = s/a g ?= EVALUATE: An observer on the cart sees the block pinned there, with no reason for a horizontal force on it because the block is at rest relative to the cart. Therefore, such an observer concludes that 0n = and thus s 0,f = and he doesn?t understand what holds the block up against the downward force of gravity. The reason for this 5-42 Chapter 5 difficulty is that m=?F a! ! does not apply in a coordinate frame attached to the cart. This reference frame is accelerated, and hence not inertial. The smaller s? is, the larger a must be to keep the block pinned against the front of the cart. 5.92. IDENTIFY: Apply m=?F a! ! to each block. SET UP: Use coordinates where x+ is directed down the incline. EXECUTE: (a) Since the larger block (the trailing block) has the larger coefficient of friction, it will need to be pulled down the plane; i . e . , the larger block will not move faster than the smaller block, and the blocks will have the same acceleration. For the smaller block, (4.00 kg) (sin30 (0.25)cos 30 ) (4.00 kg) ,g T a° ? ° ? = or 11.11 N (4.00 kg) ,T a? = and similarly for the larger, 15.44 N (8.00 kg)T a+ = . Adding these two relations, 26.55 N (12.00 kg) ,a= 22.21 m s .a = (b) Substitution into either of the above relations gives 2.27 N.T = (c) The string will be slack. The 4.00-kg block will have 22.78 m sa = and the 8.00-kg block will have 21.93 m s ,a = until the 4.00-kg block overtakes the 8.00-kg block and collides with it. EVALUATE: If the string is cut the acceleration of each block will be independent of the mass of that block and will depend only on the slope angle and the coefficient of kinetic friction. The 8.00-kg block would have a smaller acceleration even though it has a larger mass, since it has a larger k? . 5.93. IDENTIFY: Apply m=?F a! ! to the block and to the plank. SET UP: Both objects have 0a = . EXECUTE: Let Bn be the normal force between the plank and the block and An be the normal force between the block and the incline. Then, cosBn w ?= and 3 cos 4 cos .A Bn n w ? w ?= + = The net frictional force on the block is k k( ) 5 cosA Bn n w? ? ?+ = . To move at constant speed, this must balance the component of the block?s weight along the incline, so k3 sin 5 cos ,w ? w ??= and 3 3k 5 5tan tan37 0.452.?? = = ° = EVALUATE: In the absence of the plank the block slides down at constant speed when the slope angle and coefficient of friction are related by ktan? ?= . For 36.9? = ° , k 0.75? = . A smaller k? is needed when the plank is present because the plank provides an additional friction force. 5.94. IDENTIFY: Apply m?F = a! ! to the ball, to 1m and to 2m SET UP: The free-body diagrams for the ball, 1m and 2m are given in Figures 5.94a-c. All three objects have the same magnitude of acceleration. In each case take the direction of a ! to be a positive coordinate direction. EXECUTE: (a) y yF ma=? applied to the ball gives cosT mg? = . x xF ma=? applied to the ball gives sinT ma? = . Combining these two equations to eliminate T gives tan /a g? = . (b) x xF ma=? applied to 2m gives 2T m a= . y yF ma=? applied to 1m gives 1 1m g T m a? = . Combining these two equations gives 1 1 2 m a g m m ? ?= ? ?+? ? . Then 1 1 2 250 kg tan 1500 kg m m m ? = =+ and 9.46? = ° . (c) As 1m becomes much larger than 2m , a g? and tan 1? ? , so 45? ? ° . EVALUATE: The device requires that the ball is at rest relative to the platform; any motion swinging back and forth must be damped out. When 1 2m m<< the system still accelerates, but with small a and 0? ? ° . Figure 5.94a?c Applying Newton?s Laws 5-43 5.95. IDENTIFY: Apply m=?F a! ! to the automobile. SET UP: The "correct" banking angle is for zero friction and is given by 2 0tan v g R ? = , as derived in Example 5.23. Use coordinates that are vertical and horizontal, since the acceleration is horizontal. EXECUTE: For speeds larger than 0v , a frictional force is needed to keep the car from skidding. In this case, the inward force will consist of a part due to the normal force n and the friction force rad; sin cos .f n f ma? ?+ = The normal and friction forces both have vertical components; since there is no vertical acceleration, cos sin .n f mg? ?? = Using sf n?= and 22 0 rad (1.5 ) 2.25 tan , vva g R R ?= = = these two relations become ssin cos 2.25 tann n mg? ? ? ?+ = and scos sinn n mg? ? ?? = . Dividing to cancel n gives s s sin cos 2.25 tan . cos sin ? ? ? ?? ? ? + =? Solving for s? and simplifying yields s 2 1.25 sin cos 1 1.25sin ? ?? ?= + . Using 2 2 (20 m s) arctan 18.79 (9.80 m s )(120 m) ? ? ?= = °? ?? ? gives s 0.34.? = EVALUATE: If s? is insufficient, the car skids away from the center of curvature of the roadway, so the friction in inward. 5.96. IDENTIFY: Apply m=?F a! ! to the car. The car moves in the arc of a horizontal circle, so rad,=a a! ! directed toward the center of curvature of the roadway. The target variable is the speed of the car. rada will be calculated from the forces and then v will be calculated from 2rad / .a v R= (a) To keep the car from sliding up the banking the static friction force is directed down the incline. At maximum speed the static friction force has its maximum value s s .f n?= SET UP: The free-body diagram for the car is sketched in Figure 5.96a. EXECUTE: y yF ma=? scos sin 0n f mg? ?? ? = But s s ,f n?= so scos sin 0n n mg? ? ?? ? = scos sin mg n ? ? ?= ? Figure 5.96a x xF ma=? s radsin cosn n ma? ? ?+ = s rad(sin cos )n ma? ? ?+ = Use the yF? equation to replace n : s rad s (sin cos ) cos sin mg ma? ? ?? ? ? ? ? + =? ??? ? 2 2s rad s sin cos sin 25 (0.30)cos25 (9.80 m/s ) 8.73 m/s cos sin cos25 (0.30)sin 25 a g ? ? ? ? ? ? ? ? ? ?+ ° + °= = =? ? ? ?? ° ? °? ?? ? rad /a v R 2= implies 2rad (8.73 m/s )(50 m) 21 m/s.v a R= = = (b) IDENTIFY: To keep the car from sliding down the banking the static friction force is directed up the incline. At the minimum speed the static friction force has its maximum value s s .f n?= 5-44 Chapter 5 SET UP: The free-body diagram for the car is sketched in Figure 5.96b. The free-body diagram is identical to that in part (a) except that now the components of sf have opposite directions. The force equations are all the same except for the opposite sign for terms containing s.? Figure 5.96b EXECUTE: 2 2srad s sin cos sin 25 (0.30)cos25 (9.80 m/s ) 1.43 m/s cos sin cos25 (0.30)sin 25 a g ? ? ? ? ? ? ? ? ? ?? ° ? °= = =? ? ? ?+ ° + °? ?? ? 2 rad (1.43 m/s )(50 m) 8.5 m/s.v a R= = = EVALUATE: For v between these maximum and minimum values, the car is held on the road at a constant height by a static friction force that is less than s .n? When s 0,? ? rad tan .a g ?= Our analysis agrees with the result of Example 5.23 in this special case. 5.97. IDENTIFY: Apply m=?F a! ! to the car. SET UP: 1 mi/h 0.447 m/s= . The acceleration of the car is 2rad /a v r= , directed toward the center of curvature of the roadway. EXECUTE: (a) 80 mi h 35.7 m s= . The centripetal force needed to keep the car on the road is provided by friction; thus 2 s mv mg r ? = and 2 2 2 s (35.7 m s) 171 m (0.76)(9.8 m s ) v r g?= = = . (b) If s 0.20? = , 2 s (171 m) (0.20) (9.8 m/s ) 18.3 m s or about 41 mi hv r g?= = = . (c) If s 0.37? = , 2(171 m) (0.37) (9.8 m/s ) 24.9 m s or about 56 mi hv = = The speed limit is evidently designed for these conditions. EVALUATE: The maximum safe speed is proportional to s? . 0.20/ 0.76 0.51= , so the maximum safe speed for wet-ice conditions is about half what it is for a dry road. 5.98. IDENTIFY: The analysis of this problem is the same as that of Example 5.21. SET UP: From Example 5.21, 2 radtan a v g rg ? = = . EXECUTE: Solving for v in terms of ? and R, 2tan (9.80 m s ) (50.0) tan 30.0 16.8 m sv gR ?= = ° = , about 60.6 km h. EVALUATE: The greater the speed of the bus the larger will be the angle ? , so T will have a larger horizontal, inward component. 5.99. IDENTIFY and SET UP: The monkey and bananas have the same mass and the tension in the rope has the same upward value at the bananas and at the monkey. Therefore, the monkey and bananas will have the same net force and hence the same acceleration, in both magnitude and direction. EXECUTE: (a) For the monkey to move up, T mg> . The bananas also move up. (b) The bananas and monkey move with the same acceleration and the distance between them remains constant. (c) Both the monkey and bananas are in free fall. They have the same initial velocity and as they fall the distance between them doesn?t change. (d) The bananas will slow down at the same rate as the monkey. If the monkey comes to a stop, so will the bananas. EVALUATE: None of these actions bring the monkey any closer to the bananas. Applying Newton?s Laws 5-45 5.100. IDENTIFY: Apply m=?F a! ! , with f kv= . SET UP: Follow the analysis that leads to Eq.(5.10), except now the initial speed is 0 t3 / 3yv mg k v= = rather than zero. EXECUTE: The separated equation of motion has a lower limit of t3v instead of 0; specifically, t ( )t t t t t3 1 1 ln ln , or 2 . 2 2 2 2 v k m t v dv v v v k t v v e v v v v m ?? ?? ? ?= = ? = ? = +? ? ? ?? ? ? ?? ?? EVALUATE: As t ? ? the speed approaches tv . The speed is always greater than tv and this limit is approached from above. 5.101. IDENTIFY: Apply m=?F a! ! to the rock. SET UP: Equations 5.9 through 5.13 apply, but with 0a rather than g as the initial acceleration. EXECUTE: (a) The rock is released from rest, and so there is initially no resistive force and 2 0 (18.0 N) (3.00 kg) 6.00 m s .a = = (b) 2(18.0 N (2.20 N s m) (3.00 m s)) (3.00 kg) 3.80 m s .? ? = (c) The net force must be 1.80 N, so 16.2 Nkv = and (16.2 N) (2.20 N s m) 7.36 m s.v = ? = (d) When the net force is equal to zero, and hence the acceleration is zero, t 18.0 Nkv = and t (18.0 N) (2.20 N s m) 8.18 m s.v = ? = (e) From Eq.(5.12), ( )((2.20 N s m) (3.00 kg))(2.00 s)3.00 kg(8.18 m s) (2.00 s) 1 7.78 m. 2.20 N s m y e ? ? ? ?= ? ? = +? ??? ? From Eq. (5.10), ((2.20 N s m) (3.00 kg))(2.00 s)(8.18 m s)[1 ] 6.29 m s.v e ? ?= ? = From Eq.(5.11), but with 0a instead of g , 2 ((2.20 N s m) (3.00 kg))(2.00 s) 2(6.00 m s ) 1.38 m s .a e ? ?= = (f) ( ) t 1 0.1 k m t v e v ?? = = and ln (10) 3.14 s.mt k = = EVALUATE: The acceleration decreases with time until it becomes zero when tv v= . The speed increases with time and approaches tv as t ? ? . 5.102. IDENTIFY: Apply m=?F a! ! to the rock. dva dt= and dxv dt= yield differential equations that can be integrated to give ( )v t and ( )x t . SET UP: The retarding force of the surface is the only horizontal force acting. EXECUTE: (a) Thus 1 2 net RF F kv dva m m m dt ?= = = = and 1 2dv k dtv m= ? . Integrating gives 0 1 2 0 v t v dv k dt v m = ?? ? and 0 1 22 vv kt v m = ? . This gives 1 2 2 2 0 0 24 v kt k t v v m m = ? + . For the rock?s position: 1 2 2 2 0 0 24 dx v kt k t v dt m m = ? + and 1 2 2 2 0 0 24 v ktdt k t dt dx v dt m m = ? + . Integrating gives 1 2 2 2 3 0 0 22 12 v kt k t x v t m m = ? + . (b) 1 2 2 2 0 0 2 0 2 v kt k t v v m m = = ? + . This is a quadratic equation in t; from the quadratic formula we can find the single solution 1 2 02mvt k = . (c) Substituting the expression for t into the equation for x: 1 2 1 2 2 2 3 3 2 3 2 0 0 0 0 0 0 2 2 3 2 4 8 2 2 12 3 mv v k m v k m v mv x v k m k m k k = ? ? ? + ? = EVALUATE: The magnitude of the average acceleration is 1/ 2 0 0 av 1/ 2 0 1 (2 / ) 2 v v kv a t mv k m ?= ? =? . The average force is 1/ 21 av av 02F ma kv= = , which is 12 times the initial value of the force. 5-46 Chapter 5 5.103. IDENTIFY: Apply m=?F a! ! to the object, with and without including the buoyancy force. SET UP: At the terminal speed tv , 0a = . EXECUTE: Without buoyancy, t t , so . 0.36 s mg mg kv mg k v = = = With buoyancy included there is the additional upward buoyancy force B, so tB kv mg+ = . t 0.24 m s1 30.36 m sB mg kv mg mg ? ?= ? = ? =? ?? ? . EVALUATE: At the terminal speed, B and f kv= together equal mg . The presence of B reduces the value of f required, so the presence of B reduces the terminal speed. 5.104. IDENTIFY: The block has acceleration 2rad /a v r= , directed to the left in the figure in the problem. Apply m=?F a! ! to the block. SET UP: The block moves in a horizontal circle of radius 2 2(1.25 m) (1.00 m) 0.75 mr = ? = . Each string makes an angle ? with the vertical. 1.00 mcos 1.25 m ? = , so 36.9? = ° . The free-body diagram for the block is given in Figure 5.104. Let x+ be to the left and let y+ be upward. EXECUTE: (a) y yF ma=? gives u lcos cos 0T T mg? ?? ? = . 2 l u (4.00 kg)(9.80 m/s ) 80.0 N 31.0 N cos cos36.9 mg T T ?= ? = ? =° . (b) x xF ma=? gives 2u l( )sin vT T m r?+ = . u l( )sin (0.75 m)(80.0 N 31.0 N)sin36.9 3.53 m/s 4.00 kg r T T v m ?+ += = =° . The number of revolutions per second is 3.53 m/s 0.749 rev/s 44.9 rev/min 2 2 (0.75 m) v r? ?= = = . (c) If l 0T ? , u cosT mg? = and 2 u (4.00 kg)(9.80 m/s ) 49.0 N cos cos36.9 mg T ?= = =° . 2 u sin v T m r ? = . u sin (0.75 m)(49.0 N)sin36.9 2.35 m/s 4.00 kg rT v m ?= = =° . The number of revolutions per minute is 2.35 m/s (44.9 rev/min) 29.9 rev/min 3.53 m/s ? ? =? ?? ? EVALUATE: The tension in the upper string must be greater than the tension in the lower string so that together they produce an upward component of force that balances the weight of the block. Figure 5.104 5.105. IDENTIFY: Apply m=?F a! ! to the falling object. SET UP: Follow the steps that lead to Eq.(5.10), except now 0 0yv v= and is not zero. Applying Newton?s Laws 5-47 EXECUTE: (a) Newton?s 2nd law gives ,y ydvm mg kvdt = ? where t mg v k = . 0 t 0 yv t y yv dv k dt v v m = ??? ? . This is the same expression used in the derivation of Eq. (5.10), except the lower limit in the velocity integral is the initial speed 0v instead of zero. Evaluating the integrals and rearranging gives 0 t (1 ) kt m kt mv v e v e? ?= + ? . Note that at 0t = this expression says 0yv v= and at t ?? it says t .yv v? (b) The downward gravity force is larger than the upward fluid resistance force so the acceleration is downward, until the fluid resistance force equals gravity when the terminal speed is reached. The object speeds up until tyv v= . Take y+ to be downward. The graph is sketched in Figure 5.105a. (c) The upward resistance force is larger than the downward gravity force so the acceleration is upward and the object slows down, until the fluid resistance force equals gravity when the terminal speed is reached. Take y+ to be downward. The graph is sketched in Figure 5.105b. (d) When 0 tv v= the acceleration at 0t = is zero and remains zero; the velocity is constant and equal to the terminal velocity. EVALUATE: In all cases the speed becomes tv as t ? ? . Figure 5.105a, b 5.106. IDENTIFY: Apply m=?F a! ! to the rock. SET UP: At the maximum height, 0yv = . Let y+ be upward. Suppress the y subscripts on v and a. EXECUTE: (a) To find the maximum height and time to the top without fluid resistance: 2 20 02 ( )v v a y y= + ? and 2 2 2 0 0 2 0 (6.0 m s) 1.84 m 2 2( 9.8 m s ) v v y y a ? ?? = = =? . 0 2 0 6.0 m s 0.61 s 9.8 m s v v t a ? ?= = =? . (b) Starting from Newton?s Second Law for this situation dv m mg kv dt = ? . We rearrange and integrate, taking downward as positive as in the text and noting that the velocity at the top of the rock?s flight is zero: 0 t v dv k t v v m = ??? . 0 tt t 2.0 m s ln( ) ln ln ln(0.25) 1.386 6.0 m s 2.0 m sv v v v v v ? ?? = = = = ?? ? ? From Eq.(5.9), 2 2t (2.0 m s ) (9.8 m s ) 0.204 s,m k v g= = = and ( 1.386) (0.204 s) (1.386) 0.283 smt k= ? ? = = to the top. Equation 5.10 in the text gives us ( ) ( )t t t(1 ) k m t k m tdx v e v v e dt ? ?= ? = ? . ( ) ( )t t t t 0 0 0 ( 1) x t t k m t k m tv mx dx v dt v e dt v t e k ? ?= = ? = + ?? ? ? . 1.387(2.0 m s) (0.283 s) (2.0 m s) (0.204 s)(e 1) 0.26 mx ?= + ? = . EVALUATE: With fluid resistance present the maximum height is much less and the time to reach it is less. 5.107. IDENTIFY: Apply m=?F a! ! to the car. SET UP: The forces on the car are the air drag force 2Df Dv= and the rolling friction force r .mg? Take the velocity to be in the x+ -direction. The forces are opposite in direction to the velocity. EXECUTE: (a) x xF ma? = gives 2 rDv mg ma?? ? = . We can write this equation twice, once with 32 m sv = and 2 0.42 m sa = ? and once with 24 m sv = and 20.30 m/s .a = ? Solving these two simultaneous equations in the unknowns D and r? gives r 0.015? = and 2 20.36 N s m .D = ? (b) cosn mg ?= and the component of gravity parallel to the incline is sinmg ? , where 2.2 .? = ° For constant speed, 2rsin 2.2 cos2.2 0.mg mg Dv?° ? ° ? = Solving for v gives 29 m s.v = 5-48 Chapter 5 (c) For angle 2r, sin cos 0mg mg Dv? ? ? ?? ? = and r(sin cos )mgv D ? ? ??= . The terminal speed for a falling object is derived from 2t 0,Dv mg? = so t .v mg D= t rsin cosv v ? ? ?= ? . And since r t0.015, sin (0.015) cosv v? ? ?= = ? . EVALUATE: In part (c), tv v? as 90? ? ° , since in that limit the incline becomes vertical. 5.108. IDENTIFY: Apply m=?F a! ! to the person and to the cart. SET UP: The apparent weight, appw , which is the same as the upward force on the person exerted by the car seat. EXECUTE: (a) The apparent weight is the actual weight of the person minus the centripetal force needed to keep him moving in its circular path: 2 2 2 app (12 m s) (70 kg) (9.8 m s ) 434 N 40 m mv w mg R ? ?= ? = ? =? ?? ? . (b) The cart will lose contact with the surface when its apparent weight is zero; i.e., when the road no longer has to exert any upward force on it: 2 0 mv mg R ? = . 2(40 m) (9.8 m/s ) 19.8 m sv Rg= = = . The answer doesn?t depend on the cart?s mass, because the centripetal force needed to hold it on the road is proportional to its mass and so to its weight, which provides the centripetal force in this situation. EVALUATE: At the speed calculated in part (b), the downward force needed for circular motion is provided by gravity. For speeds greater than this more, downward force is needed and there is no source for it and the cart leaves the circular path. For speeds less than this, less downward force than gravity is needed, so the roadway must exert an upward vertical force. 5.109. (a) IDENTIFY: Use the information given about Jena to find the time t for one revolution of the merry-go-round. Her acceleration is rad ,a directed in toward the axis. Let 1F ! be the horizontal force that keeps her from sliding off. Let her speed be 1v and let 1R be her distance from the axis. Apply m=?F a! ! to Jena, who moves in uniform circular motion. SET UP: The free-body diagram for Jena is sketched in Figure 5.109a EXECUTE: x xF ma=? 1 radF ma= 2 1 1 1 , v F m R = 1 11 1.90 m/sR Fv m= = Figure 5.109a The time for one revolution is 1 1 1 1 1 2 2 . R m t R v R F ? ?= = Jackie goes around once in the same time but her speed 2( )v and the radius of her circular path 2( )R are different. 2 1 1 2 1 1 2 2 1 1 2 1 2 . 2 R R F R R F v R t R m R m ? ? ? ? ?= = =? ?? ? IDENTIFY: Now apply m=?F a! ! to Jackie. She also moves in uniform circular motion. SET UP: The free-body diagram for Jackie is sketched in Figure 5.109b. EXECUTE: x xF ma=? 2 radF ma= Figure 5.109b 2 2 2 2 1 1 2 2 12 2 2 1 1 3.60 m (60.0 N) 1.80 m v m R R F R F m F R R R m R ? ?? ? ? ?? ? ? ?= = = =? ?? ? ? ?? ? ? ?? ? ? ?? ?? ? ? ? 120.0 N= (b) 2 2 2 2 , v F m R = so 2 22 (120.0 N)(3.60 m) 3.79 m/s30.0 kg F R v m = = = Applying Newton?s Laws 5-49 EVALUATE: Both girls rotate together so have the same period T . By Eq.(5.16), rada is larger for Jackie so the force on her is larger. Eq.(5.15) says 1 1 2 2/ /R v R v= so 2 1 2 1( / );v v R R= this agrees with our result in (a). 5.110. IDENTIFY: Apply m=?F a! ! to the passenger. The passenger has acceleration rada , directed inward toward the center of the circular path. SET UP: The passenger?s velocity is 2 8.80 m s.v ? R t= = The vertical component of the seat?s force must balance the passenger?s weight and the horizontal component must provide the centripetal force. EXECUTE: (a) seat sin 833 NF mg? = = and 2 seat cos 188 N mv F R ? = = . Therefore tan (833 N) (188 N) 4.43;? = = 77.3? = ° above the horizontal. The magnitude of the net force exerted by the seat (note that this is not the net force on the passenger) is 2 2 seat (833 N) (188 N) 854 NF = + = (b) The magnitude of the force is the same, but the horizontal component is reversed. EVALUATE: At the highest point in the motion, 2 seat 645 N v F mg m R = ? = . At the lowest point in the motion, 2 seat 1021 N v F mg m R = + = . The result in parts (a) and (b) lies between these extreme values. 5.111. IDENTIFY: Apply m=?F a! ! to the person. The person moves in a horizontal circle so his acceleration is 2 rad / ,a v R= directed toward the center of the circle. The target variable is the coefficient of static friction between the person and the surface of the cylinder. 2 2 (2.5 m) (0.60 rev/s) (0.60 rev/s) 9.425 m/s 1 rev 1 rev R v ? ?? ? ? ?= = =? ? ? ?? ? ? ? (a) SET UP: The problem situation is sketched in Figure 5.111a. Figure 5.111a The free-body diagram for the person is sketched in Figure 5.111b. The person is held up against gravity by the static friction force exerted on him by the wall. The acceleration of the person is rad ,a directed in towards the axis of rotation. Figure 5.111b (b) EXECUTE: To calculate the minimum s? required, take sf to have its maximum value, s s .f n?= y yF ma=? s 0f mg? = s n mg? = x xF ma=? 2 /n mv R= Combine these two equations to eliminate n : 2 s /mv R mg? = 2 s 2 2 (2.5 m)(9.80 m/s ) 0.28 (9.425 m/s) Rg v ? = = = 5-50 Chapter 5 (c) EVALUATE: No, the mass of the person divided out of the equation for s.? Also, the smaller s? is, the larger v must be to keep the person from sliding down. For smaller s? the cylinder must rotate faster to make n larger enough. 5.112. IDENTIFY: Apply m=?F a! ! to the combined object of motorcycle plus rider. SET UP: The object has acceleration 2rad /a v r= , directed toward the center of the circular path. EXECUTE: (a) For the tires not to lose contact, there must be a downward force on the tires. Thus, the (downward) acceleration at the top of the sphere must exceed mg , so 2 , v m mg R > and 2(9.80 m s ) (13.0 m) 11.3 m s.v gR> = = (b) The (upward) acceleration will then be 4g , so the upward normal force must be 25 5(110 kg) (9.80 m s ) 5390 N.mg = = EVALUATE: At any nonzero speed the normal force at the bottom of the path exceeds the weight of the object. 5.113. IDENTIFY: Apply m=?F a! ! to your friend. Your friend moves in the arc of a circle as the car turns. (a) Turn to the right. The situation is sketched in Figure 5.113a. As viewed in an inertial frame, in the absence of sufficient friction your friend doesn?t make the turn completely and you move to the right toward your friend. Figure 5.113a (b) The maximum radius of the turn is the one that makes rada just equal to the maximum acceleration that static friction can give to your friend, and for this situation sf has its maximum value s s .f n?= SET UP: The free-body diagram for your friend, as viewed by someone standing behind the car, is sketched in Figure 5.113b. EXECUTE: y yF ma=? 0n mg? = n mg= Figure 5.113b x xF ma=? s radf ma= 2 s /n mv R? = 2 s /mg mv R? = 2 2 2 s (20 m/s) 120 m (0.35)(9.80 m/s ) v R g?= = = EVALUATE: The larger s? is, the smaller the radius R must be. 5.114. IDENTIFY: The tension F in the string must be the same as the weight of the hanging block, and must also provide the resultant force necessary to keep the block on the table in uniform circular motion. SET UP: The acceleration of the block is 2rad /a v r= , directed toward the hole. EXECUTE: 2 , v Mg F m r = = so .v gr M m= EVALUATE: The larger M is the greater must be the speed v, if r remains the same. Applying Newton?s Laws 5-51 5.115. IDENTIFY: Apply m=?F a! ! to the circular motion of the bead. Also use Eq.(5.16) to relate rada to the period of rotation T . SET UP: The bead and hoop are sketched in Figure 5.115a. The bead moves in a circle of radius sin .R r ?= The normal force exerted on the bead by the hoop is radially inward. Figure 5.115a The free-body diagram for the bead is sketched in Figure 5.115b. EXECUTE: y yF ma=? cos 0n mg? ? = / cosn mg ?= x xF ma=? radsinn ma? = Figure 5.115b Combine these two equations to eliminate n : radsincos mg ma?? ? ? =? ?? ? radsin cos a g ? ? = 2 rad /a v R= and 2 / ,v R T?= so 2 2rad 4 / ,a R T?= where T is the time for one revolution. sin ,R r ?= so 2 rad 2 4 sinr a T ? ?= Use this in the above equation: 2 2 sin 4 sin cos r T g ? ? ? ? = This equation is satisfied by sin 0,? = so 0,? = or by 2 2 1 4 , cos r T g ? ? = which gives 2 2cos 4 T g r ? ?= (a) 4.00 rev/s implies (1/ 4.00) s 0.250 sT = = Then 2 2 2 (0.250 s) (9.80 m/s ) cos 4 (0.100 m) ? ?= and 81.1 .? = ° (b) This would mean 90 .? = ° But cos90 0,° = so this requires 0.T ? So ? approaches 90° as the hoop rotates very fast, but 90? = ° is not possible. (c) 1.00 rev/s implies 1.00 sT = The 2 2cos 4 T g r ? ?= equation then says 2 2 2 (1.00 s) (9.80 m/s ) cos 2.48, 4 (0.100 m) ? ?= = which is not possible. The only way to have the m=?F a! ! equations satisfied is for sin 0.? = This means 0;? = the bead sits at the bottom of the hoop. 5-52 Chapter 5 EVALUATE: 90? ? ° as 0T ? (hoop moves faster). The largest value T can have is given by 2 2/(4 ) 1T g r? = so 2 / 0.635 s.T r g?= = This corresponds to a rotation rate of (1/ 0.635) rev/s 1.58 rev/s.= For a rotation rate less than 1.58 rev/s, 0? = is the only solution and the bead sits at the bottom of the hoop. Part (c) is an example of this. 5.116. IDENTIFY: 2 2x d x a dt = and 2 2y d y a dt = . Then apply m=?F a! ! to calculate the components of the net force. SET UP: The components of F! determine its magnitude and direction. EXECUTE: (a) Differentiating twice, 6xa ?t= ? and 2 ,ya ?= ? so (2.20 kg) ( 0.72 N s) (1.58 N/s)x xF ma t t= = ? = ? and 2(2.20 kg) ( 2.00 m s ) 4.40 Ny yF ma= = ? = ? . (b) The graph is given in Figure 5.116. (c) At 3.00 s, 4.75 N and 4.40 N,x yt F F= = ? = ? so 2 2( 4.75 N) ( 4.40 N) 6.48 NF = ? + ? = at an angle of ( )4.40arctan 223 .4.75? = °? EVALUATE: yF is constant and negative. xF is zero at 0t = and becomes increasingly more negative as t increases. Figure 5.116 5.117. IDENTIFY: The velocity is tangent to the path. The acceleration has a tangential component when the speed is changing and a radial component when the path is curving. SET UP: rada! is toward the center of curvature of the path. tana! is parallel to v! when the speed is increasing and antiparallel to v ! when the speed is decreasing. The net force F ! is proportional to a ! . EXECUTE: The diagram is sketched in Figure 5.117. EVALUATE: v! , a! , and F! all change during the motion. Figure 5.117 5.118. IDENTIFY: Apply m=?F a! ! to the car. It has acceleration rada! , directed toward the center of the circular path. SET UP: The analysis is the same as in Example 5.24. EXECUTE: (a) 2 2 2 (12.0 m/s)(1.60 kg) 9.80 m/s 61.8 N. 5.00 mA v F m g R ? ? ? ?= + = + =? ? ? ?? ? ? ? (b) 2 2 2 (12.0 m/s)(1.60 kg) 9.80 m/s 30.4 N. 5.00 mB v F m g R ? ? ? ?= ? = ? = ?? ? ? ?? ? ? ? , where the minus sign indicates that the track pushes down on the car. The magnitude of this force is 30.4 N. EVALUATE: A BF F> . 2AF mg? . Applying Newton?s Laws 5-53 5.119. IDENTIFY: The analysis is the same as for Problem 5.96. SET UP: The speed is related to the period by 2 2 (tan ) /v R T h T? ? ?= = , or 2 (tan ) /T h v? ?= . EXECUTE: The maximum and minimum speeds are the same as those found in Problem 5.96, s max s cos sin tan sin cos v gh ? ? ?? ? ? ? += ? and s min s cos sin tan sin cos v gh ? ? ?? ? ? ? ?= + . The minimum and maximum values of the period T are then s min s tan sin cos 2 cos sin h T g ? ? ? ?? ? ? ? ?= + and s max s tan sin cos 2 cos sin h T g ? ? ? ?? ? ? ? += ? . EVALUATE: For s 0? = the results for the speeds reduce to min maxv v gh= = . tan R h ?= . The result for v then agrees with the result in Example 5.23, if we take into account that in this problem ? is measured from the vertical whereas in Example 5.23 it is measured relative to the horizontal. 5.120. IDENTIFY: Apply m=?F a! ! to the block and to the wedge. SET UP: For both parts, take the x-direction to be horizontal and positive to the right, and the y-direction to be vertical and positive upward. The normal force between the block and the wedge is n ; the normal force between the wedge and the horizontal surface will not enter, as the wedge is presumed to have zero vertical acceleration. The horizontal acceleration of the wedge is A, and the components of acceleration of the block are xa and ya . EXECUTE: (a) The equations of motion are then sinMA n ?= ? , sinxma n ?= and cosyma n mg?= ? . Note that the normal force gives the wedge a negative acceleration; the wedge is expected to move to the left. These are three equations in four unknowns, A, , x ya a and n . Solution is possible with the imposition of the relation between A, xa and ya . An observer on the wedge is not in an inertial frame, and should not apply Newton?s laws, but the kinematic relation between the components of acceleration are not so restricted. To such an observer, the vertical acceleration of the block is ,ya but the horizontal acceleration of the block is .xa A? To this observer, the block descends at an angle ,? so the relation needed is tan .y x a ? a A = ?? At this point, algebra is unavoidable. A possible approach is to eliminate xa by noting that x M a A m = ? , using this in the kinematic constraint to eliminate ya and then eliminating n . The results are: ( ) tan ( tan ) gm A M m M? ? ?= + + ( ) tan ( tan )x gM a M m M? ?= + + ( m) tan ( ) tan ( tan )y g M a M m M ? ? ? ? += + + (b) When , 0,M m A>> ? as expected (the large block won?t move). Also, 2 tan sin cos tan (1 tan ) tan 1x g a g g ? ? ?? ? ?? = =+ + which is the acceleration of the block ( sing ? in this case), with the factor of cos? giving the horizontal component. Similarly, 2sinya g ?? ? . (c) The trajectory is a spiral. EVALUATE: If m M>> , our general results give 0xa = and ya g= ? . The massive block accelerates straight downward, as if it were in free-fall. 5.121. IDENTIFY: Apply m=?F a! ! to the block and to the wedge. SET UP: From Problem 5.120, sinxma n ?= and cosyma n mg?= ? for the block. 0ya = gives tanxa g ?= . EXECUTE: If the block is not to move vertically, both the block and the wedge have this horizontal acceleration and the applied force must be ( ) ( ) tanF M m a M m g ?= + = + . EVALUATE: 0F ? as 0? ? and F ? ? as ? ? ? . 5-54 Chapter 5 5.122. IDENTIFY: Apply m=?F a! ! . SET UP: Let x+ be directed up the ramp. EXECUTE: The normal force that the ramp exerts on the box will be cos sinn w T? ?= ? . The rope provides a force of cosT ? up the ramp, and the component of the weight down the ramp is sinw ? . Thus, the net force up the ramp is k k kcos sin ( cos sin ) (cos sin ) (sin cos )F T w w T T w? ? ? ? ? ? ? ? ? ? ?= ? ? ? = + ? + The acceleration will be the greatest when the first term in parentheses is greatest and this occurs when ktan .? ?= EVALUATE: Small ? means F is more nearly in the direction of the motion. But 90? ? °means F is directed to reduce the normal force and thereby reduce friction. The optimum value of ? is somewhere in between and depends on k? . When k 0? = , the optimum value of ? is 0? = ° . 5.123. IDENTIFY: Use the results of Problem 5.44. SET UP: ( )f x is a minimum when 0df dx = and 2 2 0 d f dx > . EXECUTE: (a) k k/(cos sin )F w? ? ? ?= + (b) The graph of F versus ? is given in Figure 5.123. (c) F is minimized at ktan .? ?= For k 0.25? = , 14.0? = ° . EVALUATE: Small ? means F is more nearly in the direction of the motion. But 90? ? °means F is directed to reduce the normal force and thereby reduce friction. The optimum value of ? is somewhere in between and depends on k? . Figure 5.123 5.124. IDENTIFY: Apply m=?F a! ! to the ball. At the terminal speed, 0a = . SET UP: For convenience, take the positive direction to be down, so that for the baseball released from rest, the acceleration and velocity will be positive, and the speed of the baseball is the same as its positive component of velocity. Then the resisting force, directed against the velocity, is upward and hence negative. EXECUTE: (a) The free-body diagram for the falling ball is sketched in Figure 5.124. (b) Newton?s Second Law is then 2.ma mg Dv= ? Initially, when 0,v = the acceleration is g , and the speed increases. As the speed increases, the resistive force increases and hence the acceleration decreases. This continues as the speed approaches the terminal speed. (c) At terminal velocity, 0,a = so t mgv D= in agreement with Eq. (5.13). (d) The equation of motion may be rewritten as 2 2t2 t ( ) dv g v v dt v = ? . This is a separable equation and may be expressed as 2 2 2 t t dv g dt v v v =?? ? or 2t t t 1 arctanh . v gt v v v ? ? =? ?? ? ( )t ttanh .v v gt v= EVALUATE: tanh x x x x e e x e e ? ? ?= + . At 0t ? , ttanh( / ) 0gt v ? and 0v ? . At t ? ? , ttanh( / )gt v ? ? and tv v? . Figure 5.124 Applying Newton?s Laws 5-55 5.125. IDENTIFY: Apply m=?F a! ! to each of the three masses and to the pulley B. SET UP: Take all accelerations to be positive downward. The equations of motion are straightforward, but the kinematic relations between the accelerations, and the resultant algebra, are not immediately obvious. If the acceleration of pulley B is ,Ba then 3,Ba a= ? and Ba is the average of the accelerations of masses 1 and 2, or 1 2 32 2 .Ba a a a+ = = ? EXECUTE: (a) There can be no net force on the massless pulley B, so 2 .C AT T= The five equations to be solved are then 1 1 1Am g T m a? = , 2 2 2Am g T m a? = , 3 3 3Cm g T m a? = , 1 2 32 0a a a+ + = and 2 0A CT T? = . These are five equations in five unknowns, and may be solved by standard means. The accelerations 1a and 2a may be eliminated by using 3 1 2 1 22 ( ) (2 ((1 ) (1 ))).Aa a a g T m m= ? + = ? ? + The tension AT may be eliminated by using 3 3(1 2) (1 2) ( ).A CT T m g a= = ? Combining and solving for 3a gives 1 2 2 3 1 3 3 1 2 2 3 1 3 4 . 4 mm m m mm a g m m m m mm ? + += + + (b) The acceleration of the pulley B has the same magnitude as 3a and is in the opposite direction. (c) 31 3 1 1 1 ( ). 2 2 A CT T ma g g g g a m m m = ? = ? = ? ? Substituting the above expression for 3a gives 1 2 2 3 1 3 1 1 2 2 3 1 3 4 3 . 4 mm m m mm a g m m m m mm ? += + + (d) A similar analysis (or, interchanging the labels 1 and 2) gives 1 2 1 3 2 32 1 2 2 3 1 3 4 3 . 4 mm mm m m a g m m m m mm ? += + + (e), (f) Once the accelerations are known, the tensions may be found by substitution into the appropriate equation of motion, giving 1 2 3 1 2 3 1 2 2 3 1 3 1 2 2 3 1 3 4 8 , . 4 4A C m m m mm m T g T g m m m m mm mm m m mm = =+ + + + (g) If 1 2m m m= = and 3 2 ,m m= all of the accelerations are zero, 2CT mg= and .AT mg= All masses and pulleys are in equilibrium, and the tensions are equal to the weights they support, which is what is expected. EVALUATE: It is useful to consider special cases. For example, when 1 2 3m m m= >> our general result gives 1 2a a g= = + and 3a g= ? . 5.126. IDENTIFY: Apply m=?F a! ! to each block. The tension in the string is the same at both ends. If T w< for a block, that block remains at rest. SET UP: In all cases, the tension in the string will be half of F. EXECUTE: (a) 2 62 N,F = which is insufficient to raise either block; 1 2 0.a a= = (b) 2 62 N.F = The larger block (of weight 196 N) will not move, so 1 0,a = but the smaller block, of weight 98 N, has a net upward force of 49 N applied to it, and so will accelerate upwards with 22 49 N 4.9 m s . 10.0 kg a = = (c) 2 212 N,F = so the net upward force on block A is 16 N and that on block B is 114 N, so 2 1 16 N 0.8 m s 20.0 kg a = = and 22 114 N 11.4 m s .10.0 kga = = EVALUATE: The two blocks need not have accelerations with the same magnitudes. 5.127. IDENTIFY: Apply m=?F a! ! to the ball at each position. SET UP: When the ball is at rest, 0a = . When the ball is swinging in an arc it has acceleration component 2 rad v a R = , directed inward. EXECUTE: Before the horizontal string is cut, the ball is in equilibrium, and the vertical component of the tension force must balance the weight, so cosAT w? = or cosAT w ?= . At point B, the ball is not in equilibrium; its speed is instantaneously 0, so there is no radial acceleration, and the tension force must balance the radial component of the weight, so cosBT w ?= and the ratio 2( ) cosB AT T ?= . EVALUATE: At point B the net force on the ball is not zero; the ball has a tangential acceleration. 6-1 W ORK AND K INETIC ENERGY 6.1. IDENTIFY: Apply Eq.(6.2). SET UP: The bucket rises slowly, so the tension in the rope may be taken to be the bucket?s weight. EXECUTE: (a) 2(6.75 kg) (9.80 m /s )(4.00 m) 265 J.W Fs mgs= = = = (b) Gravity is directed opposite to the direction of the bucket? s motion, so Eq.(6.2) gives the negative of the result of part (a), or 265 J? . (c) The total work done on the bucket is zero. EVALUATE: When the force is in the direction of the displacement, the force does positive work. When the force is directed opposite to the displacement, the force does negative work. 6.2. IDENTIFY: In each case the forces are constant and the displacement is along a straight line, so cosW Fs ?= . SET UP: In part (a), when the cable pulls horizontally 0? = ° and when it pulls at 35.0° above the horizontal 35.0? = ° . In part (b), if the cable pulls horizontally 180? = ° . If the cable pulls on the car at 35.0° above the horizontal it pulls on the truck at 35.0° below the horizontal and 145.0? = ° . For the gravity force 90? = ° , since the force is vertical and the displacement is horizontal. EXECUTE: (a) When the cable is horizontal, 3 6(850 N)(5.00 10 m)cos0 4.25 10 JW = × = ×° . When the cable is 35.0° above the horizontal, 3 6(850 N)(5.00 10 m)cos35.0 3.48 10 JW = × = ×° . (b) cos180 cos0= ?° ° and cos145.0 cos35.0= ?° ° , so the answers are 64.26 10 J? × and 63.48 10 J? × . (c) Since cos cos90 0? = =° , 0W = in both cases. EVALUATE: If the car and truck are taken together as the system, the tension in the cable does no net work. 6.3. IDENTIFY: Each force can be used in the relation ( cos )W F s F s?= =? for parts (b) through (d). For part (e), apply the net work relation as net worker grav .n fW W W W W= + + + SET UP: In order to move the crate at constant velocity, the worker must apply a force that equals the force of friction, worker k k .F f n?= = EXECUTE: (a) The magnitude of the force the worker must apply is: ( )( )( )2worker k k k 0.25 30.0 kg 9.80 m/s 74 NF f n mg? ?= = = = = (b) Since the force applied by the worker is horizontal and in the direction of the displacement, 0? = ° and the work is: ( ) ( )( )[ ]( )worker worker cos 74 N cos0 4.5 m 333 JW F s?= = = +° (c) Friction acts in the direction opposite of motion, thus 180? = ° and the work of friction is: ( ) ( )( )[ ]( )k cos 74 N cos180 4.5 m 333 JfW f s?= = = ?° (d) Both gravity and the normal force act perpendicular to the direction of displacement. Thus, neither force does any work on the crate and grav 0.0 J.nW W= = (e) Substituting into the net work relation, the net work done on the crate is: net worker grav 333 J 0.0 J 0.0 J 333 J 0.0 Jn fW W W W W= + + + = + + + ? = EVALUATE: The net work done on the crate is zero because the two contributing forces, worker and ,fF F are equal in magnitude and opposite in direction. 6.4. IDENTIFY: The forces are constant so Eq.(6.2) can be used to calculate the work. Constant speed implies 0.a = We must use m=?F a? ? applied to the crate to find the forces acting on it. 6 6-2 Chapter 6 (a) SET UP: The free-body diagram for the crate is given in Figure 6.4. EXECUTE: y yF ma=? sin30 0n mg F? ? ° = sin30n mg F= + ° k k k k sin30f n mg F? ? ?= = + ° Figure 6.4 x xF ma=? kcos30 0F f° ? = k kcos30 sin30 0F mg F? ?° ? ? ° = 2 k k 0.25(30.0 kg)(9.80 m/s ) 99.2 N cos30 sin30 cos30 (0.25)sin30 mg F ? ?= = =° ? ° ° ? ° (b) ( cos ) (99.2 N)(cos30 )(4.5 m) 387 JFW F s?= = ° = ( cos30F ° is the horizontal component of ;F? the work done by F? is the displacement times the component of F? in the direction of the displacement.) (c) We have an expression for kf from part (a): 2 k k ( sin30 ) (0.250)[(30.0 kg)(9.80 m/s ) (99.2 N)(sin30 )] 85.9 Nf mg F?= + ° = + ° = 180? = ° since kf is opposite to the displacement. Thus k( cos ) (85.9 N)(cos180 )(4.5 m) 387 JfW f s?= = ° = ? (d) The normal force is perpendicular to the displacement so 90? = ° and 0.nW = The gravity force (the weight) is perpendicular to the displacement so 90? = ° and 0wW = (e) tot 387 J ( 387 J) 0F f n wW W W W W= + + + = + + ? = EVALUATE: Forces with a component in the direction of the displacement do positive work, forces opposite to the displacement do negative work and forces perpendicular to the displacement do zero work. The total work, obtained as the sum of the work done by each force, equals the work done by the net force. In this problem, net 0F = since 0a = and tot 0,W = which agrees with the sum calculated in part (e). 6.5. IDENTIFY: The gravity force is constant and the displacement is along a straight line, so cosW Fs ?= . SET UP: The displacement is upward along the ladder and the gravity force is downward, so 180.0 30.0 150.0? = ? =° ° ° . 735 Nw mg= = . EXECUTE: (a) (735 N)(2.75 m)cos150.0 1750 JW = = ?° . (b) No, the gravity force is independent of the motion of the painter. EVALUATE: Gravity is downward and the vertical component of the displacement is upward, so the gravity force does negative work. 6.6. IDENTIFY and SET UP: ( cos ) ,FW F s?= since the forces are constant. We can calculate the total work by summing the work done by each force. The forces are sketched in Figure 6.6. EXECUTE: 1 1 1cosW F s ?= 6 3 1 (1.80 10 N)(0.75 10 m)cos14W = × × ° 9 1 1.31 10 JW = × 2 2 2 1cosW F s W?= = Figure 6.6 9 9 tot 1 2 2(1.31 10 J) 2.62 10 JW W W= + = × = × EVALUATE: Only the component cosF ? of force in the direction of the displacement does work. These components are in the direction of s ? so the forces do positive work. Work and Kinetic Energy 6-3 6.7. IDENTIFY: All forces are constant and each block moves in a straight line. so cosW Fs ?= . The only direction the system can move at constant speed is for the 12.0 N block to descend and the 20.0 N block to move to the right. SET UP: Since the 12.0 N block moves at constant speed, 0a = for it and the tension T in the string is 12.0 NT = . Since the 20.0 N block moves to the right at constant speed the friction force kf on it is to the left and k 12.0 Nf T= = . EXECUTE: (a) (i) 0? = ° and (12.0 N)(0.750 m)cos0 9.00 JW = =° . (ii) 180? = ° and (12.0 N)(0.750 m)cos180 9.00 JW = = ?° . (b) (i) 90? = ° and 0W = . (ii) 0? = ° and (12.0 N)(0.750 m)cos0 9.00 JW = =° . (iii) 180? = ° and (12.0 N)(0.750 m)cos180 9.00 JW = = ?° . (iv) 90? = ° and 0W = . (c) tot 0W = for each block. EVALUATE: For each block there are two forces that do work, and for each block the two forces do work of equal magnitude and opposite sign. When the force and displacement are in opposite directions, the work done is negative. 6.8. IDENTIFY: Apply Eq.(6.5). SET UP: ? ? ? ? 1? = ? =i i j j and ? ? ? ? 0? = ? =i j j i EXECUTE: The work you do is ? ? ? ?((30 N) (40 N) ) (( 9.0 m) (3.0 m) )? = ? ? ? ?F s i j i j? ? (30 N)( 9.0 m) ( 40 N)( 3.0 m) 270 N m 120 N m 150 J? = ? + ? ? = ? ? + ? = ?F s? ? . EVALUATE: The x-component of F? does negative work and the y-component of F? does positive work. The total work done by F ? is the sum of the work done by each of its components. 6.9. IDENTIFY: Apply Eq.(6.2) or (6.3). SET UP: The gravity force is in the -directiony? , so 2 1( )mg mg y y? = ? ?F s ? ? EXECUTE: (a) (i) Tension force is always perpendicular to the displacement and does no work. (ii) Work done by gravity is 2 1( ).mg y y? ? When 1 2y y= , 0mgW = . (b) (i) Tension does no work. (ii) Let l be the length of the string. 2 1( ) (2 ) 25.1 JmgW mg y y mg l= ? ? = ? = ? EVALUATE: In part (b) the displacement is upward and the gravity force is downward, so the gravity force does negative work. 6.10. IDENTIFY: 212K mv= SET UP: 65 mi/h 29.1 m/s= EXECUTE: (a) 2 512 (750 kg)(29.1 m/s) 3.18 10 JK = = × (b) 211 12K mv= . 212 22K mv= , with 2 1 / 2v v= , so 2 21 1 12 1 1 12 4 2( / 2) ( ) / 4K m v mv K= = = . The change in kinetic energy is a decrease of 3 14 K . (c) 12 12K K= . 2 constant2 K m v = = , so 1 22 2 1 2 K K v v = . 12 1 2 1 1 12/ (65 mi/h) / 46 mi/hv v K K K K= = = . EVALUATE: Since 2K v? , to have half the kinetic energy the speed must be less than half of the original speed. 6.11. IDENTIFY: 212K mv= . Since the meteor comes to rest the energy it delivers to the ground equals its original kinetic energy. SET UP: 412 km/s 1.2 10 m/sv = = × . A 1.0 megaton bomb releases 154.184 10 J× of energy. EXECUTE: (a) 8 4 2 1612 (1.4 10 kg)(1.2 10 m/s) 1.0 10 JK = × × = × . (b) 16 15 1.0 10 J 2.4 4.184 10 J × =× . The energy is equivalent to 2.4 one-megaton bombs. EVALUATE: Part of the energy transferred to the ground lifts soil and rocks into the air and creates a large crater. 6.12. IDENTIFY: 212K mv= . Use the equations for free-fall to find the speed of the weight when it reaches the ground. SET UP: Estimate that a person has speed 2 m/s when walking and 6 m/s when running. The mass of an electron is 319.11 10 kg?× . In part (c) take y+ downward, so 29.80 m/sya = + . Estimate a shoulder height of 1.6 m. EXECUTE: (a) Walking: 212 (75 kg)(2 m/s) 150 JK = = . Running: 212 (75 kg)(6 m/s) 1400 JK = = . (b) 31 6 2 1812 (9.11 10 kg)(2.19 10 m/s) 2.2 10 JK ? ?= × × = × . (c) 2 20 02 ( )y y yv v a y y= + ? gives 22(9.80 m/s )(1.6 m) 5.6 m/syv = = . 212 (1.0 kg)(5.6 m/s) 16 JK = = . 6-4 Chapter 6 (d) 2 2(100 J) 2.6 m/s 30 kg K v m = = = . Yes, this is reasonable. EVALUATE: A walking speed of 2 m/s corresponds to walking a mile in about 13 min. A running speed of 6 m/s corresponds to running a 100 m dash in about 17 s. 6.13. IDENTIFY: 212K mv= . Set up a ratio that relates K , m and v. SET UP: p e1836m m= EXECUTE: (a) p eK K= gives 2 2e e p pm v m v= . e p p e/ 1836 42.85v v m m V V= = = . (b) p ev v= gives p e p e K K m m = . p e p e( / ) 1836K K m m K= = . EVALUATE: The electron has less mass so must travel faster to have the same kinetic energy. And with equal speeds the proton has more kinetic energy. 6.14. IDENTIFY: Only gravity does work on the watermelon, so tot gravW W= . totW K= ? and 212K mv= . SET UP: Since the watermelon is dropped from rest, 1 0K = . EXECUTE: (a) 2grav (4.80 kg)(9.80 m/s )(25.0 m) 1180 JW mgs= = = (b) tot 2 1W K K= ? so 2 1180 JK = . 22 2(1180 J) 22.2 m/s4.80 kg K v m = = = . (c) The work done by gravity would be the same. Air resistance would do negative work and totW would be less than gravW . The answer in (a) would be unchanged and both answers in (b) would decrease. EVALUATE: The gravity force is downward and the displacement is downward, so gravity does positive work. 6.15. IDENTIFY: tot 2 1W K K= ? . In each case calculate totW from what we know about the force and the displacement. SET UP: The gravity force is mg , downward. The friction force is k k kf n mg? ?= = and is directed opposite to the displacement. The mass of the object isn't given, so we expect that it will divide out in the calculation. EXECUTE: (a) 1 0K = . tot gravW W mgs= = . 21 22mgs mv= and 22 2 2(9.80 m/s )(95.0 m) 43.2 m/sv gs= = = . (b) 2 0K = (at the maximum height). tot gravW W mgs= = ? . 21 12mgs mv? = ? and 2 1 2 2(9.80 m/s )(525 m) 101 m/sv gs= = = . (c) 211 12K mv= . 2 0K = . tot kfW W mgs?= = ? . 21k 12mgs mv?? = ? . 2 2 1 2 k (5.00 m/s) 5.80 m 2 2(0.220)(9.80 m/s ) v s g?= = = . (d) 211 12K mv= . 212 22K mv= . tot kfW W mgs?= = ? . 2 tot 1K W K= + . 2 21 12 k 12 2mv mgs mv?= ? + 2 2 2 2 1 k2 (5.00 m/s) 2(0.220)(9.80 m/s )(2.90 m) 3.53 m/sv v gs?= ? = ? = . (e) 211 12K mv= . 2 0K = . grav 2W mgy= ? , where 2y is the vertical height. 212 12mgy mv? = ? and 2 2 1 2 2 (12.0 m/s) 7.35 m 2 2(9.80 m/s ) v y g = = = . EVALUATE: In parts (c) and (d), friction does negative work and the kinetic energy is reduced. In part (a), gravity does positive work and the speed increases. In parts (b) and (e), gravity does negative work and the speed decreases. The vertical height in part (e) is independent of the slope angle of the hill. 6.16. IDENTIFY: From the work-energy relation, grav rockW W K= = ? . SET UP: As the rock rises, the gravitational force, ,F mg= does work on the rock. Since this force acts in the direction opposite to the motion and displacement, s , the work is negative. Let h be the vertical distance the rock travels. EXECUTE: (a) Applying grav 2 1W K K= ? we obtain 2 21 12 12 2mgh mv mv? = ? . Dividing by m and solving for 1v , 2 1 2 2v v gh= + . Substituting 215.0 m and 25.0 m/s,h v= = ( ) ( )( )2 2 1 25.0 m/s 2 9.80 m/s 15.0 m 30.3 m/sv = + = (b) Solve the same work-energy relation for h. At the maximum height 2 0v = . 2 21 1 2 12 2mgh mv mv? = ? and ( ) ( ) ( ) 2 22 2 1 2 2 30.3 m/s 0.0 m/s 46.8 m 2 2 9.80 m/s v v h g ? ?= = = . Work and Kinetic Energy 6-5 EVALUATE: Note that the weight of 20 N was never used in the calculations because both gravitational potential and kinetic energy are proportional to mass, m. Thus any object, that attains 25.0 m/s at a height of 15.0 m, must have an initial velocity of 30.3 m/s. As the rock moves upward gravity does negative work and this reduces the kinetic energy of the rock. 6.17. IDENTIFY and SET UP: Apply Eq.(6.6) to the box. Let point 1 be at the bottom of the incline and let point 2 be at the skier. Work is done by gravity and by friction. Solve for 1K and from that obtain the required initial speed. EXECUTE: tot 2 1W K K= ? 21 1 02 ,K mv= 2 0K = Work is done by gravity and friction, so tot .mg fW W W= + 2 1( )mgW mg y y mgh= ? ? = ? .fW fs= ? The normal force is cosn mg ?= and / sin ,s h ?= where s is the distance the box travels along the incline. k k( cos )( / sin ) / tanfW mg h mgh? ? ? ? ?= ? = ? Substituting these expressions into the work-energy theorem gives 21 k 02/ tan .mgh mgh mv? ?? ? = ? Solving for 0v then gives 0 k2 (1 / tan ).v gh ? ?= + EVALUATE: The result is independent of the mass of the box. As 90 ,? ? ° h s= and 0 2 ,v gh= the same as throwing the box straight up into the air. For 90? = ° the normal force is zero so there is no friction. 6.18. IDENTIFY: Apply cosW Fs ?= and totW K= ? . SET UP: Parallel to incline: force component sinW mg ?=? , down incline; displacement sins h ?= / , down incline. Perpendicular to the incline: 0s = . EXECUTE: (a) || ( sin )( / sin )W mg h mgh? ?= = . 0W? = , since there is no displacement in this direction. ||mgW W W mgh?= + = , same as falling height h. (b) tot 2 1W K K= ? gives 212mgh mv= and 2v gh= , same as if had been dropped from height h. The work done by gravity depends only on the vertical displacement of the object. When the slope angle is small, there is a small force component in the direction of the displacement but a large displacement in this direction. When the slope angle is large, the force component in the direction of the displacement along the incline is larger but the displacement in this direction is smaller. (c) 15.0 mh = , so 2 17.1 sv gh= = . EVALUATE: The acceleration and time of travel are different for an object sliding down an incline and an object in free-fall, but the final velocity is the same in these two cases. 6.19. IDENTIFY: tot 2 1W K K= ? with tot fW W= . The car stops, so 2 0K = . In each case identify what is constant and set up a ratio. SET UP: fW fs= ? , so 21 02fs mv? = ? . EXECUTE: (a) 0 03b av v= . as D= . f is constant. 2 0 2 constant v f s m = = , so 2 2 0 0a b a b v v s s = . 2 20 0 (3) 9bb a a v s s D D v ? ?= = =? ?? ? . (b) 3b af f= . 0v is constant. 21 02 constantfs mv= = , so a a b bf s f s= . /3ab a b f s s D f ? ?= =? ?? ? . EVALUATE: The stopping distance is proportional to the square of the initial speed. When the friction force increases, the stopping distance decreases. 6.20. IDENTIFY and SET UP: Apply Eq.(6.6). The relation between the speeds 1v and 2v tells us the relation between 1K and 2.K EXECUTE: (a) 2 1W K K= ? 21 1 12 ,K mv= 212 22K mv= 1 2 14v v= gives that ( ) ( )2 21 1 1 1 12 1 1 12 4 16 2 16K m v mv K= = = 151 2 1 1 1 116 16W K K K K K= ? = ? = ? (b) EVALUATE: K depends only on the magnitude of v? not on its direction, so the answer for W in part (a) does not depend on the final direction of the electron?s motion. The electron slows down, so its kinetic energy decreases and the total work done on it is negative. 6-6 Chapter 6 6.21. IDENTIFY: Apply cosW Fs ?= and totW K= ? . SET UP: 0? = ° EXECUTE: From Equations (6.1), (6.5) and (6.6), and solving for F, 2 2 2 21 1 2 12 2( ) (8.00 kg)((6.00 m /s) (4.00 m /s) ) 32.0 N. (2.50 m) m v vK F s s ? ??= = = = EVALUATE: The force is in the direction of the displacement, so the force does positive work and the kinetic energy of the object increases. 6.22. IDENTIFY and SET UP: Use Eq.(6.6) to calculate the work done by the foot on the ball. Then use Eq.(6.2) to find the distance over which this force acts. EXECUTE: tot 2 1W K K= ? 2 21 1 1 12 2 (0.420 kg)(2.00 m/s) 0.84 JK mv= = = 2 21 1 2 22 2 (0.420 kg)(6.00 m/s) 7.56 JK mv= = = tot 2 1 7.56 J 0.84 J 6.72 JW K K= ? = ? = The 40.0 N force is the only force doing work on the ball, so it must do 6.72 J of work. ( cos )FW F s?= gives that 6.72 J 0.168 m cos (40.0 N)(cos0) W s F ?= = = EVALUATE: The force is in the direction of the motion so positive work is done and this is consistent with an increase in kinetic energy. 6.23. IDENTIFY: Apply totW K= ? . SET UP: 1 0v = , 2v v= . k kf mg?= and kf does negative work. The force 36.0 NF = is in the direction of the motion and does positive work. EXECUTE: (a) If there is no work done by friction, the final kinetic energy is the work done by the applied force, and solving for the speed, 2 2 2(36.0 N)(1.20 m) 4.48 m /s. (4.30 kg) W Fs v m m = = = = (b) The net work is k k( )Fs f s F mg s?? = ? , so 2 k2( ) 2(36.0 N (0.30)(4.30 kg)(9.80 m /s ))(1.20 m) 3.61 m/s (4.30 kg) F mg s v m ?? ?= = = EVALUATE: The total work done is larger in the absence of friction and the final speed is larger in that case. 6.24. IDENTIFY: Apply cosW Fs ?= and totW K= ? SET UP: The gravity force has magnitude mg and is directed downward. EXECUTE: (a) On the way up, gravity is opposed to the direction of motion, and so 2(0.145 kg)(9.80 m /s )(20.0 m) 28.4 JW mgs= ? = ? = ? . (b) 2 22 1 2( 28.4 J) 2 (25.0 m /s) 15.3 m /s (0.145 kg) W v v m ?= + = + = . (c) No; in the absence of air resistance, the ball will have the same speed on the way down as on the way up. On the way down, gravity will have done both negative and positive work on the ball, but the net work at this height will be the same. EVALUATE: As the baseball moves upward, gravity does negative work and the speed of the baseball decreases. 6.25. (a) IDENTIFY and SET UP: Use Eq.(6.2) to find the work done by the positive force. Then use Eq.(6.6) to find the final kinetic energy, and then 212 22K mv= gives the final speed. EXECUTE: tot 2 1,W K K= ? so 2 tot 1K W K= + 2 21 1 1 12 2 (7.00 kg)(4.00 m/s) 56.0 JK mv= = = The only force that does work on the wagon is the 10.0 N force. This force is in the direction of the displacement so 0? = ° and the force does positive work: ( cos ) (10.0 N)(cos0)(3.0 m) 30.0 JFW F s?= = = Then 2 tot 1 30.0 J 56.0 J 86.0 J.K W K= + = + = 21 2 22 ;K mv= 22 2 2(86.0 J) 4.96 m/s7.00 kg K v m = = = Work and Kinetic Energy 6-7 (b) IDENTIFY: Apply m=?F a? ? to the wagon to calculate a. Then use a constant acceleration equation to calculate the final speed. The free-body diagram is given in Figure 6.25. SET UP: EXECUTE: x xF ma=? xF ma= 210.0 N 1.43 m/s 7.00 kgx F a m = = = Figure 6.25 2 2 2 1 2 02 ( )x xv v a x x= + ? 2 2 2 2 1 02 ( ) (4.00 m/s) 2(1.43 m/s )(3.0 m) 4.96 m/sx x xv v a x x= + ? = + = EVALUATE: This agrees with the result calculated in part (a). The force in the direction of the motion does positive work and the kinetic energy and speed increase. In part (b), the equivalent statement is that the force produces an acceleration in the direction of the velocity and this causes the magnitude of the velocity to increase. 6.26. IDENTIFY: Apply tot 2 1W K K= ? . SET UP: 1 0K = . The normal force does no work. The work W done by gravity is W mgh= , where sinh L ?= is the vertical distance the block has dropped when it has traveled a distance L down the incline and ? is the angle the plane makes with the horizontal. EXECUTE: The work-energy theorem gives 2 2 2 2 sinK Wv gh gL m m ?= = = = . Using the given numbers, 22(9.80 m /s )(0.75 m)sin36.9 2.97 m /s.v = ° = EVALUATE: The final speed of the block is the same as if it had been dropped from a height h. 6.27. IDENTIFY: tot 2 1W K K= ? . Only friction does work. SET UP: ktot kf W W mgs?= = ? . 2 0K = (car stops). 211 02K mv= . EXECUTE: (a) tot 2 1W K K= ? gives 21k 02mgs mv?? = ? . 2 0 k2 v s g?= . (b) (i) k k2b a? ?= . 2 0 k constant2 v s g ? = = so k ka a b bs s? ?= . k k / 2ab a a b s s s ? ? ? ?= =? ?? ? . The minimum stopping distance would be halved. (ii) 0 02b av v= . 2 0 k 1 constant 2 s v g?= = , so 2 20 0 a b a b s s v v = . 2 0 0 4bb a a a v s s s v ? ?= =? ?? ? . The stopping distance would become 4 times as great. (iii) 0 02b av v= , k k2b a? ?= . k2 0 1 constant 2 s v g ? = = , so k k2 2 0 0 a a b b a b s s v v ? ?= . 2 2k 0 k 0 1 (2) 2 2 a b b a a a b a v s s s s v ? ? ? ?? ? ? ?= = =? ?? ? ? ?? ?? ?? ? . The stopping distance would double. EVALUATE: The stopping distance is directly proportional to the square of the initial speed and indirectly proportional to the coefficient of kinetic friction. 6.28. IDENTIFY: The work that must be done to move the end of a spring from 1x to 2x is 2 21 12 12 2W kx kx= ? . The force required to hold the end of the spring at displacement x is xF kx= . SET UP: When the spring is at its unstretched length, 0x = . When the spring is stretched, 0x > , and when the spring is compressed, 0x < . EXECUTE: (a) 1 0x = and 21 22W kx= . 42 2 2 2 2(12.0 J) 2.67 10 N/m (0.0300 m) W k x = = = × . (b) 4(2.67 10 N/m)(0.0300 m) 801 NxF kx= = × = . (c) 1 0x = , 2 0.0400 mx = ? . 4 212 (2.67 10 N/m)( 0.0400 m) 21.4 JW = × ? = . 4(2.67 10 N/m)(0.0400 m) 1070 NxF kx= = × = . EVALUATE: When a spring, initially unstretched, is either compressed or stretched, positive work is done by the force that moves the end of the spring. 6-8 Chapter 6 6.29. IDENTIFY and SET UP: Use Eq.(6.8) to calculate k for the spring. Then Eq.(6.10), with 1 0,x = can be used to calculate the work done to stretch or compress the spring an amount 2.x EXECUTE: Use the information given to calculate the force constant of the spring. xF kx= gives 160 N 3200 N/m0.050 m xFk x = = = (a) (3200 N/m)(0.015 m) 48 NxF kx= = = (3200 N/m)( 0.020 m) 64 NxF kx= = ? = ? (magnitude 64 N) (b) 2 21 12 2 (3200 N/m)(0.015 m) 0.36 JW kx= = = 2 21 1 2 2 (3200 N/m)( 0.020 m) 0.64 JW kx= = ? = Note that in each case the work done is positive. EVALUATE: The force is not constant during the displacement so Eq.(6.2) cannot be used. A force in the x+ direction is required to stretch the spring and a force in the opposite direction to compress it. The force xF is in the same direction as the displacement, so positive work is done in both cases. 6.30. IDENTIFY: The magnitude of the work can be found by finding the area under the graph. SET UP: The area under each triangle is 1/2 base height× . 0xF > , so the work done is positive when x increases during the displacement. EXECUTE: (a) 1/ 2 (8 m)(10 N) 40 J= . (b) 1/ 2 (4 m)(10 N) 20 J= . (c) 1/ 2 (12 m)(10 N) 60 J= . EVALUATE: The sum of the answers to parts (a) and (b) equals the answer to part (c). 6.31. IDENTIFY: Use the work-energy theorem and the results of Problem 6.30. SET UP: For 0x = to 8.0 mx = , tot 40 JW = . For 0x = to 12.0 mx = , tot 60 JW = . EXECUTE: (a) (2)(40 J) 2.83 m /s 10 kg v = = (b) (2)(60 J) 3.46 m /s 10 kg v = = . EVALUATE: F? is always in the -direction.x+ For this motion F? does positive work and the speed continually increases during the motion. 6.32. IDENTIFY: The force has only an x-component and the motion is along the x-direction, so 2 1 x xx W F dx= ? . SET UP: 1 0x = and 2 6.9 mx = . EXECUTE: The work you do with your changing force is 2 2 2 2 2 1 1 1 1 1 2( ) ( 20.0 N) (3.0 N/m) ( 20.0 N) | (3.0 N/m)( /2) | x x x x x x xx x x W F x dx dx xdx x x= = ? ? = ? ?? ? ? 138 N m 71.4 N m 209 JW = ? ? ? ? = ? . EVALUATE: The work is negative because the cow continues to move forward (in the -directionx+ ) as you vainly attempt to push her backward. 6.33. IDENTIFY: Apply Eq.(6.6) to the box. SET UP: Let point 1 be just before the box reaches the end of the spring and let point 2 be where the spring has maximum compression and the box has momentarily come to rest. EXECUTE: tot 2 1W K K= ? 21 1 02 ,K mv= 2 0K = Work is done by the spring force. 21tot 22 ,W kx= ? where 2x is the amount the spring is compressed. 2 21 1 2 02 2kx mv? = ? and 2 0 / (3.0 m/s) (6.0 kg)/(7500 N/m) 8.5 cmx v m k= = = EVALUATE: The compression of the spring increases when either 0v or m increases and decreases when k increases (stiffer spring). 6.34. IDENTIFY: The force applied to the springs is xF kx= . The work done on a spring to move its end from 1x to 2x is 2 21 1 2 12 2W kx kx= ? . Use the information that is given to calculate k . SET UP: When the springs are compressed 0.200 m from their uncompressed length, 1 0x = and 2 0.200 mx = ? . When the platform is moved 0.200 m farther, 2x becomes 0.400 m? . Work and Kinetic Energy 6-9 EVALUATE: (a) 2 2 2 2 1 2 2(80.0 J) 4000 N/m (0.200 m) 0 W k x x = = =? ? . (4000 N/m)( 0.200 m) 800 NxF kx= = ? = ? . The magnitude of force that is required is 800 N. (b) To compress the springs from 1 0x = to 2 0.400 mx = ? , the work required is 2 2 21 1 1 2 12 2 2 (4000 N/m)( 0.400 m) 320 JW kx kx= ? = ? = . The additional work required is 320 J 80 J 240 J? = . For 0.400 mx = ? , 1600 NxF kx= = ? . The magnitude of force required is 1600 N. EVALUATE: More work is required to move the end of the spring from 0.200 mx = ? to 0.400 mx = ? than to move it from 0x = to 0.200 mx = ? , even though the displacement of the platform is the same in each case. The magnitude of the force increases as the compression of the spring increases. 6.35. IDENTIFY: Apply m=?F a? ? to calculate the s? required for the static friction force to equal the spring force. SET UP: (a) The free-body diagram for the glider is given in Figure 6.35. EXECUTE: y yF ma=? 0n mg? = n mg= s sf mg?= Figure 6.35 x xF ma=? s spring 0f F? = s 0mg kd? ? = s 2 (20.0 N/m)(0.086 m) 1.76 (0.100 kg)(9.80 m/s ) kd mg ? = = = (b) IDENTIFY and SET UP: Apply m=?F a? ? to find the maximum amount the spring can be compressed and still have the spring force balanced by friction. Then use tot 2 1W K K= ? to find the initial speed that results in this compression of the spring when the glider stops. EXECUTE: smg kd? = 2 s (0.60)(0.100 kg)(9.80 m/s ) 0.0294 m 20.0 N/m mg d k ?= = = Now apply the work-energy theorem to the motion of the glider: tot 2 1W K K= ? 21 1 12 ,K mv= 2 0K = (instantaneously stops) 21 tot spring fric k2W W W kd mgd?= + = ? ? (as in Example 6.8) 2 21 tot 2 (20.0 N/m)(0.0294 m) 0.47(0.100 kg)(9.80 m/s )(0.0294 m) 0.02218 JW = ? ? = ? Then tot 2 1W K K= ? gives 21 120.02218 J .mv? = ? 1 2(0.02218 J) 0.67 m/s 0.100 kg v = = EVALUATE: In Example 6.8 an initial speed of 1.50 m/s compresses the spring 0.086 m and in part (a) of this problem we found that the glider doesn?t stay at rest. In part (b) we found that a smaller displacement of 0.0294 m when the glider stops is required if it is to stay at rest. And we calculate a smaller initial speed (0.67 m/s) to produce this smaller displacement. 6.36. IDENTIFY: For the spring, 2 21 11 22 2W kx kx= ? . Apply tot 2 1W K K= ? . SET UP: 1 0.025 mx = ? and 2 0x = . EXECUTE: (a) 2 21 112 2 (200 N / m)( 0.025 m) 0.060 JW kx= = ? = . (b) The work-energy theorem gives 2 2 2(0.060 J) 0.18 m /s. (4.0 kg) W v m = = = EVALUATE: The block moves in the direction of the spring force, the spring does positive work and the kinetic energy of the block increases. 6-10 Chapter 6 6.37. IDENTIFY and SET UP: The magnitude of the work done by xF equals the area under the xF versus x curve. The work is positive when xF and the displacement are in the same direction; it is negative when they are in opposite directions. EXECUTE: (a) xF is positive and the displacement x? is positive, so 0.W > 1 2 (2.0 N)(2.0 m) (2.0 N)(1.0 m) 4.0 JW = + = + (b) During this displacement 0,xF = so 0.W = (c) xF is negative, x? is positive, so 0.W < 12 (1.0 N)(2.0 m) 1.0 JW = ? = ? (d) The work is the sum of the answers to parts (a), (b), and (c), so 4.0 J 0 1.0 J 3.0 JW = + ? = + (e) The work done for 7.0 mx = to 3.0 mx = is 1.0 J.+ This work is positive since the displacement and the force are both in the -direction.x? The magnitude of the work done for 3.0 mx = to 2.0 mx = is 2.0 J, the area under xF versus x. This work is negative since the displacement is in the -directionx? and the force is in the -direction.x+ Thus 1.0 J 2.0 J 1.0 JW = + ? = ? EVALUATE: The work done when the car moves from 2.0 mx = to 0x = is 12 (2.0 N)(2.0 m) 2.0 J.? = ? Adding this to the work for 7.0 mx = to 2.0 mx = gives a total of 3.0 JW = ? for 7.0 mx = to 0.x = The work for 7.0 mx = to 0x = is the negative of the work for 0x = to 7.0 m.x = 6.38. IDENTIFY: Apply tot 2 1W K K= ? . SET UP: 1 0K = . From Exercise 6.37, the work for 0x = to 3.0 mx = is 4.0 J. W for 0x = to 4.0 mx = is also 4.0 J. For 0x = to 7.0 mx = , 3.0 JW = . EXECUTE: (a) 4.0 JK = , so 2 2(4.0 J) (2.0 kg) 2.00 m /sv K m= = = . (b) No work is done between 3.0 mx = and 4.0 mx = , so the speed is the same, 2.00 m/s. (c) 3.0 JK = , so 2 / 2(3.0 J) /(2.0 kg) 1.73 m /sv K m= = = . EVALUATE: In each case the work done by F is positive and the car gains kinetic energy. 6.39. IDENTIFY and SET UP: Apply Eq.(6.6). Let point 1 be where the sled is released and point 2 be at 0x = for part (a) and at 0.200 mx = ? for part (b). Use Eq.(6.10) for the work done by the spring and calculate 2.K Then 212 22K mv= gives 2.v EXECUTE: (a) tot 2 1W K K= ? so 2 1 totK K W= + 1 0K = (released with no initial velocity), 212 22K mv= The only force doing work is the spring force. Eq.(6.10) gives the work done on the spring to move its end from 1x to 2.x The force the spring exerts on an object attached to it is ,F kx= ? so the work the spring does is ( )2 2 2 21 1 1 1spr 2 1 1 22 2 2 2 .W kx kx kx kx= ? ? = ? Here 1 0.375 mx = ? and 2 0.x = Thus 21spr 2 (4000 N/m)( 0.375 m) 0 281 J.W = ? ? = 2 1 tot 0 281 J 281 JK K W= + = + = Then 212 22K mv= implies 22 2 2(281 J) 2.83 m/s.70.0 kg K v m = = = (b) 2 1 totK K W= + 1 0K = 2 21 1 tot spr 1 22 2 .W W kx kx= = ? Now 2 0.200 m,x = so 2 21 1 spr 2 2(4000 N/m)( 0.375 m) (4000 N/m)( 0.200 m) 281 J 80 J 201 JW = ? ? ? = ? = Thus 2 0 201 J 201 JK = + = and 212 22K mv= gives 22 2 2(201 J) 2.40 m/s.70.0 kg K v m = = = EVALUATE: The spring does positive work and the sled gains speed as it returns to 0.x = More work is done during the larger displacement in part (a), so the speed there is larger than in part (b). 6.40. IDENTIFY: xF kx= SET UP: When the spring is in equilibrium, the same force is applied to both ends of any segment of the spring. EXECUTE: (a) When a force F is applied to each end of the original spring, the end of the spring is displaced a distance x. Each half of the spring elongates a distance hx , where h / 2x x= . Since F is also the force applied to each half of the spring, F kx= and h hF k x= . h hkx k x= and h h 2 x k k k x ? ?= =? ?? ? . Work and Kinetic Energy 6-11 (b) The same reasoning as in part (a) gives seg 3k k= , where segk is the force constant of each segment. EVALUATE: For half of the spring the same force produces less displacement than for the original spring. Since /k F x= , smaller x for the same F means larger k . 6.41. IDENTIFY and SET UP: Apply Eq.(6.6) to the glider. Work is done by the spring and by gravity. Take point 1 to be where the glider is released. In part (a) point 2 is where the glider has traveled 1.80 m and 2 0.K = There two points are shown in Figure 6.41a. In part (b) point 2 is where the glider has traveled 0.80 m. EXECUTE: (a) tot 2 1 0.W K K= ? = Solve for 1,x the amount the spring is initially compressed. tot spr 0wW W W= + = So spr wW W= ? (The spring does positive work on the glider since the spring force is directed up the incline, the same as the direction of the displacement.) Figure 6.41a The directions of the displacement and of the gravity force are shown in Figure 6.41b. ( cos ) ( cos130.0 )wW w s mg s?= = ° 2(0.0900 kg)(9.80 m/s )(cos130.0 )(1.80 m) 1.020 JwW = ° = ? (The component of w parallel to the incline is directed down the incline, opposite to the displacement, so gravity does negative work.) Figure 6.41b spr 1.020 JwW W= ? = + 21 spr 12W kx= so spr1 2 2(1.020 J) 0.0565 m 640 N/m W x k = = = (b) The spring was compressed only 0.0565 m so at this point in the motion the glider is no longer in contact with the spring. Points 1 and 2 are shown in Figure 6.41c. tot 2 1W K K= ? 2 1 totK K W= + 1 0K = Figure 6.41c tot spr wW W W= + From part (a), spr 1.020 JW = and 2( cos130.0 ) (0.0900 kg)(9.80 m/s )(cos130.0 )(0.80 m) 0.454 JwW mg s= ° = ° = ? Then 2 spr 1.020 J 0.454 J 0.57 J.wK W W= + = + ? = + EVALUATE: The kinetic energy in part (b) is positive, as it must be. In part (a), 2 0x = since the spring force is no longer applied past this point. In computing the work done by gravity we use the full 0.80 m the glider moves. 6.42. IDENTIFY: Apply tot 2 1W K K= ? to the brick. Work is done by the spring force and by gravity. SET UP: At the maximum height. 0v = . Gravity does negative work, gravW mgh= ? . The work done by the spring is 212 kd , where d is the distance the spring is compressed initially. EXECUTE: The initial and final kinetic energies of the brick are both zero, so the net work done on the brick by the spring and gravity is zero, so 2(1 2) 0kd mgh? = , or 22 / 2(1.80 kg)(9.80 m /s )(3.6 m) /(450 N / m) 0.53 m.d mgh k= = = The spring will provide an upward force while the spring and the brick are in contact. When this force goes to zero, the spring is at its uncompressed length. But when the spring reaches its uncompressed length the brick has an upward velocity and leaves the spring. EVALUATE: Gravity does negative work because the gravity force is downward and the brick moves upward. The spring force does positive work on the brick because the spring force is upward and the brick moves upward. 6-12 Chapter 6 6.43. IDENTIFY: Apply the relation between energy and power. SET UP: Use WP t = ? to solve for W, the energy the bulb uses. Then set this value equal to 21 2 mv and solve for the speed. EXECUTE: 5(100 W)(3600 s) 3.6 10 JW P t= ? = = × 53.6 10 JK = × so 52 2(3.6 10 J) 100 m/s 70 kg K v m ×= = = EVALUATE: Olympic runners achieve speeds up to approximately 36 m/s, or roughly one third the result calculated. 6.44. IDENTIFY: Energy is power times time. SET UP: 1 W 1 J/s= . 71 yr 3.16 10 s= × . EXECUTE: (a) 19 11 7 (1.0 10 J / yr) 3.2 10 W. (3.16 10 s / yr) × = ×× (b) 11 8 3.2 10 W 1.1 kW/person. 3.0 10 folks × =× (c) 11 8 2 2 3 2 3.2 10 W 8.0 10 m 800 km . (0.40)1.0 10 W / m A ×= = × =× EVALUATE: The area in part (c) corresponds to a square about 28 km on a side, which is about 18 miles. The space required is not an impediment. 6.45. IDENTIFY: av WP t ?= ? . W? is the energy released. SET UP: W? is to be the same. 71 y 3.156 10 s= × . EXECUTE: av constantP t W? = ? = , so av-sun sun av-m mP t P t? = ? . 5 7 13sun av-m av-sun m [2.5 10 y][3.156 10 s/y] 3.9 10 0.20 s t P P P t ? ? ? ?? × ×= = = ×? ? ? ?? ? ?? ? . EVALUATE: Since the power output of the magnetar is so much larger than that of our sun, the mechanism by which it radiates energy must be quite different. 6.46. IDENTIFY: The thermal energy is produced as a result of the force of friction, k .F mg?= The average thermal power is thus the average rate of work done by friction or avP F v= ? . SET UP: 2 1av 8.00 m/s 0 4.00 m/s2 2 v v v + +? ?= = =? ?? ? EXECUTE: ( )( )( ) ( )2av 0.200 20.0 kg 9.80 m/s 4.00 m/s 157 WP Fv ? ?= = =? ? EVALUATE: The power could also be determined as the rate of change of kinetic energy, ,K t? where the time is calculated from f iv v at= + and a is calculated from a force balance, k .F ma mg?= =? 6.47. IDENTIFY: Use the relation P F v= ? to relate the given force and velocity to the total power developed. SET UP: 1 hp 746 W= EXECUTE: The total power is ( )( ) 3165 N 9.00 m/s 1.49 10 W.P F v= = = ×? Each rider therefore contributes ( )3 each rider 1.49 10 W / 2 745 W 1 hp.P = × = ? EVALUATE: The result of one horsepower is very large; a rider could not sustain this output for long periods of time. 6.48. IDENTIFY and SET UP: Calculate the power used to make the plane climb against gravity. Consider the vertical motion since gravity is vertical. EXECUTE: The rate at which work is being done against gravity is 2(700 kg)(9.80 m/s )(2.5 m/s) 17.15 kW.P Fv mgv= = = = This is the part of the engine power that is being used to make the airplane climb. The fraction this is of the total is 17.15 kW/75 kW 0.23.= EVALUATE: The power we calculate for making the airplane climb is considerably less than the power output of the engine. Work and Kinetic Energy 6-13 6.49. IDENTIFY: av WP t ?= ? . The work you do in lifting mass m a height h is mgh . SET UP: 1 hp 746 W= EXECUTE: (a) The number per minute would be the average power divided by the work (mgh ) required to lift one box, 2 (0.50 hp) (746 W hp) 1.41 s, (30 kg) (9.80 m s ) (0.90 m) = or 84.6 min. (b) Similarly, 2 (100 W) 0.378 s, (30 kg) (9.80 m s ) (0.90 m) = or 22.7 min. EVALUATE: A 30-kg crate weighs about 66 lbs. It is not possible for a person to perform work at this rate. 6.50. IDENTIFY and SET UP: Use Eq.(6.15) to relate the power provided and the amount of work done against gravity in 16.0 s. The work done against gravity depends on the total weight which depends on the number of passengers. EXECUTE: Find the total mass that can be lifted: av , W mgh P t t ?= =? so avP tm g h = 4 av 746 W (40 hp) 2.984 10 W 1 hp P ? ?= = ×? ?? ? 4 3av 2 (2.984 10 W)(16.0 s) 2.436 10 kg (9.80 m/s )(20.0 m) P t m gh ×= = = × This is the total mass of elevator plus passengers. The mass of the passengers is 3 32.436 10 kg 600 kg 1.836 10 kg.× ? = × The number of passengers is 31.836 10 kg 28.2. 65.0 kg × = 28 passengers can ride. EVALUATE: Typical elevator capacities are about half this, in order to have a margin of safety. 6.51. IDENTIFY: Calculate the gallons of gasoline consumed and from that the energy consumed. Find the time t? for the trip and use av W P t ?= ? , where W? is the energy consumed. SET UP: 200 km 124 mi= EXECUTE: (a) The gallons of gasoline consumed is 124 mi 4.13 gal 30 mi/gal = . The energy consumed is 9 9(4.13 gal)(1.3 10 J/gal) 5.4 10 J× = × . (b) The time for the trip is 124 mi 2.07 h 7450 s 60 mi/h = = . 9 5 av 5.4 10 J 7.2 10 W 720 kW 7450 s W P t ? ×= = = × =? . EVALUATE: The rate of energy consumption is 3720 10 W 970 hp 746 W/hp × = . 6.52. IDENTIFY: Apply P F v= ? . F? is the force F of water resistance. SET UP: 1 hp 746 W= . 1 km/h 0.228 m/s= EXECUTE: 6(0.70) (0.70) (280,000 hp)(746 W hp) 8.1 10 N. (65 km h) ((0.228 m/s) (1 km/h)) P F v = = = × EVALUATE: The power required depends on speed, because of the factor of v in P F v= ? and also because the resistive force increases with speed. 6.53. IDENTIFY: To lift the skiers, the rope must do positive work to counteract the negative work developed by the component of the gravitational force acting on the total number of skiers, rope sinF Nmg ?= . SET UP: ropeP F v F v= =? EXECUTE: ( )rope rope cosP F v Nmg v?? ?= = +? ? . ( )( )( )( ) ( )2rope 1 m/s50 riders 70.0 kg 9.80 m/s cos75.0 12.0 km/h 3.60 km/hP ? ?? ?? ?= ? ?? ? ? ?? ?? ?? . 4 rope 2.96 10 W 29.6 kWP = × = . EVALUATE: Some additional power would be needed to give the riders kinetic energy as they are accelerated from rest. 6-14 Chapter 6 6.54. IDENTIFY: Relate power, work and time. SET UP: Work done in each stroke is W Fs= and avP W t= . EXECUTE: 100 strokes per second means av 100P Fs t= with 1.00 s, 2t F mg= = and 0.010 m.s = av 0.20 W.P = EVALUATE: For a 70 kg person to apply a force of twice his weight through a distance of 0.50 m for 100 times per second, the average power output would be 57.0 10 W× . This power output is very far beyond the capability of a person. 6.55. IDENTIFY: For mass dm located a distance x from the axis and moving with speed v, the kinetic energy is 21 2 ( )K dm v= . Follow the procedure specified in the hint. SET UP: The bar and an infinitesimal mass element along the bar are sketched in Figure 6.55. Let total massM = and time for one revolutionT = . 2?xv T = . EXECUTE: 21 ( ) 2 K dm v= ? . Mdm dxL= , so 2 2 2 3 2 2 2 2 2 2 0 0 1 2 1 4 1 4 2 2 2 2 3 3 L LM ?x M ? M ? L K dx x dx ? ML T L T L T L T ? ? ? ? ? ?? ? ? ? ? ? ? ?= = = =? ? ? ? ? ?? ? ? ? ? ? ? ?? ? ? ? ? ? ? ?? ? ? ? ? ?? ? There are 5 revolutions in 3 seconds, so 3 5 s 0.60 sT = = 2 2 22 (12.0 kg) (2.00 m) (0.60 s) 877 J. 3 K ?= = EVALUATE: If a point mass 12.0 kg is 2.00 m from the axis and rotates at the same rate as the bar, 2 (2.00 m) 20.9 m/s 0.60 s v ?= = and 2 2 31 12 2 (12 kg)(20.9 m/s) 2.62 10 JK mv= = = × . K for the bar is smaller by a factor of 0.33. The speed of a segment of the bar decreases toward the axis. Figure 6.55 6.56. IDENTIFY: Density is mass per unit volume, /m V? = , so we can calculate the mass of the asteroid. 212K mv= . Since the asteroid comes to rest, the kinetic energy it delivers equals its initial kinetic energy. SET UP: The volume of a sphere is related to its diameter by 31 6 V d?= . EXECUTE: (a) 3 7 3(320 m) 1.72 10 m 6 V ?= = × . 3 7 3 10(2600 kg/m )(1.72 10 m ) 4.47 10 kgm V?= = × = × . 2 10 3 2 181 1 2 2 (4.47 10 kg)(12.6 10 m/s) 3.55 10 JK mv= = × × = × . (b) The yield of a Castle/Bravo device is 15 16(1 s)(4.184 10 J) 6.28 10 J× = × . 18 16 3.55 10 J 56.5 devices 6.28 10 J × =× . EVALUATE: If such an asteroid were to hit the earth the effect would be catastrophic. 6.57. IDENTIFY and SET UP: Since the forces are constant, Eq.(6.2) can be used to calculate the work done by each force. The forces on the suitcase are shown in Figure 6.57a. Figure 6.57a In part (f ) , Eq.(6.6) is used to relate the total work to the initial and final kinetic energy. EXECUTE: (a) ( cos )FW F s?= Both F ? and s ? are parallel to the incline and in the same direction, so 90? = ° and (140 N)(3.80 m) 532 JFW Fs= = = Work and Kinetic Energy 6-15 (b) The directions of the displacement and of the gravity force are shown in Figure 6.57b. ( cos )wW w s?= 115 ,? = ° so (196 N)(cos115 )(3.80 m)wW = ° 315 JwW = ? Figure 6.57b Alternatively, the component of w parallel to the incline is sin 25 .w ° This component is down the incline so its angle with s ? is 180 .? = ° sin 25 (196 Nsin 25 )(cos180 )(3.80 m) 315 J.wW ° = ° ° = ? The other component of w, cos25 ,w ° is perpendicular to s ? and hence does no work. Thus sin 25 315 J,w wW W °= = ? which agrees with the above. (c) The normal force is perpendicular to the displacement ( 90 ),? = ° so 0.nW = (d) cos25n w= ° so k k k cos25 (0.30)(196 N)cos25 53.3 Nf n w? ?= = ° = ° = k( cos ) (53.3 N)(cos180 )(3.80 m) 202 JfW f x?= = ° = ? (e) tot 532 J 315 J 0 202 J 15 JF w n fW W W W W= + + + = + ? + ? = (f ) tot 2 1,W K K= ? 1 0,K = so 2 totK W= 21 2 tot2 mv W= so tot2 2 2(15 J) 1.2 m/s20.0 kg W v m = = = EVALUATE: The total work done is positive and the kinetic energy of the suitcase increases as it moves up the incline. 6.58. IDENTIFY: The work he does to lift his body a distance h is W mgh= . The work per unit mass is ( ) .W m gh= SET UP: The quantity gh has units of N/kg. EXECUTE: (a) The man does work, (9.8 N kg) (0.4 m) 3.92 J kg.= (b) (3.92 J kg) (70 J kg) 100 5.6%.× = (c) The child does work (9.8 N kg)(0.2 m) 1.96 J kg.= (1.96 J kg) (70 J kg) 100 2.8%.× = (d) If both the man and the child can do work at the rate of 70 J kg, and if the child only needs to use 1.96 J kg instead of 3.92 J kg, the child should be able to do more chin-ups. EVALUATE: Since the child has arms half the length of his father?s arms, the child must lift his body only 0.20 m to do a chin-up. 6.59. IDENTIFY: Apply the definitions of IMA and AMA given in the problem. SET UP: When the object moves a distance L along the ramp, it rises a vertical distance sinL ? . EXECUTE: (a) in out, sin ,s L s L ?= = so 1sin IMA ?= . (b) If out in in out, ( ) ( )AMA IMA F F s s= = and so out out in in( ) ( ) ( ) ( )F s F s= , or out in .W W= (c) The pulley is sketched in Figure 6.59. (d) out out out out in in in in in out ( )( ) ( )( ) W F s F F AMA e W F s s s IMA = = = = . EVALUATE: in sinF w ?= and outF w= . in in( )( ) ( sin )F s w L?= . out out( )( ) (sin )F s w L?= . Therefore, in in out out( )( ) ( )( )F s F s= . A smaller force acting over a larger distance does the same amount of work as a larger force acting over a smaller distance. Figure 6.59 6.60. IDENTIFY: Apply m?F = a? to each block to find the tension in the string. Each force is constant and cos .W Fs ?= SET UP: The free-body diagram for each block is given in Figure 6.60. 20.0 N 2.04 kgAm g= = and 12.0 N 1.22 kgBm g = = . 6-16 Chapter 6 EXECUTE: k AT f m a? = . B Bw T m a? = . k ( )B A Bw f m m a? = + . k 0f = . B A B w a m m ? ?= ? ?+? ? and 7.50 NA AB B A B A B m w T w w m m w w ? ? ? ?= = =? ? ? ?+ +? ? ? ? . 20.0 N block: tot (7.50 N)(0.750 m) 5.62 JW Ts= = = . 12.0 N block: tot ( ) (12.0 N 7.50 N)(0.750 m) 3.38 JBW w T s= ? = ? = (b) k k 6.50 NAf w?= = . kB A A B w w a m m ??= + . k k k k( ) ( ) A A B A A B A A B A B m w T f w w w w w m m w w ? ? ?? ? ? ?= + ? = + ?? ? ? ?+ +? ? ? ? . 6.50 N (5.50 N)(0.625) 9.94 NT = + = . 20.0 N block: tot k( ) (9.94 N 6.50 N)(0.750 m) 2.58 JW T f s= ? = ? = . 12.0 N block: tot ( ) (12.0 N 9.94 N)(0.750 m) 1.54 JBW w T s= ? = ? = . EVALUATE: Since the two blocks move with equal speeds, for each block tot 2 1W K K= ? is proportional to the mass (or weight) of that block. With friction the gain in kinetic energy is less, so the total work on each block is less. Figure 6.60 6.61. IDENTIFY: 212K mv= . Find the speed of the shuttle relative to the earth and relative to the satellite. SET UP: Velocity is distance divided by time. For one orbit the shuttle travels a distance 2 R? . EXECUTE: (a) 22 6 2 121 1 2 1 2 (6.66 10 m)(86,400 kg) 2.59 10 J. 2 2 2 (90.1 min) (60 s min) ?R ? mv m T ? ?×? ?= = = ×? ?? ?? ? ? ? (b) 2 2 3(1 2) (1 2) (86,400 kg) ((1.00 m) (3.00 s)) 4.80 10 J.mv = = × EVALUATE: The kinetic energy of an object depends on the reference frame in which it is measured. 6.62. IDENTIFY: cosW Fs ?= . tot 2 1W K K= ? . SET UP: k kf n?= . The normal force is cosn mg ?= , with 12.0? = ° . The component of the weight parallel to the incline is sinmg ? . EXECUTE: (a) 180? = ° and 2k k( cos ) (0.31)(5.00 kg)(9.80 m s )(cos 12.0 (1.50 m) 22.3 JfW f s ? mg ? s= ? = ? = ? = ?°) (b) 2(5.00 kg)(9.80 m s )(sin12.0 )(1.50 m) 15.3 J.° = (c) The normal force does no work. (d) tot 15.3 J 22.3 J 7.0 J.W = ? = ? (e) 22 1 tot (1 2)(5.00 kg)(2.2 m s) 7.0 J 5.1 JK K W= + = ? = , and so 2 2(5.1 J) /(5.00 kg) 1.4 m /sv = = . EVALUATE: Friction does negative work and gravity does positive work. The net work is negative and the kinetic energy of the object decreases. 6.63. IDENTIFY: The effective force constant is defined by eff /k F x= , where F is the force applied to each end of the spring combination and x is the amount the spring combination is stretched. SET UP: Consider a force F applied to each end of the combination. Then 1F and 2F are the forces applied to each spring and 1 2F F F= + . Each spring stretches the same amount x. EXECUTE: (a) effF k x= . 1 2 1 2F F F k x k x= + = + . Equating the two expressions for F gives eff 1 2k k k= + . (b) The same procedure as in part (a) gives eff 1 2 Nk k k k= + + +? . EVALUATE: The effective force constant of the configuration is greater than any of the force constants of the individual springs. More force is required to stretch the parallel combination that is required to stretch each separate spring the same amount. Work and Kinetic Energy 6-17 6.64. IDENTIFY: The effective force constant is defined by eff /k F x= , where F is the force applied to each end of the spring combination and x is the amount the spring combination is stretched. SET UP: Consider a force F applied to each end of the combination. The same force F is applied to each spring. Spring 1 stretches a distance 1x and spring 2 stretches a distance 2x , where 1 1/x F k= and 2 2/x F k= . The total distance the combination stretches is 1 2x x x= + . EXECUTE: (a) 1 2x x x= + gives eff 1 2 F F F k k k = = and eff 1 2 1 1 1 k k k = + . (b) The same procedure as in part (a) gives eff 1 2 1 1 1 1 Nk k k k = + + +? . EVALUATE: For two springs the result in part (a) can be written as 1 2eff 1 2 k k k k k = + . The effective force constant for the two springs in series is less than the force constant for each individual spring. It takes less force to stretch the combination an amount x than to stretch either separate spring an amount x. 6.65. IDENTIFY: Apply Eq.(6.7). SET UP: 2 1dxx x= ?? . EXECUTE: (a) 2 2 2 1 1 1 2 2 1 1 1 1 . x x x xx x x dx W F dx k k k x x x x ? ?? ?= = ? = ? ? = ?? ?? ?? ? ? ?? ? The force is given to be attractive, so 0xF < , and k must be positive. If 2 1 2 1 1 1 , x x x x > < , and 0W < . (b) Taking ?slowly? to be constant speed, the net force on the object is zero. The force applied by the hand is opposite xF , and the work done is negative of that found in part (a), or 1 2 1 1 k x x ? ??? ?? ? , which is positive if 2 1x x> . (c) The answers have the same magnitude but opposite signs; this is to be expected, in that the net work done is zero. EVALUATE: Your force is directed away from the origin, so when the object moves away from the origin your force does positive work. 6.66. IDENTIFY: Apply Eq.(6.6) to the motion of the asteroid. SET UP: Let point 1 be at a great distance and let point 2 be at the surface of the earth. Assume 1 0.K = From the information given about the gravitational force its magnitude as a function of distance r from the center of the earth must be 2E( / ) .F mg R r= This force is directed in the ??r direction since it is a ?pull?. F is not constant so Eq.(6.7) must be used to calculate the work it does. EXECUTE: ( )E E22 2E E E21 (1/ )R RmgRW F ds dr mgR r mgRr ?? ? ?= ? = ? = ? ? =? ?? ?? ? tot 2 1,W K K= ? 1 0K = This gives 122 E 1.25 10 JK mgR= = × 21 2 22K mv= so 2 22 / 11,000 m/sv K m= = EVALUATE: Note that 2 E2 ,v gR= the impact speed is independent of the mass of the asteroid. 6.67. IDENTIFY: Calculate the work done by friction and apply tot 2 1W K K= ? . Since the friction force is not constant, use Eq.(6.7) to calculate the work. SET UP: Let x be the distance past P . Since k? increases linearly with x, k 0.100 Ax? = + . When 12.5 mx = , k 0.600? = , so 0.500/(12.5 m) 0 0400 mA . /= = EXECUTE: (a) tot 2 1W K K K= ? = ? gives 2k 110 2? mgdx mv? = ?? . Using the above expression for k? , 2 2 10 1 (0 100 ) 2 x g . Ax dx v+ =? and 2 222 11(0.100) 2 2xg x A v? ?+ =? ?? ? . 2 2 2f f 1 (9.80 m/s ) (0.100) (0.0400/m) (4.50 m/s) 2 2 x x ? ?+ =? ?? ? . Solving for 2x gives 2 5.11 mx = . (b) k 0.100 (0.0400/m)(5 11 m) 0.304.? = + = 6-18 Chapter 6 (c) tot 2 1W K K= ? gives 2k 2 110 2? mgx mv? = ? . 2 2 1 2 2 k (4.50 m/s) 10.3 m 2 2(0.100)(9.80 m/s ) v x ? g = = = . EVALUATE: The box goes farther when the friction coefficient doesn?t increase. 6.68. IDENTIFY: Use Eq.(6.7) to calculate W. SET UP: 1 0x = . In part (a), 2 0.050 mx = . In part (b), 2 0.050 mx = ? . EXECUTE: (a) 2 2 2 3 2 3 42 2 20 0 ( ) 2 3 4 x x k b c W Fdx kx bx cx dx x x x= = ? + = ? +? ? . 2 2 3 3 4 2 2 2(50.0 N / m) (233 N / m ) (3000 N / m )W x x x= ? + . When 2 0.050 mx = , 0.12 JW = . (b) When 2 0.050 m,x = ? 0.17 JW = . (c) It?s easier to stretch the spring; the quadratic 2bx? term is always in the x? -direction, and so the needed force, and hence the needed work, will be less when 2 0x > . EVALUATE: When 0.050 mx = , 4.75 NxF = . When 0.050 mx = ? , 8.25 NxF = . 6.69. IDENTIFY and SET UP: Use m=?F a? ? to find the tension force T . The block moves in uniform circular motion and rad.=a a? (a) The free-body diagram for the block is given in Figure 6.69. EXECUTE: x xF ma=? 2v T m R = 2(0.70 m/s) (0.120 kg) 0.15 N 0.40 m T = = Figure 6.69 (b) 2 2(2.80 m/s) (0.120 kg) 9.4 N 0.10 m v T m R = = = (c) SET UP: The tension changes as the distance of the block from the hole changes. We could use 2 1 x xx W F dx= ? to calculate the work. But a much simpler approach is to use tot 2 1.W K K= ? EXECUTE: The only force doing work on the block is the tension in the cord, so tot .TW W= 2 21 1 1 12 2 (0.120 kg)(0.70 m/s) 0.0294 JK mv= = = 2 21 1 2 22 2 (0.120 kg)(2.80 m/s) 0.470 JK mv= = = tot 2 1 0.470 J 0.029 J 0.44 JW K K= ? = ? = This is the amount of work done by the person who pulled the cord. EVALUATE: The block moves inward, in the direction of the tension, so T does positive work and the kinetic energy increases. 6.70. IDENTIFY: Use Eq.(6.7) to find the work done by F. Then apply tot 2 1W K K= ? . SET UP: 2 1dxx x= ?? . EXECUTE: 2 1 2 1 2 1 1x x W dx x x x ? ? ? ?= = ?? ?? ?? . 26 2 1 9 1 17(2.12 10 N m )((0.200 m ) (1.25 10 m )) 2.65 10 JW ? ? ? ?= × ? ? × = ? × . Note that 1x is so large compared to 2x that the term 11/ x is negligible. Then, using Eq. (6.13)) and solving for 2v , 17 2 5 2 5 2 1 27 2 2( 2.65 10 J) (3.00 10 m/s) 2.41 10 m/s. (1.67 10 kg) W v v m ? ? ? ×= + = × + = ×× (b) With 2 10,K W K= = ? . Using 2 W x ?= ? , 26 2 10 2 2 27 5 2 1 1 2 2(2.12 10 N m ) 2.82 10 m. (1.67 10 kg)(3.00 10 m/s) x K mv ? ? ? ? ? × ?= = = = ×× × Work and Kinetic Energy 6-19 (c) The repulsive force has done no net work, so the kinetic energy and hence the speed of the proton have their original values, and the speed is 53.00 10 m/s× . EVALUATE: As the proton moves toward the uranium nucleus the repulsive force does negative work and the kinetic energy of the proton decreases. As the proton moves away from the uranium nucleus the repulsive force does positive work and the kinetic energy of the proton increases. 6.71. IDENTIFY and SET UP: Use /xv dx dt= and / .x xa dv dt= Use m=?F a? ? to calculate F? from .a? EXECUTE: (a) 2( ) ,x t t t? ? 3= + 2( ) 2 3x dxv t t tdt ? ?= = + 4.00 s:t = 2 3 22(0.200 m/s )(4.00 s) 3(0.0200 m/s )(4.00 s) 2.56 m/s.xv = + = (b) ( ) 2 6xx dv a t t dt ? ?= = + (2 6 )x xF ma m t? ?= = + 4.00 s:t = 2 36.00 kg(2(0.200 m/s ) 6(0.0200 m/s )(4.00 s)) 5.28 NxF = + = (c) IDENTIFY and SET UP: Use Eq.(6.6) to calculate the work. EXECUTE: tot 2 1W K K= ? At 1 0,t = 1 0v = so 1 0.K = tot FW W= 2 21 1 2 22 2 (6.00 kg)(2.56 m/s) 19.7 JK mv= = = Then tot 2 1W K K= ? gives that 19.7 JFW = EVALUATE: v increases with t so the kinetic energy increases and the work done is positive. We can also calculate FW directly from Eq.(6.7), by writing dx as xv dt and performing the integral. 6.72. IDENTIFY: Since the capsule comes to rest, the amount of work the capsule does on the ground equals its original kinetic energy. Use constant acceleration kinematic equations to calculate the stopping time t; t t? = . SET UP: 311 km/h 86.4 m/s= . Let y+ be the direction the capsule is traveling before the crash. EXECUTE: 2 2 51 11 12 2 (210 kg)(86.4 m/s) 7.84 10 JW K mv? = = = = × . 0 0.810 my y? = , 0 86.4 m/syv = and 0yv = . 0 0 2 y yv vy y t +? ?? = ? ?? ? gives 0 0 2( ) 2(0.810 m) 0.01875 s 86.4 m/sy y y t v ?= = = . 5 77.84 10 J 4.18 10 W 0.01875 s W t ? ×= = ×? EVALUATE: A large amount of work is done in a very small amount of time. 6.73. IDENTIFY and SET UP: Use Eq.(6.6). You do positive work and gravity does negative work. Let point 1 be at the base of the bridge and point 2 be at the top of the bridge. EXECUTE: (a) tot 2 1W K K= ? 2 21 1 1 12 2 (80.0 kg)(5.00 m/s) 1000 JK mv= = = 2 21 1 2 22 2 (80.0 kg)(1.50 m/s) 90 JK mv= = = tot 90 J 1000 J 910 JW = ? = ? (b) Neglecting friction, work is done by you (with the force you apply to the pedals) and by gravity: tot you gravity.W W W= + The gravity force is 2(80.0 kg)(9.80 m/s ) 784 N,w mg= = = downward. The displacement is 5.20 m, upward. Thus 180? = ° and gravity ( cos ) (784 N)(5.20 m)cos180 4077 JW F s?= = ° = ? Then tot you gravityW W W= + gives you tot gravity 910 J ( 4077 J) 3170 JW W W= ? = ? ? ? = + EVALUATE: The total work done is negative and you lose kinetic energy. 6.74. IDENTIFY: Use Eq.(6.7) to calculate W. SET UP: ( 1)1 1 n nx dx x n ? ? ?= ? ?? EXECUTE: (a) 0 0 1 1 0 . ( 1)) ( 1)n n nx x b b b W dx x n x n x ? ? ? ?= = =? ?? Note that for this part, for 11, 0nn x ?> ? as x ? ? . 6-20 Chapter 6 (b) When 0 1n< < , the improper integral must be used, 2 1 1 2 0lim ( ) ,( 1) n n x b W x x n ? ? ?? ? ?= ?? ??? ? and because the exponent on the 12 nx ? is positive, the limit does not exist, and the integral diverges. This is interpreted as the force F doing an infinite amount of work, even though 0F ? as 2 .x ? ? EVALUATE: The work-energy theorem says that an object gains an infinite amount of kinetic energy when an infinite amount of work is done on it. 6.75. IDENTIFY: The negative work done by the spring equals the change in kinetic energy of the car. SET UP: The work done by a spring when it is compressed a distance x from equilibrium is 212 kx? . 2 0K = . EXECUTE: 21 2 12 kx K K? = ? gives 2 21 1 12 2kx mv= and ( ) ( )( ) ( )2 22 2 51 1200 kg 0.65 m/s 0.070 m 1.0 10 N/mk mv x ? ?= = = ×? ? . EVALUATE: When the spring is compressed, the spring force is directed opposite to the displacement of the object and the work done by the spring is negative. 6.76. IDENTIFY: Apply tot 2 1W K K= ? . SET UP: Let 0x be the initial distance the spring is compressed. The work done by the spring is 2 21 102 2kx kx? , where x is the final distance the spring is compressed. EXECUTE: (a) Equating the work done by the spring to the gain in kinetic energy, 2 21 102 2kx mv= , so 0 400 N / m (0.060 m) 6.93 m/s. 0.0300 kg k v x m = = = (b) totW must now include friction, so 2 21 1 tot 0 02 2mv W kx fx= = ? , where f is the magnitude of the friction force. Then, 2 2 0 0 2 400 N/m 2(6.00 N) (0.060 m) (0.060 m) 4.90 m/s. 0.0300 kg (0.0300 kg) k f v x x m m = ? = ? = (c) The greatest speed occurs when the acceleration (and the net force) are zero. Let x be the amount the spring is still compressed, so the distance the ball has moved is 0x x? . 6.00 N, 0.0150 m400 N/m f kx f x k = = = = . To find the speed, the net work is 2 21tot 0 02 ( ) ( )W k x x f x x= ? ? ? , so the maximum speed is 2 2max 0 02( ) ( )k fv x x x xm m= ? ? ? . 2 2 max 400 N / m 2(6.00 N) ((0.060 m) (0.0150 m) ) (0.060 m 0.0150 m) 5.20 m/s (0.0300 kg) (0.0300 kg) v = ? ? ? = EVALUATE: The maximum speed with friction present (part (c)) is larger than the result of part (b) but smaller than the result of part (a). 6.77. IDENTIFY and SET UP: Use Eq.(6.6). Work is done by the spring and by gravity. Let point 1 be where the textbook is released and point 2 be where it stops sliding. 2 0x = since at point 2 the spring is neither stretched nor compressed. The situation is sketched in Figure 6.77. EXECUTE: tot 2 1W K K= ? 1 0,K = 2 0K = tot fric sprW W W= + Figure 6.77 21 spr 12 ,W kx= where 1 0.250 mx = (Spring force is in direction of motion of block so it does positive work.) fric kW mgd?= ? Then tot 2 1W K K= ? gives 21 1 k2 0kx mgd?? = 2 2 1 2 k (250 N/m)(0.250 m) 1.1 m, 2 2(0.30)(2.50 kg)(9.80 m/s ) kx d mg?= = = measured from the point where the block was released. EVALUATE: The positive work done by the spring equals the magnitude of the negative work done by friction. The total work done during the motion between points 1 and 2 is zero and the textbook starts and ends with zero kinetic energy. Work and Kinetic Energy 6-21 6.78. IDENTIFY: Apply tot 2 1W K K= ? to the cat. SET UP: Let point 1 be at the bottom of the ramp and point 2 be at the top of the ramp. EXECUTE: The work done by gravity is g sinW mgL ?= ? (negative since the cat is moving up), and the work done by the applied force is FL, where F is the magnitude of the applied force. The total work is 2 tot (100 N)(2.00 m) (7.00 kg)(9.80 m/s )(2.00 m)sin30 131.4 JW = ? ° = . The cat?s initial kinetic energy is 2 21 112 2 (7.00 kg)(2.40 m/s) 20.2 Jmv = = , and 1 2 2( ) 2(20.2 J 131.4 J) 6.58 m/s. (7.00 kg) K W v m + += = = EVALUATE: The net work done on the cat is positive and the cat gains speed. Without your push, tot grav 68.6 JW W= = ? and the cat wouldn?t have enough initial kinetic energy to reach the top of the ramp. 6.79. IDENTIFY: Apply tot 2 1W K K= ? to the vehicle. SET UP: Call the bumper compression x and the initial speed 0v . The work done by the spring is 212 kx? and 2 0K = . EXECUTE: (a) The necessary relations are 2 201 1 , 5 .2 2kx mv kx mg= < Combining to eliminate k and then x, the two inequalities are 2 2 2 and 25 .5 v mg x k g v > < Using the given numerical values, 2 2 (20.0 m/s) 8.16 m 5(9.80 m/s ) x > = and 2 2 4 2 (1700 kg)(9.80 m/s ) 25 1.02 10 N/m. (20.0 m/s) k < = × (b) A distance of 8 m is not commonly available as space in which to stop a car. Also, the car stops only momentarily and then returns to its original speed when the spring returns to its original length. EVALUATE: If k were doubled, to 42.04 10 N/m× , then 5.77 mx = . The stopping distance is reduced by a factor of 1/ 2 , but the maximum acceleration would then be 2/ 69.2 m/skx m = , which is 7.07 g . 6.80. IDENTIFY: Apply tot 2 1W K K= ? . cosW Fs ?= . SET UP: The students do positive work, and the force that they exert makes an angle of 30.0° with the direction of motion. Gravity does negative work, and is at an angle of 120.0 ° with the chair?s motion, EXECUTE: The total work done is 2tot ((600 N)cos30.0 (85.0 kg)(9.80 m/s )cos120.0 )(2.50 m) 257.8 JW = ° + ° = , and so the speed at the top of the ramp is 2 2tot2 1 2 2(257.8 J) (2.00 m/s) 3.17 m/s. (85.0 kg) W v v m = + = + = EVALUATE: The component of gravity down the incline is sin30 417 Nmg =° and the component of the push up the incline is (600 N)cos30 520 N=° . The force component up the incline is greater than the force component down the incline, the net work done is positive and the speed increases. 6.81. IDENTIFY: Apply tot 2 1W K K= ? to the blocks. SET UP: If X is the distance the spring is compressed, the work done by the spring is 212 kX? . At maximum compression, the spring (and hence the block) is not moving, so the block has no kinetic energy and 2 0x = . EXECUTE: (a) The work done by the block is equal to its initial kinetic energy, and the maximum compression is found from 2 21 1 02 2kX mv= and 5.00 kg (6.00 m/s) 0.600 m.500 N/m m X v k = = = (b) Solving for 0v in terms of a known X, 0 500 N/m (0.150 m) 1.50 m/s. 5.00 kg k v X m = = = EVALUATE: The negative work done by the spring removes the kinetic energy of the block. 6.82. IDENTIFY: Apply tot 2 1W K K= ? to the system of the two blocks. The total work done is the sum of that done by gravity (on the hanging block) and that done by friction (on the block on the table). SET UP: Let h be the distance the 6.00 kg block descends. The work done by gravity is (6.00 kg)gh and the work done by friction is k (8.00 kg)? gh? . 6-22 Chapter 6 EXECUTE: 2tot (6.00 kg (0.25)(8.00 kg)) (9.80 m/s ) (1.50 m) 58.8 J.W = ? = This work increases the kinetic energy of both blocks: 2tot 1 2 1 ( ) , 2 W m m v= + so 2(58.8 J) 2.90 m/s. (14.00 kg) v = = EVALUATE: Since the two blocks are connected by the rope, they move the same distance h and have the same speed v. 6.83. IDENTIFY and SET UP: Apply tot 2 1W K K= ? to the system consisting of both blocks. Since they are connected by the cord, both blocks have the same speed at every point in the motion. Also, when the 6.00-kg block has moved downward 1.50 m, the 8.00-kg block has moved 1.50 m to the right. The target variable, k ,? will be a factor in the work done by friction. The forces on each block are shown in Figure 6.83. EXECUTE: 2 2 21 1 11 1 1 12 2 2 ( )A B A BK m v m v m m v= + = + 2 0K = Figure 6.83 The tension T in the rope does positive work on block B and the same magnitude of negative work on block A, so T does no net work on the system. Gravity does work mg AW m gd= on block A, where 2.00 m.d = (Block B moves horizontally, so no work is done on it by gravity.) Friction does work fric k BW m gd?= ? on block B. Thus tot fric k .mg A BW W W m gd m gd?= + = ? Then tot 2 1W K K= ? gives 21k 12 ( )A B A Bm gd m gd m m v?? = ? + and 2 21 12 k 2 ( ) 6.00 kg (6.00 kg 8.00 kg)(0.900 m/s) 0.786 8.00 kg 2(8.00 kg)(9.80 m/s )(2.00 m) A BA B B m m vm m m gd ? + += + = + = EVALUATE: The weight of block A does positive work and the friction force on block B does negative work, so the net work is positive and the kinetic energy of the blocks increases as block A descends. Note that 1K includes the kinetic energy of both blocks. We could have applied the work-energy theorem to block A alone, but then totW includes the work done on block A by the tension force. 6.84. IDENTIFY: Apply tot 2 1W K K= ? . The work done by the force from the bow is the area under the graph of xF versus the draw length. SET UP: One possible way of estimating the work is to approximate the F versus x curve as a parabola which goes to zero at 0x = and 0,x x= and has a maximum of 0F at 0 / 2x x= , so that 20 0 0( ) (4 / ) ( ).F x F x x x x= ? This may seem like a crude approximation to the figure, but it has the advantage of being easy to integrate. EXECUTE: 0 0 2 3 20 0 0 0 0 0 0 02 20 0 0 0 4 4 2 ( ) 2 3 3 x xF F x x Fdx x x x dx x F x x x ? ?= ? = ? =? ?? ?? ? . With 0 200 NF = and 0 0.75 m,x = 100 J.W = The speed of the arrow is then 2 2(100 J) 89 m/s (0.025 kg) W m = = . EVALUATE: We could alternatively represent the area as that of a rectangle 180 N by 0.55 m. This gives 99 JW = , in close agreement with our more elaborate estimate. 6.85. IDENTIFY: Apply Eq.(6.6) to the skater. SET UP: Let point 1 be just before she reaches the rough patch and let point 2 be where she exits from the patch. Work is done by friction. We don?t know the skater?s mass so can?t calculate either friction or the initial kinetic energy. Leave her mass m as a variable and expect that it will divide out of the final equation. EXECUTE: k 0.25f mg= so tot (0.25 ) ,fW W mg s= = ? where s is the length of the rough patch. tot 2 1W K K= ? 21 1 02 ,K mv= ( )2 2 21 1 12 2 0 02 2 2(0.45 ) 0.2025K mv m v mv= = = The work-energy relation gives ( ) 21 02(0.25 ) 0.2025 1mg s mv? = ? The mass divides out, and solving gives 1.5 m.s = EVALUATE: Friction does negative work and this reduces her kinetic energy. Work and Kinetic Energy 6-23 6.86. IDENTIFY: av avP F v= ? . Use F ma= to calculate the force. SET UP: av 0 6.00 m/s 3.00 m/s2v += = EXECUTE: Your friend?s average acceleration is 20 6.00 m/s 2.00 m/s 3.00 s v v a t ?= = = . Since there are no other horizontal forces acting, the force you exert on her is given by 2net (65.0 kg)(2.00 m/s ) 130 NF ma= = = . av (130 N)(3.00 m/s) 390 WP = = . EVALUATE: We could also use the work-energy theorem: 212 1 2 (65.0 kg)(6.00 m/s) 1170 JW K K= ? = = . av 1170 J 390 W 3.00 s W P t = = = , the same as obtained by our other approach. 6.87. IDENTIFY: To lift a mass m a height h requires work W mgh= . To accelerate mass m from rest to speed v requires 21 2 1 2W K K mv= ? = . av WP t ?= ? . SET UP: 60 st = EXECUTE: (a) 2 5(800 kg)(9.80 m/s )(14.0 m) 1.10 10 J= × (b) 2 5(1/ 2)(800 kg)(18.0 m/s ) 1.30 10 J.= × (c) 5 51.10 10 J 1.30 10 J 3.99 kW. 60 s × + × = EVALUATE: Approximately the same amount of work is required to lift the water against gravity as to accelerate it to its final speed. 6.88. IDENTIFY: P F v= ? and F ma=? . SET UP: From Problem 6.71, 22 3v t t? ?= + and 2 6a t? ?= + . EXECUTE: 2 2 2 2 3(2 6 )(2 3 ) (4 18 18 )P F v mav m ?t t ?t m t ?t ? t? ? ? ?= = = + + = + +? . 2 2 3 3(0.96 N/s) (0.43 N/s ) (0.043 N/s )P t t t= + + . At 4.00 s,t = the power output is 13.5 W. EVALUATE: P increases in time because v increase and because a increases. 6.89. IDENTIFY and SET UP: Energy is av .P t The total energy expended in one day is the sum of the energy expended in each type of activity. EXECUTE: 41 day 8.64 10 s= × Let walkt be the time she spends walking and othert be the time she spends in other activities; 4 other walk8.64 10 s .t t= × ? The energy expended in each activity is the power output times the time, so 7 walk other(280 W) (100 W) 1.1 10 JE Pt t t= = + = × 4 7 walk walk(280 W) (100 W)(8.64 10 s ) 1.1 10 Jt t+ × ? = × 6 walk(180 W) 2.36 10 Jt = × 4 walk 1.31 10 s 218 min 3.6 h.t = × = = EVALUATE: Her average power for one day is 7(1.1 10 J)/([24][3600 s]) 127 W.× = This is much closer to her 100 W rate than to her 280 W rate, so most of her day is spent at the 100 W rate. 6.90. IDENTIFY and SET UP: W Pt= EXECUTE: (a) The hummingbird produces energy at a rate of 0.7 J/s to 1.75 J/s. At 10 beats/s, the bird must expend between 0.07 J/beat and 0.175 J/beat. (b) The steady output of the athlete is (500 W)/(70 kg) 7 W/kg,= which is below the 10 W/kg necessary to stay aloft. Though the athlete can expend 1400 W/70 kg 20 W/kg= for short periods of time, no human-powered aircraft could stay aloft for very long. EVALUATE: Movies of early attempts at human-powered flight bear out our results. 6.91. IDENTIFY and SET UP: Use Eq.(6.15). The work done on the water by gravity is mgh , where 170 m.h = Solve for the mass m of water for 1.00 s and then calculate the volume of water that has this mass. 6-24 Chapter 6 EXECUTE: The power output is 9av 2000 MW 2.00 10 W.P = = × av WP t ?= ? and 92% of the work done on the water by gravity is converted to electrical power output, so in 1.00 s the amount of work done on the water by gravity is 9 9av (2.00 10 W)(1.00 s) 2.174 10 J 0.92 0.92 P t W ? ×= = = × ,W mgh= so the mass of water flowing over the dam in 1.00 s must be 9 6 2 2.174 10 J 1.30 10 kg (9.80 m/s )(170 m) W m gh ×= = = × density m V = so 6 3 3 3 3 1.30 10 kg 1.30 10 m . density 1.00 10 kg/m m V ×= = = ×× EVALUATE: The dam is 1270 m long, so this volume corresponds to about a 3m flowing over each 1 m length of the dam, a reasonable amount. 6.92. IDENTIFY: WP t = and 212W mv= , if the object starts from rest. dva dt= and 0x x vdt? = ? . SET UP: 1/ 2 1/ 212d t tdt ?= . 1/ 2 3 / 223t dt t=? . EXECUTE: (a) The power P is related to the speed by 212 ,Pt K mv= = so 2 Ptv m= . (b) 2 2 2 1 . 22 dv d Pt P d P P a t dt dt m m dt m mtt = = = = = (c) 3 31 2 2 2 0 2 2 2 8 . 3 9 P P P x x v dt t dt t t m m m ? = = = =? ? EVALUATE: v, a, and 0x x? at a particular time are all proportional to 1/ 2P . The result in part (b) could also be obtained from P Fv= and /a F m= , so Pa vm = . 6.93. IDENTIFY and SET UP: For part (a) calculate m from the volume of blood pumped by the heart in one day. For part (b) use W calculated in part (a) in Eq.(6.15). EXECUTE: (a) ,W mgh= as in Example 6.11. We need the mass of blood lifted; we are given the volume 3 3 31 10 m(7500 L) 7.50 m . 1 L V ?? ?×= =? ?? ? 3 3 3 3density volume (1.05 10 kg/m )(7.50 m ) 7.875 10 kgm = × = × = × Then 3 2 5(7.875 10 kg)(9.80 m/s )(1.63 m) 1.26 10 J.W mgh= = × = × (b) 5 av 1.26 10 J 1.46 W. (24 h)(3600 s/h) W P t ? ×= = =? EVALUATE: Compared to light bulbs or common electrical devices, the power output of the heart is rather small. 6.94. IDENTIFY: P F v Mav= =? . To overcome gravity on a slope that is at an angle ? above the horizontal, ( sin ) .P Mg v?= SET UP: 61 MW 10 W= . 31 kN 10 N= . When ? is small, tan sin? ?? . EXECUTE: (a) The number of cars is the total power available divided by the power needed per car, 6 3 13.4 10 W 177, (2.8 10 N)(27 m/s) × =× rounding down to the nearest integer. (b) To accelerate a total mass M at an acceleration a and speed v, the extra power needed is Mav. To climb a hill of angle ? , the extra power needed is ( sin ) .Mg v? This will be nearly the same if ~ sin ;a g ? if 2sin ~ tan ~ 0.10 m/s ,g g? ? the power is about the same as that needed to accelerate at 20.10 m/s . (c) ( sin )P Mg v?= , where M is the total mass of the diesel units. 6 2(1.10 10 kg)(9.80 m/s )(0.010)(27 m/s) 2.9 MW.P = × = (d) The power available to the cars is 13.4 MW, minus the 2.9 MW needed to maintain the speed of the diesel units on the incline. The total number of cars is then 6 6 3 4 2 13.4 10 W 2.9 10 W 36, (2.8 10 N (8.2 10 kg)(9.80 m/s )(0.010))(27 m/s) × ? × =× + × rounding to the nearest integer. Work and Kinetic Energy 6-25 EVALUATE: For a single car, 4 2 3sin (8.2 10 kg)(9.80 m/s )(0.010) 8.0 10 NMg ? = × = × , which is over twice the 2.8 kN required to pull the car at 27 m/s on level tracks. Even a slope as gradual as 1.0% greatly increases the power requirements, or for constant power greatly decreases the number of cars that can be pulled. 6.95. IDENTIFY: P F v= ? . The force required to give mass m an acceleration a is F ma= . For an incline at an angle ? above the horizontal, the component of mg down the incline is sinmg ? . SET UP: For small ? , sin tan? ?? . EXECUTE: (a) 30 (53 10 N)(45 m/s) 2.4 MW.P Fv= = × = (b) 5 21 (9.1 10 kg)(1.5 m/s )(45 m/s) 61 MW.P mav= = × = (c) Approximating sin ,? by tan ,? and using the component of gravity down the incline as sin ,mg ? 5 2 2 ( sin ) (9.1 10 kg)(9.80 m/s )(0.015)(45 m/s) 6.0 MW.P mg v?= = × = EVALUATE: From Problem 6.94, we would expect that a 20.15 m/s acceleration and a 1.5% slope would require the same power. We found that a 21.5 m/s acceleration requires ten times more power than a 1.5% slope, which is consistent. 6.96. IDENTIFY: 2 1 x xx W F dx= ? , and xF depends on both x and y. SET UP: In each case, use the value of y that applies to the specified path. 212xdx x=? . 2 313x dx x=? EXECUTE: (a) Along this path, y is constant, with the value 3.00 my = . 2 1 2 2.00 m(2.50 N/m )(3.00 m) 15.0 J 2 x x W ?y xdx ? ?= = =? ?? ?? , since 1 0x = and 2 2.00 mx = . (b) Since the force has no y-component, no work is done moving in the y-direction. (c) Along this path, y varies with position along the path, given by 1.5 ,y x= so 2(1.5 ) 1.5 ,xF x x x? ?= = and 2 2 1 1 3 2 2 (2.00 m)1.5 1.5(2.50 N/m ) 10.0 J. 3 x x x x W Fdx x dx?= = = =? ? EVALUATE: The force depends on the position of the object along its path. 6.97. IDENTIFY and SET UP: Use Eq.(6.18) to relate the forces to the power required. The air resistance force is 21 air 2 ,F CA v?= where C is the drag coefficient. EXECUTE: (a) tot ,P F v= with tot roll airF F F= + 2 3 3 21 1 air 2 2 (1.0)(0.463 m )(1.2 kg/m )(12.0 m/s) 40.0 NF CA v?= = = roll r r (0.0045)(490 N 118 N) 2.74 NF n w? ?= = = + = roll air( ) (2.74 N 40.0 N)(12.0 s) 513 WP F F v= + = + = (b) 2 3 3 21 1air 2 2 (0.88)(0.366 m )(1.2 kg/m )(12.0 m/s) 27.8 NF CA v?= = = roll r r (0.0030)(490 N 88 N) 1.73 NF n w? ?= = = + = roll air( ) (1.73 N 27.8 N)(12.0 s) 354 WP F F v= + = + = (c) 2 3 3 21 1air 2 2 (0.88)(0.366 m )(1.2 kg/m )(6.0 m/s) 6.96 NF CA v?= = = roll r 1.73 NF n?= = (unchanged) roll air( ) (1.73 N 6.96 N)(6.0 s) 52.1 WP F F v= + = + = EVALUATE: Since airF is proportional to 2v and ,P Fv= reducing the speed greatly reduces the power required. 6.98. IDENTIFY: P F v= ? SET UP: 1 m/s 3.6 km/h= EXECUTE: (a) 3 328.0 10 W 1.68 10 N. (60.0 km/h)((1 m/s)/(3.6 km/h)) P F v ×= = = × (b) The speed is lowered by a factor of one-half, and the resisting force is lowered by a factor of (0.65 0.35/ 4),+ and so the power at the lower speed is (28.0 kW)(0.50)(0.65 0.35/4) 10.3 kW 13.8 hp.+ = = (c) Similarly, at the higher speed, (28.0 kW)(2.0)(0.65 0.35 4) 114.8 kW 154 hp.+ × = = EVALUATE: At low speeds rolling friction dominates the power requirement but at high speeds air resistance dominates. 6-26 Chapter 6 6.99. IDENTIFY and SET UP: Use Eq.(6.18) to relate F and P . In part (a), F is the retarding force. In parts (b) and (c), F includes gravity. EXECUTE: (a) ,P Fv= so / .F P v= 746 W (8.00 hp) 5968 W 1 hp P ? ?= =? ?? ? 1000 m 1 h (60.0 km/h) 16.67 m/s 1 km 3600 s v ? ?? ?= =? ?? ?? ?? ? 5968 W 358 N. 16.67 m/s P F v = = = (b) The power required is the 8.00 hp of part (a) plus the power gP required to lift the car against gravity. The situation is sketched in Figure 6.99. 10 m tan 0.10 100 m ? = = 5.71? = ° Figure 6.99 The vertical component of the velocity of the car is sin (16.67 m/s)sin5.71 1.658 m/s.v ? = ° = Then 2 4( sin ) sin (1800 kg)(9.80 m/s )(1.658 m/s) 2.92 10 WgP F v a mgv ?= = = = × 4 1 hp2.92 10 W 39.1 hp 746 Wg P ? ?= × =? ?? ? The total power required is 8.00 hp 39.1 hp 47.1 hp.+ = (c) The power required from the engine is reduced by the rate at which gravity does positive work. The road incline angle ? is given by tan 0.0100,? = so 0.5729 .? = ° 2 3( sin ) (1800 kg)(9.80 m/s )(16.67 m/s)sin 0.5729 2.94 10 W 3.94 hp.gP mg v ?= = ° = × = The power required from the engine is then 8.00 hp 3.94 hp 4.06 hp.? = (d) No power is needed from the engine if gravity does work at the rate of 8.00 hp 5968 WgP = = sin ,gP mgv ?= so 25968 Wsin 0.02030(1800 kg)(9.80 m/s )(16.67 m/s) gP mgv ? = = = 1.163? = ° and tan 0.0203,? = a 2.03% grade. EVALUATE: More power is required when the car goes uphill and less when it goes downhill. In part (d), at this angle the component of gravity down the incline is sin 358 Nmg ? = and this force cancels the retarding force and no force from the engine is required. The retarding force depends on the speed so it is the same in parts (a), (b), and (c). 6.100. IDENTIFY: Apply tot 2 1W K K= ? to relate the initial speed 0v to the distance x along the plank that the box moves before coming to rest. SET UP: The component of weight down the incline is sinmg ? , the normal force is cosmg ? and the friction force is cosf mg? ?= . EXECUTE: 20 0 1 0 and ( sin cos ) . 2 x K mv W mg ?mg dx? ?? = ? = ? ?? Then, 2 0 (sin cos ) , sin cos . 2 x Ax W mg Ax dx W mg x? ? ? ?? ?= ? + = ? +? ?? ?? Set W K= ? : 2 2 0 1 sin cos . 2 2 Ax mv mg x? ?? ?? = ? +? ?? ? To eliminate x, note that the box comes to a rest when the force of static friction balances the component of the weight directed down the plane. So, sin cosmg Ax mg? ?= . Solve this for x and substitute into the previous equation: sin . cos x A ? ?= Then, 2 2 0 1 sin 1 sin sin cos , 2 cos 2 cos v g A A A ? ?? ?? ? ? ?? ?= + +? ?? ?? ?? ?? ? and upon canceling factors and collecting terms, 2 2 0 3 sin . cos g v A ? ?= The box will remain stationary whenever 2 2 0 3 sin . cos g v A ? ?? Work and Kinetic Energy 6-27 EVALUATE: If 0v is too small the box stops at a point where the friction force is too small to hold the box in place. sin? increases and cos? decreases as ? increases, so the 0v required increases as ? increases. 6.101. IDENTIFY: In part (a) follow the steps outlined in the problem. For parts (b), (c) and (d) apply the work-energy theorem. SET UP: 2 313x dx x=? EXECUTE: (a) Denote the position of a piece of the spring by l; 0l = is the fixed point and l L= is the moving end of the spring. Then the velocity of the point corresponding to l, denoted u , is ( ) (I/ )u l v L= (when the spring is moving, l will be a function of time, and so u is an implicit function of time). The mass of a piece of length dl is ( / ) ,dm M L dl= and so 2 2 2 3 1 1 ( ) , 2 2 Mv dK dm u l dl L = = and 2 2 2 3 02 6 LMv Mv K dK l dl L = = =? ? . (b) 2 21 12 2 ,kx mv= so 2( / ) (3200 N/m)/(0.053 kg)(2.50 10 m) 6.1 m/s.v k m x ?= = × = (c) With the mass of the spring included, the work that the spring does goes into the kinetic energies of both the ball and the spring, so 2 2 21 1 12 2 6 .kx mv Mv= + Solving for v, 2(3200 N/m) (2.50 10 m) 3.9 m/s. /3 (0.053 kg) (0.243 kg)/3 k v x m M ?= = × =+ + (d) Algebraically, 2 21 (1/2) 0.40 J 2 (1 /3 ) kx mv M m = =+ and 2 21 (1/2) 0.60 J. 6 (1 3 / ) kx Mv m M = =+ EVALUATE: For this ball and spring, ball spring 3 0.053 kg 3 0.65 0.243 kg K m K M ? ?= = =? ?? ? . The percentage of the final kinetic energy that ends up with each object depends on the ratio of the masses of the two objects. As expected, when the mass of the spring is a small fraction of the mass of the ball, the fraction of the kinetic energy that ends up in the spring is small. 6.102. IDENTIFY: In both cases, a given amount of fuel represents a given amount of work 0W that the engine does in moving the plane forward against the resisting force. Write 0W in terms of the range R and speed v and in terms of the time of flight T and v. SET UP: In both cases assume v is constant, so 0W RF= and R vT= . EXECUTE: In terms of the range R and the constant speed v, 20 2 .W RF R v v ??? ?= = +? ?? ? In terms of the time of flight , ,T R vt= so 30 .?W vTF T v v? ? ?= = +? ?? ? (a) Rather than solve for R as a function of v, differentiate the first of these relations with respect to v, setting 0 0 dW dv = to obtain 0.dR dFF R dv dv + = For the maximum range, 0,dR dv = so 0.dF dv = Performing the differentiation, 32 2 / 0, dF v ? v dv ?= ? = which is solved for 1 41 4 5 2 2 2 2 3.5 10 N m /s 32.9 m/s 118 km/h. 0.30 N s /m v ? ? ? ?× ?? ?= = = =? ?? ? ?? ? ? ? (b) Similarly, the maximum time is found by setting ( ) 0; d Fv dv = performing the differentiation, 2 23 / 0v v? ?? = . 1/41/4 5 2 2 2 2 3.5 10 N m /s 25 m/s 90 km/h. 3 3(0.30 N s /m ) ? v ? ? ?× ?? ?= = = =? ?? ? ?? ? ? ? EVALUATE: When 1/4( / )v ? ?= , airF has its minimum value air 2F ??= . For this v, 01 (0.50) WR ??= and 1/ 4 3/ 4 1 (0.50)T ? ?? ?= . When 1/4( /3 )v ? ?= , air 2.3F ??= . For this v, 02 (0.43) WR ??= and 1/ 4 3/ 4 2 (0.57)T ? ?? ?= . 1 2R R> and 2 1T T> , as they should be. 6.103. IDENTIFY: For each speed, calculate the time. Then use the graph to find the oxygen consumption and from that the energy consumption. SET UP: /t d v= 6-28 Chapter 6 EXECUTE: (a) The walk will take one-fifth of an hour, 12 min. From the graph, the oxygen consumption rate appears to be about 312 cm /kg min,? and so the total energy is 3 3 5(12 cm /kg min) (70 kg) (12 min) (20 J/cm ) 2.0 10 J.? = × (b) The run will take 6 min. Using an estimation of the rate from the graph of about 333 cm /kg min? gives an energy consumption of about 52.8 10 J.× (c) The run takes 4 min, and with an estimated rate of about 350 cm /kg min,? the energy used is about 52.8 10 J.× (d) Walking is the most efficient way to go. In general, the point where the slope of the line from the origin to the point on the graph is the smallest is the most efficient speed; about 5 km/h. EVALUATE: In an exercise program, for a fixed distance, running burns more energy than walking. 6.104. IDENTIFY: Write equations similar to (6.11) for each component. Eq.(6.12) will now involve the sum of three integrals, one for each component. SET UP: 2 2 2 2x y zv v v v= + + EXECUTE: From , ,x x y ym F ma F ma= = =F a ? ? and .z zF ma= The generalization of Eq. (6.11) is then , , .yx zx x y y z z dvdv dv a v a v a v dx dy dz = = = The total work is then 2 2 2 2 2 2 1 1 1 1 1 1 ( , , ) tot ( , , ) x y z x y zyx z x y z x y zx y z x y z dvdv dv W F dx F dy F dz m v dx v dy v dz dx dy dz ? ?= + + = + +? ?? ?? ? ? ? . 2 2 2 1 1 1 2 2 2 2 2 2 2 2 tot 2 1 2 1 2 1 2 1 1 1 1 ( . 2 2 2 x y z x y z v v v x x y y z z x x y y z zv v v W m v dv v dv v dv m v v v v v v mv mv? ?= + + = ? + ? + ? = ?? ?? ?? ? ? EVALUATE: F? and d l? are vectors and have components. W and K are scalars and we never speak of their components. 7-1 POTENTIAL ENERGY AND ENERGY CONSERVATION 7.1. IDENTIFY: gravU mgy= so grav 2 1( )U mg y y? = ? SET UP: y+ is upward. EXECUTE: (a) 2 5(75 kg)(9.80 m/s )(2400 m 1500 m) 6.6 10 JU? = ? = + × (b) 2 5(75 kg)(9.80 m/s )(1350 m 2400 m) 7.7 10 JU? = ? = ? × EVALUATE: gravU increases when the altitude of the object increases. 7.2. IDENTIFY: Apply m=?F a! ! to the sack to find the force. cosW Fs ?= . SET UP: The lifting force acts in the same direction as the sack?s motion, so 0? = ° EXECUTE: (a) For constant speed, the net force is zero, so the required force is the sack?s weight, 2(5.00 kg)(9.80 m/s ) 49.0 N.= (b) (49.0 N) (15.0 m) 735 JW = = . This work becomes potential energy. EVALUATE: The results are independent of the speed. 7.3. IDENTIFY: Use the free-body diagram for the bag and Newton's first law to find the force the worker applies. Since the bag starts and ends at rest, 2 1 0K K? = and tot 0W = . SET UP: A sketch showing the initial and final positions of the bag is given in Figure 7.3a. 2.0 msin 3.5 m ? = and 34.85? = ° . The free-body diagram is given in Figure 7.3b. F! is the horizontal force applied by the worker. In the calculation of gravU take y+ upward and 0y = at the initial position of the bag. EXECUTE: (a) 0yF =? gives cosT mg? = and 0xF =? gives sinF T ?= . Combining these equations to eliminate T gives 2tan (120 kg)(9.80 m/s ) tan34.85 820 NF mg ?= = =° . (b) (i) The tension in the rope is radial and the displacement is tangential so there is no component of T in the direction of the displacement during the motion and the tension in the rope does no work. (ii) tot 0W = so 2 worker grav grav,2 grav,1 2 1( ) (120 kg)(9.80 m/s )(0.6277 m) 740 JW W U U mg y y= ? = ? = ? = = . EVALUATE: The force applied by the worker varies during the motion of the bag and it would be difficult to calculate workerW directly. Figure 7.3 7.4. IDENTIFY: Only gravity does work on him from the point where he has just left the board until just before he enters the water, so Eq.(7.4) applies. SET UP: Let point 1 be just after he leaves the board and point 2 be just before he enters the water. y+ is upward and 0y = at the water. 7 7-2 Chapter 7 EXECUTE: (a) 1 0K = . 2 0y = . 1 3.25 my = . 1 grav,1 2 grav,2K U K U+ = + gives grav,1 2U K= and 211 22mgy mv= . 2 2 12 2(9.80 m/s )(3.25 m) 7.98 m/sv gy= = = . (b) 1 2.50 m/sv = , 2 0y = , 1 3.25 my = . 1 grav,1 2K U K+ = and 2 21 11 1 22 2mv mgy mv+ = . 2 2 2 2 1 12 (2.50 m/s) 2(9.80 m/s )(3.25 m) 8.36 m/sv v gy= + = + = . (c) 1 2.5 m/sv = and 2 8.36 m/sv = , the same as in part (b). EVALUATE: Kinetic energy depends only on the speed, not on the direction of the velocity. 7.5. IDENTIFY and SET UP: Use energy methods. (a) 1 1 other 2 2.K U W K U+ + = + Solve for 2K and then use 212 22K mv= to obtain 2.v other 0W = (The only force on the ball while it is in the air is gravity.) 21 1 12 ;K mv= 212 22K mv= 1 1,U mgy= 1 22.0 my = 2 2 0,U mgy= = since 2 0y = for our choice of coordinates. Figure 7.5 EXECUTE: 2 21 11 1 22 2mv mgy mv+ = 2 2 2 2 1 12 (12.0 m/s) 2(9.80 m/s )(22.0 m) 24.0 m/sv v gy= + = + = EVALUATE: The projection angle of 53.1° doesn?t enter into the calculation. The kinetic energy depends only on the magnitude of the velocity; it is independent of the direction of the velocity. (b) Nothing changes in the calculation. The expression derived in part (a) for 2v is independent of the angle, so 2 24.0 m/s,v = the same as in part (a). (c) The ball travels a shorter distance in part (b), so in that case air resistance will have less effect. 7.6. IDENTIFY: The normal force does no work, so only gravity does work and Eq.(7.4) applies. SET UP: 1 0K = . The crate?s initial point is at a vertical height of sind ? above the bottom of the ramp. EXECUTE: (a) 2 0,y = 1 sin .y d ?= 1 grav,1 2 grav,2K U K U+ = + gives grav,1 2.U K= 21 22sinmgd mv? = and 2 2 sin .v gd ?= (b) 1 0y = , 2 siny d ?= ? . 1 grav,1 2 grav,2K U K U+ = + gives 2 grav,20 K U= + . 21 220 ( sin )mv mgd ?= + ? and 2 2 sinv gd ?= , the same as in part (a). (c) The normal force is perpendicular to the displacement and does no work. EVALUATE: When we use gravU mgy= we can take any point as 0y = but we must take y+ to be upward. 7.7. IDENTIFY: Apply Eq.(7.7) to points 2 and 3. Take results from Example 7.6. other ,W fs= ? the work done by friction. SET UP: As in Example 7.6, 2 20, 94 J,K U= = and 3 0.U = EXECUTE: The work done by friction is (35 N) (1.6 m) 56 J? = ? . 3 38 J,K = and 3 2(38 J) 2.5 m/s.12 kgv = = EVALUATE: The value of 3v we obtained is the same as calculated in Example 7.6. For the motion from point 2 to point 3, gravity does positive work, friction does negative work and the net work is positive. 7.8. IDENTIFY and SET UP: Apply Eq.(7.7) and consider how each term depends on the mass. EXECUTE: The speed is v and the kinetic energy is 4K . The work done by friction is proportional to the normal force, and hence to the mass, and so each term in Eq. (7.7) is proportional to the total mass of the crate, and the speed at the bottom is the same for any mass. The kinetic energy is proportional to the mass, and for the same speed but four times the mass, the kinetic energy is quadrupled. EVALUATE: The same result is obtained if we apply m=?F a! ! to the motion. Each force is proportional to m and m divides out, so a is independent of m. Potential Energy and Energy Conservation 7-3 7.9. IDENTIFY: tot B AW K K= ? . The forces on the rock are gravity, the normal force and friction. SET UP: Let 0y = at point B and let y+ be upward. 0.50 mAy R= = . The work done by friction is negative; 0.22 JfW = ? . 0AK = . The free-body diagram for the rock at point B is given in Figure 7.9. The acceleration of the rock at this point is 2rad /a v R= , upward. EXECUTE: (a) (i) The normal force is perpendicular to the displacement and does zero work. (ii) 2grav grav, grav, (0.20 kg)(9.80 m/s )(0.50 m) 0.98 JA B AW U U mgy= ? = = = . (b) tot grav 0 ( 0.22 J) 0.98 J 0.76 Jn fW W W W= + + = + ? + = . tot B AW K K= ? gives 21 tot2 Bmv W= . tot2 2(0.76 J) 2.8 m/s 0.20 kgB W v m = = = . (c) Gravity is constant and equal to mg . n is not constant; it is zero at A and not zero at B. Therefore, k kf n?= is also not constant. (d) y yF ma=? applied to Figure 7.9 gives radn mg ma? = . 2 2 2 [2.8 m/s](0.20 kg) 9.80 m/s 5.1 N 0.50 m v n m g R ? ? ? ?= + = + =? ? ? ?? ? ? ? . EVALUATE: In the absence of friction, the speed of the rock at point B would be 2 3.1 m/sgR = . As the rock slides through point B, the normal force is greater than the weight 2.0 Nmg = of the rock. Figure 7.9 7.10. IDENTIFY: Only gravity does work, so Eq.(7.4) applies. SET UP: Let point 1 be just after the rock leaves the thrower and point 2 be at the maximum height. Let 1 0y = and y+ be upward. 1 0v v= . At the highest point, 2 0 cosv v ?= . 2 2sin cos 1? ?+ = . EXECUTE: 1 grav,1 2 grav,2K U K U+ = + gives 2 21 10 0 22 2 ( cos )mv m v mgy?= + . 2 2 2 20 0 2 sin (1 cos ) 2 2 v v y g g ??= ? = , was to be shown. EVALUATE: The initial kinetic energy is independent of the angle ? but the kinetic energy at the maximum height depends on ? , so the maximum height depends on ? . 7.11. IDENTIFY: Apply Eq.(7.7) to the motion of the car. SET UP: Take 0y = at point A. Let point 1 be A and point 2 be B. 1 1 other 2 2K U W K U+ + = + EXECUTE: 1 0,U = 2 (2 ) 28,224 J,U mg R= = other fW W= 21 1 12 37,500 J,K mv= = 212 22 3840 JK mv= = The work-energy relation then gives 2 2 1 5400 J.fW K U K= + ? = ? EVALUATE: Friction does negative work. The final mechanical energy 2 2( 32,064 J)K U+ = is less than the initial mechanical energy 1 1( 37,500 J)K U+ = because of the energy removed by friction work. 7.12. IDENTIFY: Only gravity does work, so apply Eq.(7.5). SET UP: 1 0v = , so 21 2 1 22 ( )mv mg y y= ? . EXECUTE: Tarzan is lower than his original height by a distance 1 2 (cos30 cos45 )y y l? = ?° ° so his speed is 2 (cos30 cos45 ) 7.9 m/s,v gl= °? ° = a bit quick for conversation. EVALUATE: The result is independent of Tarzan?s mass. 7-4 Chapter 7 7.13. 1 0y = 2 (8.00 m)sin36.9y = ° 2 4.80 my = Figure 7.13a (a) IDENTIFY and SET UP: F! is constant so Eq.(6.2) can be used. The situation is sketched in Figure 7.13a. EXECUTE: ( cos ) (110 N)(cos0 )(8.00 m) 880 JFW F s?= = ° = EVALUATE: F! is in the direction of the displacement and does positive work. (b) IDENTIFY and SET UP: Calculate W using Eq.(6.2) but first must calculate the friction force. Use the free- body diagram for the oven sketched in Figure 7.13b to calculate the normal force n ; then the friction force can be calculated from k k .f n?= For this calculation use coordinates parallel and perpendicular to the incline. EXECUTE: y yF ma=? cos36.9 0n mg? ° = cos36.9n mg= ° k k k cos36.9f n mg? ?= = ° 2 k (0.25)(10.0 kg)(9.80 m/s )cos36.9 19.6 Nf = ° = Figure 7.13b k( cos ) (19.6 N)(cos180 )(8.00 m) 157 JfW f s?= = ° = ? EVALUATE: Friction does negative work. (c) IDENTIFY and SET UP: ;U mgy= take 0y = at the bottom of the ramp. EXECUTE: 22 1 2 1( ) (10.0 kg)(9.80 m/s )(4.80 m 0) 470 JU U U mg y y? = ? = ? = ? = EVALUATE: The object moves upward and U increases. (d) IDENTIFY and SET UP: Use Eq.(7.7). Solve for .K? EXECUTE: 1 1 other 2 2K U W K U+ + = + 2 1 1 2 otherK K K U U W? = ? = ? + otherK W U? = ? ? other 880 J 157 J 723 JF fW W W= + = ? = 470 JU? = Thus 723 J 470 J 253 J.K? = ? = EVALUATE: otherW is positive. Some of otherW goes to increasing U and the rest goes to increasing K . (e) IDENTIFY: Apply m=?F a! ! to the oven. Solve for a! and then use a constant acceleration equation to calculate 2.v SET UP: We can use the free-body diagram that is in part (b): x xF ma=? k sin36.9F f mg ma? ? ° = EXECUTE: k sin36.9F f mga m ? ? °= = 2 2110 N 19.6 N (10 kg)(9.80 m/s )sin36.9 3.16 m/s 10.0 kg ? ? ° = SET UP: 1 0,xv = 23.16 m/s ,xa = 0 8.00 m,x x? = 2 ?xv = 2 2 2 1 02 ( )x x xv v a x x= + ? EXECUTE: 2 22 02 ( ) 2(3.16 m/s )(8.00 m) 7.11 m/sx xv a x x= ? = = Then 2 21 12 1 22 2 (10.0 kg)(7.11 m/s) 253 J.K K K mv? = ? = = = EVALUATE: This agrees with the result calculated in part (d) using energy methods. Potential Energy and Energy Conservation 7-5 7.14. IDENTIFY: Only gravity does work, so apply Eq.(7.4). Use m=?F a! ! to calculate the tension. SET UP: Let 0y = at the bottom of the arc. Let point 1 be when the string makes a 45° angle with the vertical and point 2 be where the string is vertical. The rock moves in an arc of a circle, so it has radial acceleration 2rad /a v r= EXECUTE: (a) At the top of the swing, when the kinetic energy is zero, the potential energy (with respect to the bottom of the circular arc) is (1 cos ),mgl ?? where l is the length of the string and ? is the angle the string makes with the vertical. At the bottom of the swing, this potential energy has become kinetic energy, so 21 2(1 cos ) ,mgl ? mv? = or 22 (1 cos ) 2(9 80 m/s ) (0 80 m) (1 cos45 ) 2.1 m/sv gl ? . . = ? = ? ° = . (b) At 45° from the vertical, the speed is zero, and there is no radial acceleration; the tension is equal to the radial component of the weight, or 2cos (0.12 kg) (9.80 m/s ) cos 45 0.83 N.mg ? = ° = (c) At the bottom of the circle, the tension is the sum of the weight and the mass times the radial acceleration, 2 2 (1 2(1 cos45 )) 1.9 Nmg mv l mg+ = + ? ° = EVALUATE: When the string passes through the vertical, the tension is greater than the weight because the acceleration is upward. 7.15. IDENTIFY: Apply 21el 2U kx= . SET UP: kx F= , so 12U Fx= ,where F is the magnitude of force required to stretch or compress the spring a distance x. EXECUTE: (a) (1 2)(800 N)(0.200 m) 80.0 J.= (b) The potential energy is proportional to the square of the compression or extension; 2(80.0 J) (0.050 m 0.200 m) 5.0 J.= EVALUATE: We could have calculated 800 N 4000 N/m 0.200 m F k x = = = and then used 21el 2U kx= directly. 7.16. IDENTIFY: Use the information given in the problem with F kx= to find k . Then 21el 2U kx= . SET UP: x is the amount the spring is stretched. When the weight is hung from the spring, F mg= . EXECUTE: 2(3.15 kg)(9.80 m/s ) 2205 N/m 0.1340 m 0.1200 m F mg k x x = = = =? . el2 2(10.0 J) 0.0952 m 9.52 cm 2205 N/m U x k = ± = ± = ± = ± . The spring could be either stretched 9.52 cm or compressed 9.52 cm. If it were stretched, the total length of the spring would be 12.00 cm 9.52 cm 21.52 cm+ = . If it were compressed, the total length of the spring would be 12.00 cm 9.52 cm 2.48 cm? = . EVALUATE: To stretch or compress the spring 9.52 cm requires a force 210 NF kx= = . 7.17. IDENTIFY: Apply 21el 2U kx= . SET UP: 210 02U kx= . x is the distance the spring is stretched or compressed. EXECUTE: (a) (i) 02x x= gives 2 21 1el 0 0 02 2(2 ) 4( ) 4U k x kx U= = = . (ii) 0 / 2x x= gives 2 21 1 1 el 0 0 02 4 2( / 2) ( ) / 4U k x kx U= = = . (b) (i) 02U U= gives 2 21 1 02 22( )kx kx= and 0 2x x= . (ii) 0 / 2U U= gives 2 21 1 1 02 2 2( )kx kx= and 0 / 2x x= . EVALUATE: U is proportional to 2x and x is proportional to U . 7.18. IDENTIFY: Apply Eq.(7.13). SET UP: Initially and at the highest point, 0v = , so 1 2 0K K= = . other 0W = . EXECUTE: (a) In going from rest in the slingshot?s pocket to rest at the maximum height, the potential energy stored in the rubber band is converted to gravitational potential energy; 3 2(10 10 kg)(9.80 m/s ) (22.0 m) 2.16 J.U mgy ?= = × = (b) Because gravitational potential energy is proportional to mass, the larger pebble rises only 8.8 m. (c) The lack of air resistance and no deformation of the rubber band are two possible assumptions. EVALUATE: The potential energy stored in the rubber band depends on k for the rubber band and the maximum distance it is stretched. 7-6 Chapter 7 7.19. IDENTIFY and SET UP: Use energy methods. There are changes in both elastic and gravitational potential energy; elastic; 212 ,U kx= gravitational: .U mgy= EXECUTE: (a) 212U kx= so 2 2(3.20 J) 0.0632 m 6.32 cm1600 N/m U x k = = = = (b) Points 1 and 2 in the motion are sketched in Figure 7.19. 1 1 other 2 2K U W K U+ + = + other 0W = (Only work is that done by gravity and spring force) 1 0,K = 2 0K = 0y = at final position of book ( )1 ,U mg h d= + 212 2U kd= Figure 7.19 21 20 ( ) 0mg h d kd+ + + = The original gravitational potential energy of the system is converted into potential energy of the compressed spring. 21 2 0kd mgd mgh? ? = 21 1( ) 4 ( ) 2 d mg mg k mgh k ? ?? ?= ± +? ?? ?? ?? ?? ? d must be positive, so ( )21 ( ) 2d mg mg kmghk= + + 21 ((1.20 kg)(9.80 m/s ) 1600 N/m d = + 2 2 2((1.20 kg)(9.80 m/s )) 2(1600 N/m)(1.20 kg)(9.80 m/s )(0.80 m)+ EVALUATE: It was important to recognize that the total displacement was ;h d+ gravity continues to do work as the book moves against the spring. Also note that with the spring compressed 0.12 m it exerts an upward force (192 N) greater than the weight of the book (11.8 N). The book will be accelerated upward from this position. 7.20. IDENTIFY: Use energy methods. There are changes in both elastic and gravitational potential energy. SET UP: 1 1 other 2 2.K U W K U+ + = + Points 1 and 2 in the motion are sketched in Figure 7.20. The spring force and gravity are the only forces doing work on the cheese, so other 0W = and grav el.U U U= + Figure 7.20 EXECUTE: Cheese released from rest implies 1 0.K = At the maximum height 2 0v = so 2 0.K = 1 1,el 1,gravU U U= + 1 0y = implies 1,grav 0U = 2 21 1 1,el 12 2 (1800 N/m)(0.15 m) 20.25 JU kx= = = (Here 1x refers to the amount the spring is stretched or compressed when the cheese is at position 1; it is not the x-coordinate of the cheese in the coordinate system shown in the sketch.) 2 2,el 2,gravU U U= + 0.0074 m 0.1087 m 0.12 m 12 cmd = + = = Potential Energy and Energy Conservation 7-7 2,grav 2 ,U mgy= where 2y is the height we are solving for. 2,el 0U = since now the spring is no longer compressed. Putting all this into 1 1 other 2 2K U W K U+ + = + gives 1,el 2,gravU U= 2 2 20.25 J 20.25 J 1.72 m (1.20 kg)(9.80 m/s ) y mg = = = EVALUATE: The description in terms of energy is very simple; the elastic potential energy originally stored in the spring is converted into gravitational potential energy of the system. 7.21. IDENTIFY: Apply Eq.(7.13). SET UP: other 0W = . As in Example 7.7, 1 0K = and 1 0.0250 J.U = EXECUTE: For 2 0.20 m s,v = 2 0.0040 JK = . 212 20.0210 J ,U kx= = and 2(0.0210 J) 0.092 m.5.00 N mx = ± = ± The glider has this speed when the spring is stretched 0.092 m or compressed 0.092 m. EVALUATE: Example 7.7 showed that 0.30 m/sxv = when 0.0800 mx = . As x increases, xv decreases, so our result of 0.20 m/sxv = at 0.092 mx = is consistent with the result in the example. 7.22. IDENTIFY and SET UP: Use energy methods. The elastic potential energy changes. In part (a) solve for 2K and from this obtain 2.v In part (b) solve for 1U and from this obtain 1.x (a) 1 1 other 2 2K U W K U+ + = + point 1: the glider is at its initial position, where 1 0.100 mx = and 1 0v = point 2: the glider is at 0x = EXECUTE: 1 0K = (released from rest), 212 22K mv= 21 1 12 ,U kx= 2 0,U = other 0W = (only the spring force does work) Thus 2 21 11 22 2 .kx mv= (The initial potential energy of the stretched spring is converted entirely into kinetic energy of the glider.) 2 1 5.00 N/m (0.100 m) 0.500 m/s 0.200 kg k v x m = = = (b) The maximum speed occurs at 0,x = so the same equation applies. 2 21 1 1 22 2kx mv= 1 2 0.200 kg 2.50 m/s 0.500 m 5.00 N/m m x v k = = = EVALUATE: Elastic potential energy is converted into kinetic energy. A larger 1x gives a larger 2.v 7.23. IDENTIFY: Only the spring does work and Eq.(7.11) applies. F kxa m m ?= = , where F is the force the spring exerts on the mass. SET UP: Let point 1 be the initial position of the mass against the compressed spring, so 1 0K = and 1 11.5 JU = . Let point 2 be where the mass leaves the spring, so el,2 0U = . EXECUTE: (a) 1 el,1 2 el,2K U K U+ = + gives el,1 2U K= . 21 2 el,12 mv U= and el,12 2 2(11.5 J) 3.03 m/s2.50 kg U v m = = = . K is largest when elU is least and this is when the mass leaves the spring. The mass achieves its maximum speed of 3.03 m/s as it leaves the spring and then slides along the surface with constant speed. (b) The acceleration is greatest when the force on the mass is the greatest, and this is when the spring has its maximum compression. 21el 2U kx= so el2 2(11.5 J) 0.0959 m2500 N/m U x k = ? = ? = ? . The minus sign indicates compression. xF kx ma= ? = and 2(2500 N/m)( 0.0959 m) 95.9 m/s2.50 kgx kx a m ?= ? = ? = . EVALUATE: If the end of the spring is displaced to the left when the spring is compressed, then xa in part (b) is to the right, and vice versa. 7-8 Chapter 7 7.24. (a) IDENTIFY and SET UP: Use energy methods. Both elastic and gravitational potential energy changes. Work is done by friction. Choose point 1 as in Example 7.9 and let that be the origin, so 1 0.y = Let point 2 be 1.00 m below point 1, so 2 1.00 m.y = ? EXECUTE: 1 1 other 2 2K U W K U+ + = + 2 21 1 1 12 2 (2000 kg)(25 m/s) 625,000 J,K mv= = = 1 0U = other 2 (17,000 N)(1.00 m) 17,000 JW f y= ? = ? = ? 21 2 22K mg= 21 2 2,grav 2,el 2 22U U U mgy ky= + = + 2 5 21 2 2(2000 kg)(9.80 m/s )( 1.00 m) (1.41 10 N/m)(1.00 m)U = ? + × 2 19,600 J 70,500 J 50,900 JU = ? + = + Thus 21 22625,000 J 17,000 J 50,900 Jmv? = + 21 22 557,100 Jmv = 2 2(557,100 J) 23.6 m/s 2000 kg v = = EVALUATE: The elevator stops after descending 3.00 m. After descending 1.00 m it is still moving but has slowed down. (b) IDENTIFY: Apply m=?F a! ! to the elevator. We know the forces and can solve for .a! SET UP: The free-body diagram for the elevator is given in Figure 7.24. EXECUTE: spr ,F kd= where d is the distance the spring is compressed y yF ma=? k sprf F mg ma+ ? = kf kd mg ma+ ? = Figure 7.24 5 2 k 17,000 N (1.41 10 N/m)(1.00 m) (2000 kg)(9.80 m/s ) 2000 kg f kd mg a m + ? + × ?= = 269.2 m/s= We calculate that a is positive, so the acceleration is upward. EVALUATE: The velocity is downward and the acceleration is upward, so the elevator is slowing down at this point. Note that 7.1 ;a g= this is unacceptably high for an elevator. 7.25. IDENTIFY: Apply Eq.(7.13) and F ma= . SET UP: other 0W = . There is no change in gravU . 1 0K = , 2 0U = . EXECUTE: 2 21 12 2 xkx mv= . The relations for m, xv , k and x are 2 2 and 5 .xkx mv kx mg= = Dividing the first equation by the second gives 2 5 xvx g = , and substituting this into the second gives 2 225 x mg k v = . (a) 2 2 5 2 (1160 kg)(9.80 m/s ) 25 4.46 10 N/m (2.50 m/s) k = = × (b) 2 2 (2.50 m/s) 0.128 m 5(9.80 m/s ) x = = EVALUATE: Our results for k and x do give the required values for xa and xv : 5 2(4.46 10 N/m)(0.128 m) 49.2 m/s 5.0 1160 kgx kx a g m ×= = = = and 2.5 m/sx kv x m= = . Potential Energy and Energy Conservation 7-9 7.26. IDENTIFY: grav cosW mg ?= . SET UP: When he moves upward, 180? = ° and when he moves downward, 0? = ° . When he moves parallel to the ground, 90? = ° . EXECUTE: (a) 2grav (75 kg)(9.80 m/s )(7.0 m)cos180 5100 JW = = ?° . (b) 2grav (75 kg)(9.80 m/s )(7.0 m)cos0 5100 JW = = +° . (c) 90? = ° in each case and grav 0W = in each case. (d) The total work done on him by gravity during the round trip is 5100 J 5100 J 0? + = . (e) Gravity is a conservative force since the total work done for a round trip is zero. EVALUATE: The gravity force is independent of the position and motion of the object. When the object moves upward gravity does negative work and when the object moves downward gravity does positive work. 7.27. IDENTIFY: Apply k k cosfW f s ?= . k kf n?= . SET UP: For a circular trip the distance traveled is 2d r?= . At each point in the motion the friction force and the displacement are in opposite directions and 180? = ° . Therefore, k k k (2 )fW f d f r?= ? = ? . n mg= so k kf mg?= . EXECUTE: (a) k 2 k 2 (0.250)(10.0 kg)(9.80 m/s )(2 )(2.00 m) 308 JfW mg r? ? ?= ? = ? = ? . (b) The distance along the path doubles so the work done doubles and becomes 616 J? . (c) The work done for a round trip displacement is not zero and friction is a nonconservative force. EVALUATE: The direction of the friction force depends on the direction of motion of the object and that is why friction is a nonconservative force. 7.28. IDENTIFY and SET UP: The force is not constant so we must use Eq.(6.14) to calculate W. The properties of work done by a conservative force are described in Section 7.3. 2 1 ,W d= ?? F l!! 2 ?x?= ?F i! EXECUTE: (a) ?d dy=l j! (x is constant; the displacement is in the -directiony+ ) 0d? =F l!! (since ? ? 0)? =i j and thus 0.W = (b) ?d dx=l i! 2 2? ?( ) ( ) d x dx x dx? ?? = ? ? = ?F l i i!! 2 2 1 1 2 2 3 3 3 31 1 2 13 3 12 N/m ( ) ( ) ((0.300 m) 3 x x xx W x dx ax x x? ?= ? = ? = ? ? = ? ?? 3(0.10 m) ) 0.10 J= ? (c) ?d dx=l i! as in part (b), but now 1 0.30 mx = and 2 0.10 mx = 3 31 2 13 ( ) 0.10 JW x x?= ? ? = + (d) EVALUATE: The total work for the displacement along the x-axis from 0.10 m to 0.30 m and then back to 0.10 m is the sum of the results of parts (b) and (c), which is zero. The total work is zero when the starting and ending points are the same, so the force is conservative. EXECUTE: 1 2 3 3 3 31 1 1 2 1 1 23 3 3( )x xW x x x x? ? ?? = ? ? = ? The definition of the potential energy function is 1 2 1 2 .x xW U U? = ? Comparison of the two expressions for W gives 31 3 .U x?= This does correspond to 0U = when 0.x = EVALUATE: In part (a) the work done is zero because the force and displacement are perpendicular. In part (b) the force is directed opposite to the displacement and the work done is negative. In part (c) the force and displacement are in the same direction and the work done is positive. 7.29. IDENTIFY: Since the force is constant, use cosW Fs ?= . SET UP: For both displacements, the direction of the friction force is opposite to the displacement and 180? = ° . EXECUTE: (a) When the book moves to the left, the friction force is to the right, and the work is (1.2 N)(3.0 m) 3.6 J.? = ? (b) The friction force is now to the left, and the work is again 3.6 J.? (c) 7.2 J.? (d) The net work done by friction for the round trip is not zero, and friction is not a conservative force. EVALUATE: The direction of the friction force depends on the motion of the object. For the gravity force, which is conservative, the force does not depend on the motion of the object. 7-10 Chapter 7 7.30. IDENTIFY and SET UP: The friction force is constant during each displacement and Eq.(6.2) can be used to calculate work, but the direction of the friction force can be different for different displacements. 2 k (0.25)(1.5 kg)(9.80 m/s ) 3.675 N;f mg?= = = direction of f ! is opposite to the motion. EXECUTE: (a) The path of the book is sketched in Figure 7.30a. Figure 7.30a For the motion from you to Beth the friction force is directed opposite to the displacement s ! and 1 (3.675 N)(8.0 m) 29.4 J.W fs= ? = ? = ? For the motion from Beth to Carlos the friction force is again directed opposite to the displacement and 2 29.4 J.W = ? tot 1 2 29.4 J 29.4 J 59 JW W W= + = ? ? = ? (b) The path of the book is sketched in Figure 7.30b. 22(8.0 m) 11.3 ms = = Figure 7.30b f ! is opposite to ,s ! so (3.675 N)(11.3 m) 42 JW fs= ? = ? = ? (c) For the motion from Kim to you 29.4 JW fs= ? = ? Figure 7.30d The total work for the round trip is 29.4 J 29.4 J 59 J.? ? = ? (d) EVALUATE: Parts (a) and (b) show that for two different paths between you and Carlos, the work done by friction is different. Part (c) shows that when the starting and ending points are the same, the total work is not zero. Both these results show that the friction force is nonconservative. 7.31. IDENTIFY: The work done by a spring on an object attached to its end when the object moves from ix to fx is 2 21 1 i f2 2W kx kx= ? . This result holds for any ix and fx . SET UP: Assume for simplicity that 1x , 2x and 3x are all positive, corresponding to the spring being stretched. EXECUTE: (a) 2 21 1 22 ( )k x x? (b) 2 21 1 22 ( ).k x x? ? The total work is zero; the spring force is conservative. (c) From 1x to 3,x 2 21 3 12 ( ).W k x x= ? ? From 3x to 2x , 2 21 2 32 ( ).W k x x= ? ? The net work is 2 21 2 12 ( ).k x x? ? This is the same as the result of part (a). EVALUATE: The results of part (c) illustrate that the work done by a conservative force is path independent. For the motion from you to Kim W fs= ? (3.675 N)(8.0 m) 29.4 JW = ? = ? Figure 7.30c Potential Energy and Energy Conservation 7-11 7.32. IDENTIFY and SET UP: Use Eq.(7.17) to calculate the force from ( ).U x Use coordinates where the origin is at one atom. The other atom then has coordinate x. EXECUTE: 6 6 66 6 7 1 6 x dU d C d C F C dx dx x dx x x ? ? ? ?= ? = ? ? = + = ?? ? ? ?? ? ? ? The minus sign mean that xF is directed in the -direction,x? toward the origin. The force has magnitude 766 /C x and is attractive. EVALUATE: U depends only on x so F! is along the x-axis; it has no y or z components. 7.33. IDENTIFY: Apply Eq.(7.16). SET UP: The sign of xF indicates its direction. EXECUTE: 43 34 (4.8 J m )x dUF x xdx ?= ? = ? = ? . 4 3( 0.800 m) (4.8 J m )( 0.80 m) 2.46 N.xF ? = ? ? = The force is in the -direction.x+ EVALUATE: 0xF > when 0x < and 0xF < when 0x > , so the force is always directed towards the origin. 7.34. IDENTIFY: Apply ( )( ) dU xF x dx = ? . SET UP: 2(1/ ) 1d xdx x= ? EXECUTE: 1 2 1 21 2 2( / ) (1/ )( )x d Gm m x d x Gm mF x Gm mdx dx x ? ? ?= ? = = ?? ?? ? . The force on 2m is in the -directionx? . This is toward 1m , so the force is attractive. EVALUATE: By Newton's 3rd law the force on 1m due to 2m is 21 2 /Gm m x , in the -directionx+ (toward 2m ). The gravitational potential energy belongs to the system of the two masses. 7.35. IDENTIFY: Apply x UF x ?= ? ? and y U F y ?= ? ? . SET UP: 2 2 1/ 2( )r x y= + . 2 2 3 / 2(1/ ) ( ) r x x x y ? = ?? + and 2 2 3 / 2 (1/ ) ( ) r y y x y ? = ?? + . EXECUTE: (a) 1 2( ) Gm mU r r = ? . 1 21 2 2 2 3 / 2(1/ ) ( )x U r Gm m x F Gm m x x x y ? ?? ?= ? = + = ?? ?? ? +? ? and 1 2 1 2 2 2 3 / 2 (1/ ) ( )y U r Gm m y F Gm m y y x y ? ?? ?= ? = + = ?? ?? ? +? ? . (b) 2 2 3 / 2 3( )x y r+ = so 1 23x Gm m xF r= ? and 1 2 3y Gm m y F r = ? . 2 2 2 21 2 1 23 2x y Gm m Gm mF F F x yr r= + = + = . (c) xF and yF are negative. xF x?= and yF y?= , where ? is a constant, so F ! and the vector r ! from 1m to 2m are in the same direction. Therefore, F ! is directed toward 1m at the origin and F ! is attractive. EVALUATE: If ? is the angle between the vector r! that points from 1m to 2m , then cosxr ?= and sin y r ?= . This gives cosxF F ?= ? and sinyF F ?= ? , our more usual way of writing the components of a vector. 7.36. IDENTIFY: Apply Eq.(7.18). SET UP: 2 31 2ddx x x ? ? = ?? ?? ? and 2 3 1 2d dy y y ? ? = ?? ?? ? . EXECUTE: ? ?U U x y ? ?? ?? ?F = i j ! since U has no z-dependence. 3 3 2 2 and soU U , x yx y ? ?? ? ? ?= =? ? 3 3 3 3 2 2? ? 2 x y x y ? ? ? ?? ?? ?? = ? ?? ?? ? ? ? i j F = i + j + ! !! . EVALUATE: xF and x have the same sign and yF and y have the same sign. When 0x > , xF is in the -direction,x+ and so forth. 7-12 Chapter 7 7.37. IDENTIFY and SET UP: Use Eq.(7.17) to calculate the force from U . At equilibrium 0.F = (a) EXECUTE: The graphs are sketched in Figure 7.37. 12 6 a b U r r = ? 13 7 12 6dU a b F dr r r = ? = + ? Figure 7.37 (b) At equilibrium 0,F = so 0dU dr = 0F = implies 13 712 6 0a br r + ? = 66 12 ;br a= solution is the equilibrium distance 1/ 60 (2 / )r a b= U is a minimum at this r; the equilibrium is stable. (c) At 1/ 6(2 / ) ,r a b= 12 6 2 2/ / ( / 2 ) ( / 2 ) / 4 .U a r b r a b a b b a b a= ? = ? = ? At ,r ? ? 0.U = The energy that must be added is 2 / 4 .U b a?? = (d) 1/ 6 100 (2 / ) 1.13 10 mr a b ?= = × gives that 60 62 / 2.082 10 ma b ?= × and 59 6/ 4 2.402 10 mb a ?= × 2 18/ 4 ( / 4 ) 1.54 10 Jb a b b a ?= = × 59 6 18(2.402 10 m ) 1.54 10 Jb ? ?× = × and 78 66.41 10 J m .b ?= × ? Then 60 62 / 2.082 10 ma b ?= × gives 60 6( / 2)(2.082 10 m )a b ?= × = 78 6 60 6 138 121 2 (6.41 10 J m )(2.082 10 m ) 6.67 10 J m ? ? ?× ? × = × ? EVALUATE: As the graphs in part (a) show, ( )F r is the slope of ( )U r at each r. ( )U r has a minimum where 0.F = 7.38. IDENTIFY: Apply Eq.(7.16). SET UP: dU dx is the slope of the U versus x graph. EXECUTE: (a) Considering only forces in the x-direction, x dUF dx= ? and so the force is zero when the slope of the U vs x graph is zero, at points b and d. (b) Point b is at a potential minimum; to move it away from b would require an input of energy, so this point is stable. (c) Moving away from point d involves a decrease of potential energy, hence an increase in kinetic energy, and the marble tends to move further away, and so d is an unstable point. EVALUATE: At point b, xF is negative when the marble is displaced slightly to the right and xF is positive when the marble is displaced slightly to the left, the force is a restoring force, and the equilibrium is stable. At point d, a small displacement in either direction produces a force directed away from d and the equilibrium is unstable. 7.39. IDENTIFY: Apply m=?F a! ! to the bag and to the box. Apply Eq.(7.7) to the motion of the system of the box and bucket after the bag is removed. SET UP: Let 0y = at the final height of the bucket, so 1 2.00 my = and 2 0y = . 1 0K = . The box and the bucket move with the same speed v, so 212 box bucket2 ( )K m m v= + . other kW f d= ? , with 2.00 md = and k k boxf m g?= . Before the bag is removed, the maximum possible friction force the roof can exert on the box is 2(0.700)(80.0 kg 50.0 kg)(9.80 m/s ) 892 N+ = . This is larger than the weight of the bucket (637 N), so before the bag is removed the system is at rest. EXECUTE: (a) The friction force on the bag of gravel is zero, since there is no other horizontal force on the bag for friction to oppose. The static friction force on the box equals the weight of the bucket, 637 N. Potential Energy and Energy Conservation 7-13 (b) Eq.(7.7) gives 21bucket 1 k tot2m gy f d m v? = , with tot 145.0 kgm = . bucket 1 k box tot 2 ( )v m gy m gd m ?= ? . 2 22 (65.0 kg)(9.80 m/s )(2.00 m) (0.400)(80.0 kg)(9.80 m/s )(2.00 m) 145.0 kg v ? ?= ?? ? . 2.99 m/sv = . EVALUATE: If we apply m=?F a! ! to the box and to the bucket we can calculate their common acceleration a. Then a constant acceleration equation applied to either object gives 2.99 m/sv = , in agreement with our result obtained using energy methods. 7.40. IDENTIFY: For the system of two blocks, only gravity does work. Apply Eq.(7.5). SET UP: Call the blocks A and B, where A is the more massive one. 1 1 0A Bv v= = . Let 0y = for each block to be at the initial height of that block, so 1 1 0A By y= = . 2 1.20 mAy = ? and 2 1.20 mBy = + . 2 2 2 3.00 m/sA Bv v v= = = . EXECUTE: Eq.(7.5) gives 21 220 ( ) (1.20 m)( )A B B Am m v g m m= + + ? . 15.0 kgA Bm m+ = . 2 21 2 (15.0 kg)(3.00 m/s) (9.80 m/s )(1.20 m)(15.0 kg 2 )Am+ ? . Solving for Am gives 10.4 kgAm = . And then 4.6 kgBm = . EVALUATE: The final kinetic energy of the two blocks is 68 J. The potential energy of block A decreases by 122 J. The potential energy of block B increases by 54 J. The total decrease in potential energy is 122 J 54 J 68 J,? = and this equals the increase in kinetic energy of the system. 7.41. IDENTIFY: Apply 1 1 other 2 2K U W K U+ + = + SET UP: 1 2 2 0U U K= = = . other k with 280 ft 85.3 mfW W mgs, s?= = ? = = EXECUTE: (a) The work-energy expression gives 21 1 k2 0mv mgs?? = . 1 k2 22.4 m/s 50 mph;v gs?= = = the driver was speeding. (b) 15 mph over speed limit so $150 ticket. EVALUATE: The negative work done by friction removes the kinetic energy of the object. 7.42. IDENTIFY: Apply Eq.(7.14). SET UP: Only the spring force and gravity do work, so other 0W = . Let 0y = at the horizontal surface. EXECUTE: (a) Equating the potential energy stored in the spring to the block's kinetic energy, 2 21 12 2kx mv ,= or 400 N/m (0.220 m) 3.11 m/s. 2.00 kg k v x m = = = (b) Using energy methods directly, the initial potential energy of the spring equals the final gravitational potential energy, 212 sin ,kx mgL ?= or 2 21 1 2 2 2 (400 N/m)(0.220 m) 0.821 m. sin (2.00 kg)(9.80 m/s )sin37.0 kx L mg ?= = =° EVALUATE: The total energy of the system is constant. Initially it is all elastic potential energy stored in the spring, then it is all kinetic energy and finally it is all gravitational potential energy. 7.43. IDENTIFY: Use the work-energy theorem, Eq(7.7). The target variable k? will be a factor in the work done by friction. SET UP: Let point 1 be where the block is released and let point 2 be where the block stops, as shown in Figure 7.43. 1 1 other 2 2K U W K U+ + = + Work is done on the block by the spring and by friction, so other fW W= and el.U U= Figure 7.43 EXECUTE: 1 2 0K K= = 2 21 1 1 1,el 12 2 (100 N/m)(0.200 m) 2.00 JU U kx= = = = 2 2,el 0,U U= = since after the block leaves the spring has given up all its stored energy other k k k( cos ) (cos ) ,fW W f s mg s mgs? ? ? ?= = = = ? since 180? = ° (The friction force is directed opposite to the displacement and does negative work.) 7-14 Chapter 7 Putting all this into 1 1 other 2 2K U W K U+ + = + gives 1,el 0fU W+ = k 1,elmgs U? = 1,el k 2 200 J 0.41. (0.50 kg)(9.80 m/s )(1.00 m) U mgs ? = = = EVALUATE: 1,el 0fU W+ = says that the potential energy originally stored in the spring is taken out of the system by the negative work done by friction. 7.44. IDENTIFY: Apply Eq.(7.14). Calculate kf from the fact that the crate slides a distance 5.60 mx = before coming to rest. Then apply Eq.(7.14) again, with 2.00 mx = . SET UP: 1 el 360 JU U= = . 2 0U = . 1 0K = . other kW f x= ? . EXECUTE: Work done by friction against the crate brings it to a halt: 1 otherU W= ? . k potential energy of compressed springf x = , and k 360 J 64.29 N5.60 mf = = . The friction force working over a 2.00-m distance does work equal to k (64.29 N)(2.00 m) 128.6 J.f x? = ? = ? The kinetic energy of the crate at this point is thus 360 J 128.6 J 231.4 J,? = and its speed is found from 2 / 2 231.4 Jmv = , so 2(231.4 J) 3.04 m/s 50 0 kg v . = = . EVALUATE: The energy of the compressed spring goes partly into kinetic energy of the crate and is partly removed by the negative work done by friction. After the crate leaves the spring the crate slows down as friction does negative work on it. 7.45. IDENTIFY: At its highest point between bounces all the mechanical energy of the ball is in the form of gravitational potential energy. SET UP: E U mgh= = , where h is the height at the highest point of the motion. EXECUTE: (a) 2(0.650 kg)(9.80 m/s )(2.50 m) 15.9 Jmgh = = (b) The second height is 0.75(2.50 m) 1.875 m,= so the second 11.9 J ;mgh = it loses 15.9 J 11.9 J 4.0 J? = on first bounce. This energy is converted to thermal energy. (c) The third height is 0.75(1.875 m) 1.40 m,= , so third 8.9 J ;mgh = it loses 11.9 J 8.9 J 3.0 J? = on second bounce. EVALUATE: In each bounce the ball loses 25% of its mechanical energy. 7.46. IDENTIFY: Apply Eq.(7.14) to relate h and Bv . Apply m=?F a! ! at point B to find the minimum speed required at B for the car not to fall off the track. SET UP: At B, 2 /Ba v R= , downward. The minimum speed is when 0n ? and 2 /Bmg mv R= . The minimum speed required is Bv gR= . 1 0K = and other 0W = . EXECUTE: (a) Eq.(7.14) applied to points A and B gives 212A B BU U mv? = . The speed at the top must be at least .g R Thus, 1 5 ( 2 ) or . 2 2 mg h R mgR, h R? > > (b) Apply Eq.(7.14) to points A and C. (2.50)A C CU U Rmg K ,? = = so 2(5.00) (5.00)(9.80 m/s )(20.0 m) 31.3 m/s.Cv gR= = = The radial acceleration is 2 2 rad 49.0 m/s . Cva R = = The tangential direction is down, the normal force at point C is horizontal, there is no friction, so the only downward force is gravity, and 2tan 9.80 m/s .a g= = EVALUATE: If 52h R> , then the downward acceleration at B due to the circular motion is greater than g and the track must exert a downward normal force n . n increases as h increases and hence Bv increases. 7.47. (a) IDENTIFY: Use work-energy relation to find the kinetic energy of the wood as it enters the rough bottom. SET UP: Let point 1 be where the piece of wood is released and point 2 be just before it enters the rough bottom. Let 0y = be at point 2. EXECUTE: 1 2U K= gives 2 1 78.4 J.K mgy= = IDENTIFY: Now apply work-energy relation to the motion along the rough bottom. Potential Energy and Energy Conservation 7-15 SET UP: Let point 1 be where it enters the rough bottom and point 2 be where it stops. 1 1 other 2 2K U W K U+ + = + EXECUTE: other k ,fW W mgs?= = ? 2 1 2 0;K U U= = = 1 78.4 JK = k78.4 J 0;mgs?? = solving for s gives 20.0 m.s = The wood stops after traveling 20.0 m along the rough bottom. (b) Friction does 78.4 J? of work. EVALUATE: The piece of wood stops before it makes one trip across the rough bottom. The final mechanical energy is zero. The negative friction work takes away all the mechanical energy initially in the system. 7.48. IDENTIFY: Apply Eq.(7.14) to the rock. kother f W W= . SET UP: Let 0y = at the foot of the hill, so 1 0U = and 2U mgh= , where h is the vertical height of the rock above the foot of the hill when it stops. EXECUTE: (a) At the maximum height, 2 0K = . Eq.(7.14) gives kBottom TopfK W U+ = . 2 0 k 1 cos 2 mv mg ? d mgh?? = . sind h ?= , so 20 k1 cos2 sin h v g gh? ? ?? = . 2 2 21 cos40(15 m/s) (0.20)(9.8 m/s ) (9.8 m/s ) 2 sin 40 h h °? =° and 9.3 mh = . (b) Compare maximum static friction force to the weight component down the plane. 2 s s cos (0.75)(28 kg)(9.8 m/s )cos40 158 Nf mg? ?= = ° = . 2 ssin (28 kg)(9.8 m/s )(sin 40 ) 176 Nmg f? = ° = > , so the rock will slide down. (c) Use same procedure as in part (a), with 9.3 mh = and Bv being the speed at the bottom of the hill. kTop Bf U W K+ = . 2k B1cos sin 2 h mgh mg mv? ? ?? = and B k2 2 cos sin 11.8 m/sv gh gh? ? ?= ? = . EVALUATE: For the round trip up the hill and back down, there is negative work done by friction and the speed of the rock when it returns to the bottom of the hill is less than the speed it had when it started up the hill. 7.49. IDENTIFY: Apply Eq.(7.7) to the motion of the stone. SET UP: 1 1 other 2 2K U W K U+ + = + Let point 1 be point A and point 2 be point B. Take 0y = at point B. EXECUTE: 2 21 11 1 22 2 ,mgy mv mv+ = with 20.0 mh = and 1 10.0 m/sv = 2 2 1 2 22.2 m/sv v gh= + = EVALUATE: The loss of gravitational potential energy equals the gain of kinetic energy. (b) IDENTIFY: Apply Eq.(7.8) to the motion of the stone from point B to where it comes to rest against the spring. SET UP: Use 1 1 other 2 2 ,K U W K U+ + = + with point 1 at B and point 2 where the spring has its maximum compression x. EXECUTE: 1 2 2 0;U U K= = = 211 12K mv= with 1 22.2 m/sv = 21 other el k 2 ,fW W W mgs kx?= + = ? ? with 100 ms x= + The work-energy relation gives 1 other 0.K W+ = 2 21 1 1 k2 2 0mv mgs kx?? ? = Putting in the numerical values gives 2 29.4 750 0.x x+ ? = The positive root to this equation is 16.4 m.x = EVALUATE: Part of the initial mechanical (kinetic) energy is removed by friction work and the rest goes into the potential energy stored in the spring. (c) IDENTIFY and SET UP: Consider the forces. EXECUTE: When the spring is compressed 16.4 mx = the force it exerts on the stone is el 32.8 N.F kx= = The maximum possible static friction force is 2 s smax (0.80)(15.0 kg)(9.80 m/s ) 118 N.f mg?= = = EVALUATE: The spring force is less than the maximum possible static friction force so the stone remains at rest. 7-16 Chapter 7 7.50. IDENTIFY: Once the block leaves the top of the hill it moves in projectile motion. Use Eq.(7.14) to relate the speed Bv at the bottom of the hill to the speed Topv at the top and the 70 m height of the hill. SET UP: For the projectile motion, take y+ to be downward. 0xa = , ya g= . 0 Topxv v= , 0 0yv = . For the motion up the hill only gravity does work. Take 0y = at the base of the hill. EXECUTE: First get speed at the top of the hill for the block to clear the pit. 21 2 y gt= . 2 2120 m (9.8 m/s ) 2 t= . 2.0 st = . Then Top 40 mv t = gives Top 40 m 20 m/s2.0 sv = = . Energy conservation applied to the motion up the hill: Bottom Top TopK U K= + gives 2 2 B Top 1 1 2 2 mv mgh mv= + . 2 2 2B Top 2 (20 m/s) 2(9.8 m/s )(70 m) 42 m/sv v gh= + = + = . EVALUATE: The result does not depend on the mass of the block. 7.51. IDENTIFY: Apply 1 1 other 2 2K U W K U+ + = + to the motion of the person. SET UP: Point 1 is where he steps off the platform and point 2 is where he is stopped by the cord. Let 0y = at point 2. 1 41.0 m.y = 21other 2 ,W kx= ? where 11.0 mx = is the amount the cord is stretched at point 2. The cord does negative work. EXECUTE: 1 2 2 0,K K U= = = so 211 2 0mgy kx? = and 631 N/m.k = Now apply F kx= to the test pulls: F kx= so / 0.602 m.x F k= = EVALUATE: All his initial gravitational potential energy is taken away by the negative work done by the force exerted by the cord, and this amount of energy is stored as elastic potential energy in the stretched cord. 7.52. IDENTIFY: Apply Eq.(7.14) to the motion of the skier from the gate to the bottom of the ramp. SET UP: other 4000 JW = ? . Let 0y = at the bottom of the ramp. EXECUTE: For the skier to be moving at no more than 30.0 m/s ; his kinetic energy at the bottom of the ramp can be no bigger than 2 2(85.0 kg)(30.0 m/s) 38,250 J 2 2 mv = = . Friction does 4000 J? of work on him during his run, which means his combined U and K at the top of the ramp must be no more than 38,250 J 4000 J 42,250 J.+ = His K at the top is 2 2(85.0 kg)(2.0 m/s) 170 J 2 2 mv = = . His U at the top should thus be no more than 42,250 J 170 J 42,080 J,? = which gives a height above the bottom of the ramp of 2 42,080 J 42,080 J 50.5 m. (85.0 kg)(9.80 m/s ) h mg = = = EVALUATE: In the absence of air resistance, for this h his speed at the bottom of the ramp would be 31.5 m/s. The work done by air resistance is small compared to the kinetic and potential energies that enter into the calculation. 7.53. IDENTIFY: Use the work-energy theorem, Eq.(7.7). Solve for 2K and then for 2.v SET UP: Let point 1 be at his initial position against the compressed spring and let point 2 be at the end of the barrel, as shown in Figure 7.53. Use F kx= to find the amount the spring is initially compressed by the 4400 N force. 1 1 other 2 2K U W K U+ + = + Take 0y = at his initial position. EXECUTE: 1 0,K = 212 22K mv= other fricW W fs= = ? other (40 N)(4.0 m) 160 JW = ? = ? Figure 7.53 1,grav 0,U = 211,el 2 ,U kd= where d is the distance the spring is initially compressed. F kd= so 4400 N 4.00 m 1100 N/m F d k = = = and 211,el 2 (1100 N/m)(4.00 m) 8800 JU = = 2 2,grav 2 (60 kg)(9.80 m/s )(2.5 m) 1470 J,U mgy= = = 2,el 0U = Potential Energy and Energy Conservation 7-17 Then 1 1 other 2 2K U W K U+ + = + gives 21 228800 J 160 J 1470 Jmv? = + 21 22 7170 Jmv = and 2 2(7170 J) 15.5 m/s60 kgv = = EVALUATE: Some of the potential energy stored in the compressed spring is taken away by the work done by friction. The rest goes partly into gravitational potential energy and partly into kinetic energy. 7.54. IDENTIFY: To be at equilibrium at the bottom, with the spring compressed a distance 0x , the spring force must balance the component of the weight down the ramp plus the largest value of the static friction, or 0 sin .kx w ? f= + Apply Eq.(7.14) to the motion down the ramp. SET UP: 2 0K = , 211 2K mv= , where v is the speed at the top of the ramp. Let 2 0U = , so 1 sinU wL ?= , where L is the total length traveled down the ramp. EXECUTE: Eq.(7.14) gives 2 201 1( sin )2 2kx w f L mv?= ? + . With the given parameters, 21 02 248 Jkx = and 3 0 1.10 10 N.kx = × Solving for k gives 2440 N/m.k = EVALUATE: 0 0.451 mx = . sin 551 Nw ? = . The decrease in gravitational potential energy is only slightly larger than the amount of mechanical energy removed by the negative work done by friction. 212 243 Jmv = . The energy stored in the spring is only slightly larger than the initial kinetic energy of the crate at the top of the ramp. 7.55. IDENTIFY: Apply Eq.(7.7) to the system consisting of the two buckets. If we ignore the inertia of the pulley we ignore the kinetic energy it has. SET UP: 1 1 other 2 2.K U W K U+ + = + Points 1 and 2 in the motion are sketched in Figure 7.55. Figure 7.55 The tension force does positive work on the 4.0 kg bucket and an equal amount of negative work on the 12.0 kg bucket, so the net work done by the tension is zero. Work is done on the system only by gravity, so other 0W = and gravU U= EXECUTE: 1 0K = 2 21 1 2 ,2 ,22 2A A B BK m v m v= + But since the two buckets are connected by a rope they move together and have the same speed: ,2 ,2 2.A Bv v v= = Thus 2 212 2 22 ( ) (8.00 kg) .A BK m m v v= + = 2 1 ,1 (12.0 kg)(9.80 m/s )(2.00 m) 235.2 J.A AU m gy= = = 2 2 ,2 (4.0 kg)(9.80 m/s )(2.00 m) 78.4 J.B BU m gy= = = Putting all this into 1 1 other 2 2K U W K U+ + = + gives 1 2 2U K U= + 2 2235.2 J (8.00 kg) 78.4 Jv= + 2 235.2 J 78.4 J 4.4 m/s 8.00 kg v ?= = EVALUATE: The gravitational potential energy decreases and the kinetic energy increases by the same amount. We could apply Eq.(7.7) to one bucket, but then we would have to include in otherW the work done on the bucket by the tension T . 7-18 Chapter 7 7.56. IDENTIFY: Apply 1 1 other 2 2K U W K U+ + = + to the motion of the rocket from the starting point to the base of the ramp. otherW is the work done by the thrust and by friction. SET UP: Let point 1 be at the starting point and let point 2 be at the base of the ramp. 1 0v = , 2 50.0 m/sv = . Let 0y = at the base and take y+ upward. Then 2 0y = and 1 sin53y d= ° , where d is the distance along the ramp from the base to the starting point. Friction does negative work. EXECUTE: 1 0K = , 2 0U = . 1 other 2U W K+ = . other (2000 N) (500 N) (1500 N)W d d d= ? = . 21 22sin53 (1500 N)mgd d mv+ =° . 2 2 2 2 (1500 kg)(50.0 m/s) 142 m 2[ sin53 1500 N] 2[(1500 kg)(9.80 m/s )sin53 1500 N] mv d mg = = =+ +° ° . EVALUATE: The initial height is 1 (142 m)sin53 113 my = =° . An object free-falling from this distance attains a speed 12 47.1 m/sv gy= = . The rocket attains a greater speed than this because the forward thrust is greater than the friction force. 7.57. IDENTIFY: The force exerted by a spring is xF kx= ? . The acceleration of the object is given by x xF ma= . Apply Eq.(7.14) to relate position and speed. SET UP: Let x+ be when the spring is stretched. EXECUTE: (a) 212U kx= . Let point 1 be when the spring is initially compressed a distance 0x , so 1 0x x= ? . 1 0K = . other 0W = . 21 0 2 22 kx U K= + . The speed is maximum when 0x = , so 2 0U = . Then 2 21 10 22 2kx mv= and 2 0 /v x k m= is this maximum speed. (b) xF kx= ? and x xF ma= give x ka xm= ? . a is maximum when x is maximum, so 0 k a x m = . (c) The speed is maximum when 0x = , when the spring has returned to its natural length, and the acceleration is maximum when 0x x= ? , at the initial compression of the spring. (d) When the spring has maximum extension, 2 0v = . 2 21 102 2kx kx= and 0x x= .The magnitude of the maximum extension equals the magnitude of the maximum compression. (e) The machine part oscillates between 0x x= ? and 0x x= + and never stops permanently. EVALUATE: In any real system there are mechanical energy losses, for example due to negative work done by friction, and the object eventually comes to rest. 7.58. IDENTIFY: Conservation of energy says the decrease in potential energy equals the gain in kinetic energy. SET UP: Since the two animals are equidistant from the axis, they each have the same speed v. EXECUTE: One mass rises while the other falls, so the net loss of potential energy is 2(0.500 kg 0.200 kg)(9.80 m/s )(0.400 m) 1.176 J.? = This is the sum of the kinetic energies of the animals and is equal to 21 tot2 m v , and 2(1.176 J) 1.83 m/s. (0.700 kg) v = = EVALUATE: The mouse gains both gravitational potential energy and kinetic energy. The rat?s gain in kinetic energy is less than its decrease of potential energy, and the energy difference is transferred to the mouse. 7.59. (a) IDENTIFY and SET UP: Apply Eq.(7.7) to the motion of the potato. Let point 1 be where the potato is released and point 2 be at the lowest point in its motion, as shown in Figure 7.59a. 1 1 other 2 2K U W K U+ + = + 1 2.50 my = 2 0y = The tension in the string is at all points in the motion perpendicular to the displacement, so 0TW = The only force that does work on the potato is gravity, so other 0.W = Figure 7.59a Potential Energy and Energy Conservation 7-19 EXECUTE: 1 0,K = 212 22 ,K mv= 1 1,U mgy= 2 0U = Thus 1 2.U K= 21 1 22mgy mv= 2 2 12 2(9.80 m/s )(2.50 m) 7.00 m/sv gy= = = EVALUATE: 2v is the same as if the potato fell through 2.50 m. (b) IDENTIFY: Apply m=?F a! ! to the potato. The potato moves in an arc of a circle so its acceleration is rad ,a! where 2rad /a v R= and is directed toward the center of the circle. Solve for one of the forces, the tension T in the string. SET UP: The free-body diagram for the potato as it swings through its lowest point is given in Figure 7.59b. The acceleration rada ! is directed in toward the center of the circular path, so at this point it is upward. Figure 7.59b EXECUTE: y yF ma=? radT mg ma? = 2 2 rad( ) , v T m g a m g R ? ?= + = +? ?? ? where the radius R for the circular motion is the length L of the string. It is instructive to use the algebraic expression for 2v from part (a) rather than just putting in the numerical value: 2 12 2 ,v gy gL= = so 22 2v gL= Then 2 2 2 3 ; v gL T m g m g mg L L ? ? ? ?= + = + =? ? ? ?? ?? ? the tension at this point is three times the weight of the potato. 23 3(0.100 kg)(9.80 m/s ) 2.94 NT mg= = = EVALUATE: The tension is greater than the weight; the acceleration is upward so the net force must be upward. 7.60. IDENTIFY: Eq.(7.14) says other 2 2 1 1( )W K U K U= + ? + . otherW is the work done on the baseball by the force exerted by the air. SET UP: U mgy= . 212K mv= , where 2 2 2x yv v v= + . EXECUTE: (a) The change in total energy is the work done by the air, 2 2 other 2 2 1 1 2 1 2 1 ( ) ( ) ( ) 2 W K U K U m v v gy? ?= + ? + = ? +? ?? ? . ( )2 2 2 2other (0.145 kg) (1/ 2 (18.6 m/s) (30.0 m/s) (40.0 m/s) (9.80 m/s )(53.6 m)W ? ?= ? ? +? ? . other 80.0 JW = ? . (b) Similarly, other 3 3 2 2( ) ( )W K U K U= + ? + . ( )2 2 2 2other (0.145 kg) (1/ 2) (11.9 m/s) ( 28.7 m/s) (18.6 m/s) (9.80 m/s )(53.6 m)W ? ?= + ? ? ?? ? . other 31.3 J.W = ? (c) The ball is moving slower on the way down, and does not go as far (in the x-direction), and so the work done by the air is smaller in magnitude. EVALUATE: The initial kinetic energy of the baseball is 212 (0.145 kg)(50.0 m/s) 181 J= . For the total motion from the ground, up to the maximum height, and back down the total work done by the air is 111 J. The ball returns to the ground with 181 J 111 J 70 J? = of kinetic energy and a speed of 31 m/s, less than its initial speed of 50 m/s. 7-20 Chapter 7 7.61. IDENTIFY and SET UP: There are two situations to compare: stepping off a platform and sliding down a pole. Apply the work-energy theorem to each. (a) EXECUTE: Speed at ground if steps off platform at height h: 1 1 other 2 2K U W K U+ + = + 21 22 ,mgh mv= so 22 2v gh= Motion from top to bottom of pole: (take 0y = at bottom) 1 1 other 2 2K U W K U+ + = + 21 22mgd fd mv? = Use 22 2v gh= and get mgd fd mgh? = ( )fd mg d h= ? ( ) / (1 / )f mg d h d mg h d= ? = ? EVALUATE: For h d= this gives 0f = as it should (friction has no effect). For 0,h = 2 0v = (no motion). The equation for f gives f mg= in this special case. When f mg= the forces on him cancel and he doesn?t accelerate down the pole, which agrees with 2 0.v = (b) EXECUTE: 2(1 / ) (75 kg)(9.80 m/s )(1 1.0 m/2.5 m) 441 N.f mg h d= ? = ? = (c) Take 0y = at bottom of pole, so 1y d= and 2 .y y= 1 1 other 2 2K U W K U+ + = + 21 20 ( )mgd f d y mv mgy+ ? ? = + 21 2 ( ) ( )mv mg d y f d y= ? ? ? Using (1 / )f mg h d= ? gives 212 ( ) (1 / )( )mv mg d y mg h d d y= ? ? ? ? 21 2 ( / )( )mv mg h d d y= ? and 2 (1 / )v gh y d= ? EVALUATE: This gives the correct results for 0y = and for .y d= 7.62. IDENTIFY: Apply Eq.(7.14) to each stage of the motion. SET UP: Let 0y = at the bottom of the slope. In part (a), otherW is the work done by friction. In part (b), otherW is the work done by friction and the air resistance force. In part (c), otherW is the work done by the force exerted by the snowdrift. EXECUTE: (a) The skier?s kinetic energy at the bottom can be found from the potential energy at the top minus the work done by friction, 1 (60.0 kg)(9.8 N/kg)(65.0 m) 10,500 J,fK mgh W= ? = ? or 1 38,200 J 10,500 J 27,720 JK = ? = . Then 11 2 2(27,720 J) 30.4 m/s60 kg K v m = = = . (b) 2 1 air k air( ) 27,720 J ( ).fK K W W mgd f d?= ? + = ? + 2 27,720 J [(0.2)(588 N)(82 m) (160 N)(82 m)]K = ? + or 2 27,720 J 22,763 J 4957 JK = ? = . Then, 2 2 2(4957 J) 12.9 m/s60 kg K v m = = = (c) Use the Work-Energy Theorem to find the force. ,W K= ? / (4957 J) (2.5 m) 2000 NF K d= = = . EVALUATE: In each case, otherW is negative and removes mechanical energy from the system. 7.63. IDENTIFY and SET UP: First apply m=?F a! ! to the skier. Find the angle ? where the normal force becomes zero, in terms of the speed v2 at this point. Then apply the work-energy theorem to the motion of the skier to obtain another equation that relates v2 and .? Solve these two equations for .? Let point 2 be where the skier loses contact with the snowball, as sketched in Figure 7.63a Loses contact implies 0.n ? 1 ,y R= 2 cosy R ?= Figure 7.63a Potential Energy and Energy Conservation 7-21 First, analyze the forces on the skier when she is at point 2. The free-body diagram is given in Figure 7.63b. For this use coordinates that are in the tangential and radial directions. The skier moves in an arc of a circle, so her acceleration is 2rad / ,a v R= directed in towards the center of the snowball. EXECUTE: y yF ma=? 2 2cos /mg n mv R? ? = But 0n = so 22cos /mg mv R? = 2 2 cosv Rg ?= Figure 7.63b Now use conservation of energy to get another equation relating 2v to :? 1 1 other 2 2K U W K U+ + = + The only force that does work on the skier is gravity, so other 0.W = 1 0,K = 212 22K mv= 1 1 ,U mgy mgR= = 2 2 cosU mgy mgR ?= = Then 21 22 cosmgR mv mgR ?= + 2 2 2 (1 cos )v gR ?= ? Combine this with the y yF ma=? equation: cos 2 (1 cos )Rg gR? ?= ? cos 2 2cos? ?= ? 3cos 2? = so cos 2/3? = and 48.2? = ° EVALUATE: She speeds up and her rada increases as she loses gravitational potential energy. She loses contact when she is going so fast that the radially inward component of her weight isn?t large enough to keep her in the circular path. Note that ? where she loses contact does not depend on her mass or on the radius of the snowball. 7.64. IDENTIFY: Use conservation of energy to relate the speed at the lowest point to the speed at the highest point. Use m=?F a! ! to calculate the tension. SET UP: The rock has acceleration 2rad /a v R= , directed toward the center of the circle. EXECUTE: If the speed of the rock at the top is tv , then conservation of energy gives the speed bv at the bottom from 2 21 1b t2 2 (2 )mv mv mg R= + , R being the radius of the circle, and so 2 2b t 4v v gR= + . The tension at the top and bottom are found from 2 t t mv T mg R + = and 2 b b mv T mg R ? = , so 2 2b t b t( ) 2 6 6mT T v v mg mg wR? = ? + = = . EVALUATE: The tensions tT and bT depend on the speed of the rock and on R, but the difference b tT T? is independent of the speed of the rock and the radius of the circle. 7.65. IDENTIFY and SET UP: Ay R= 0B Cy y= = Figure 7.65 (a) Apply conservation of energy to the motion from B to C: other .B B C CK U W K U+ + = + The motion is described in Figure 7.65. EXECUTE: The only force that does work on the package during this part of the motion is friction, so other k k k(cos ) (cos180 )fW W f s mg s mgs? ? ?= = = ° = ? 21 2 ,B BK mv= 0CK = 0,BU = 0CU = 7-22 Chapter 7 Thus 0B fK W+ = 21 k2 0Bmv mgs?? = 2 2 k 2 (4.80 m/s) 0.392 2 2(9.80 m/s )(3.00 m) B gs ?? = = = EVALUATE: The negative friction work takes away all the kinetic energy. (b) IDENTIFY and SET UP: Apply conservation of energy to the motion from A to B: otherA A B BK U W K U+ + = + EXECUTE: Work is done by gravity and by friction, so other .fW W= 0,AK = 2 21 12 2 (0.200 kg)(4.80 m/s) 2.304 JB BK mv= = = 2(0.200 kg)(9.80 m/s )(1.60 m) 3.136 J,A AU mgy mgR= = = = 0BU = Thus A f BU W K+ = 2.304 J 3.136 J 0.83 Jf B AW K U= ? = ? = ? EVALUATE: fW is negative as expected; the friction force does negative work since it is directed opposite to the displacement. 7.66. IDENTIFY: Apply Eq.(7.14) to the initial and final positions of the truck. SET UP: Let 0y = at the lowest point of the path of the truck. otherW is the work done by friction. r r r cosf n mg? ? ?= = . EXECUTE: Denote the distance the truck moves up the ramp by x. 211 02K mv= , 1 sinU mgL ?= , 2 0K = , 2 sinU mgx ?= and other r cosW mgx? ?= ? . From other 2 2 1 1( ) ( )W K U K U= + ? + , and solving for x, 2 1 0 r r sin ( /2 ) sin . (sin cos ) sin cos K mgL v g L x mg ? ? ? ? ? ? ? ? + += =+ + EVALUATE: x increases when 0v increases and decreases when r? increases. 7.67. 2 ,xF x x? ?= ? ? 60.0 N/m? = and 218.0 N/m? = (a) IDENTIFY: Use Eq.(6.7) to calculate W and then use W U= ?? to identify the potential energy function ( ).U x SET UP: 2 1 1 2 ( ) x x F xx W U U F x dx= ? = ? Let 1 0x = and 1 0.U = Let 2x be some arbitrary point x, so 2 ( ).U U x= EXECUTE: 2 2 2 31 12 30 0 0( ) ( ) ( ) ( ) . x x x xU x F x dx x x dx x x dx x x? ? ? ? ? ?= ? = ? ? ? = + = +? ? ? EVALUATE: If 0,? = the spring does obey Hooke?s law, with ,k ?= and our result reduces to 212 .kx (b) IDENTIFY: Apply Eq.(7.15) to the motion of the object. SET UP: The system at points 1 and 2 is sketched in Figure 7.67. 1 1 other 2 2K U W K U+ + = + The only force that does work on the object is the spring force, so other 0.W = Figure 7.67 EXECUTE: 1 0,K = 212 22K mv= 2 3 2 2 31 1 1 1 1 1 1 12 3 2 3( ) (60.0 N/m)(1.00 m) (18.0 N/m )(1.00 m)U U x x x? ?= = + = + 36.0 J= 2 3 2 2 31 1 1 1 2 2 2 22 3 2 3( ) (60.0 N/m)(0.500 m) (18.0 N/m )(0.500 m)U U x x x? ?= = + = + 8.25 J= Thus 21 2236.0 J 8.25 Jmv= + 2 2(36.0 J 8.25 J) 7.85 m/s 0.900 kg v ?= = EVALUATE: The elastic potential energy stored in the spring decreases and the kinetic energy of the object increases. Potential Energy and Energy Conservation 7-23 7.68. IDENTIFY: Apply Eq.(7.14). otherW is the work done by F. SET UP: otherW K U= ? + ? . The distance the spring stretches is a? . 2 1 siny y a ?? = . EXECUTE: The force increases both the gravitational potential energy of the block and the potential energy of the spring. If the block is moved slowly, the kinetic energy can be taken as constant, so the work done by the force is the increase in potential energy, 212sin ( )U mga k a? ?? = + . EVALUATE: The force is kept tangent to the surface so the block will stay in contact with the surface. 7.69. IDENTIFY: Apply Eq.(7.14) to the motion of the block. SET UP: Let 0y = at the floor. Let point 1 be the initial position of the block against the compressed spring and let point 2 be just before the block strikes the floor. EXECUTE: With 2 10, 0U K= = , 2 1K U= . 2 21 122 2mv kx mgh= + . Solving for 2v , 2 2 2 2 (1900 N/m)(0.045 m) 2 2(9.80 m/s )(1.20 m) 7.01 m/s (0.150 kg) kx v gh m = + = + = . EVALUATE: The potential energy stored in the spring and the initial gravitational potential energy all go into the final kinetic energy of the block. 7.70. IDENTIFY: Apply Eq.(7.14). U is the total elastic potential energy of the two springs. SET UP: Call the two points in the motion where Eq.(7.14) is applied A and B to avoid confusion with springs 1 and 2, that have force constants 1k and 2k . At any point in the motion the distance one spring is stretched equals the distance the other spring is compressed. Let x+ be to the right. Let point A be the initial position of the block, where it is released from rest, so 1 0.150 mAx = + and 2 0.150 mAx = ? . EXECUTE: (a) With no friction, other 0W = . 0AK = and A B BU K U= + . The maximum speed is when 0BU = and this is at 1 2 0B Bx x= = , when both springs are at their natural length. 2 2 21 1 11 1 2 22 2 2A A Bk x k x mv+ = . 2 2 2 1 2 (0.150 m)A Ax x= = , so 1 2 2500 N/m 2000 N/m(0.150 m) (0.150 m) 5.81 m/s3.00 kgB k k v m + += = = . (b) At maximum compression of spring 1, spring 2 has its maximum extension and 0Bv = . Therefore, at this point A BU U= . The distance spring 1 is compressed equals the distance spring 2 is stretched, and vice versa: 1 2A Ax x= ? and 1 2B Bx x= ? . Then A BU U= gives 2 21 11 2 1 1 2 12 2( ) ( )A Bk k x k k x+ = + and 1 1 0.150 mB Ax x= ? = ? . The maximum compression of spring 1 is 15.0 cm. EVALUATE: When friction is not present mechanical energy is conserved and is continually transformed between kinetic energy of the block and potential energy in the springs. If friction is present, its work removes mechanical energy from the system. 7.71. IDENTIFY: Apply conservation of energy to relate x and h. Apply m=?F a! ! to relate a and x. SET UP: The first condition, that the maximum height above the release point is h, is expressed as 212 kx mgh= . The magnitude of the acceleration is largest when the spring is compressed to a distance x; at this point the net upward force is kx mg ma? = , so the second condition is expressed as ( / )( )x m k g a= + . EXECUTE: (a) Substituting the second expression into the first gives 2 2 21 ( )( ) , or . 2 2 m m g a k g a mgh k k gh +? ? + = =? ?? ? (b) Substituting this into the expression for x gives 2 g h x g a = + . EVALUATE: When 0a ? , our results become 2 mg k h = and 2x h= . The initial spring force is kx mg= and the net upward force approaches zero. But 212 kx mgh= and sufficient potential energy is stored in the spring to move the mass to height h. 7.72. IDENTIFY: At equilibrium the upward spring force equals the weight mg of the object. Apply conservation of energy to the motion of the fish. SET UP: The distance that the mass descends equals the distance the spring is stretched. 1 2 0K K= = , so 1 2(gravitational) (spring)U U= EXECUTE: Following the hint, the force constant k is found from mg kd= , or /k mg d= . When the fish falls from rest, its gravitational potential energy decreases by mgy ; this becomes the potential energy of the spring, which is 2 21 12 2 ( / )ky mg d y= . Equating these, 21 , or 2 .2 mg y mgy y d d = = 7-24 Chapter 7 EVALUATE: At its lowest point the fish is not in equilibrium. The upward spring force at this point is 2ky kd= , and this is equal to twice the weight. At this point the net force is mg , upward, and the fish has an upward acceleration equal to g . 7.73. IDENTIFY: Apply Eq.(7.15) to the motion of the block. SET UP: The motion from A to B is described in Figure 7.73. Figure 7.73 The normal force is cos ,n mg ?= so k k k cos .f n mg? ? ?= = 0;Ay = (60.0 m)sin30.0 3.00 mBy = ° = otherA A B BK U W K U+ + = + EXECUTE: Work is done by gravity, by the spring force, and by friction, so other fW W= and el gravU U U= + 0,AK = 2 21 12 2 (1.50 kg)(7.00 m/s) 36.75 JB BK mv= = = el, grav, el, ,A A A AU U U U= + = since grav, 0AU = 2 el, grav, 0 (1.50 kg)(9.80 m/s )(3.00 m) 44.1 JB B B BU U U mgy= + = + = = other k k k( cos ) cos (cos180 ) cosfW W f s mg s mg s? ? ? ? ?= = = ° = ? 2 other (0.50)(1.50 kg)(9.80 m/s )(cos30.0 )(6.00 m) 38.19 JW = ? ° = ? Thus el, 38.19 J 36.75 J 44.10 JAU ? = + el, 38.19 J 36.75 J 44.10 J 119 JAU = + + = EVALUATE: elU must always be positive. Part of the energy initially stored in the spring was taken away by friction work; the rest went partly into kinetic energy and partly into an increase in gravitational potential energy. 7.74. IDENTIFY: Apply Eq.(7.14) to the motion of the package. kother f W W= , the work done by the kinetic friction force. SET UP: k k k cosf n mg? ? ?= = , with 53.1? = ° . Let 4.00 mL = , the distance the package moves before reaching the spring and let d be the maximum compression of the spring. Let point 1 be the initial position of the package, point 2 be just as it contacts the spring, point 3 be at the maximum compression of the spring, and point 4 be the final position of the package after it rebounds. EXECUTE: (a) 1 0K = , 2 0U = , other k k cosW f L L? ?= ? = ? . 1 sinU mgL ?= . 212 2K mv= , where v is the speed before the block hits the spring. Eq.(7.14) applied to points 1 and 2, with 2 0y = , gives 1 other 2U W K+ = . Solving for v, 2 k2 (sin cos ) 2(9.80 m/s )(4.00 m)(sin53.1 (0.20)cos53.1 ) 7.30 m/s.v gL ? ? ?= ? = ° ? ° = (b) Apply Eq.(7.14) to points 1 and 3. Let 3 0y = . 1 3 0K K= = . 1 ( )sinU mg L d ?= + . 212 2U kd= . other k ( )W f L d= ? + . Eq.(7.14) gives 21k 2( )sin cos ( )mg L d mg L d kd? ? ?+ ? + = . This can be written as 2 k 0. 2 (sin cos ) k d d L mg ? ? ? ? ? =? The factor multiplying 2d is 14.504 m? , and use of the quadratic formula gives 1.06 md = . (c) The easy thing to do here is to recognize that the presence of the spring determines d, but at the end of the motion the spring has no potential energy, and the distance below the starting point is determined solely by how much energy has been lost to friction. If the block ends up a distance y below the starting point, then the block has moved a distance L d+ down the incline and L d y+ ? up the incline. The magnitude of the friction force is the same in both directions, k cosmg ?? , and so the work done by friction is k (2 2 ) cosL d y mg ??? + ? . This must be equal to the change in gravitational potential energy, which is sinmgy ?? . Equating these and solving for y gives k k k k 2 cos 2 ( ) ( ) . sin cos tan y L d L d ? ? ? ? ? ? ? ?= + = ++ + Using the value of d found in part (b) and the given values for k? and ? gives 1.32 my = . Potential Energy and Energy Conservation 7-25 EVALUATE: Our expression for y gives the reasonable results that 0y = when k 0? = ; in the absence of friction the package returns to its starting point. 7.75. (a) IDENTIFY and SET UP: Apply otherA A B BK U W K U+ + = + to the motion from A to B. EXECUTE: 0,AK = 212B BK mv= 0,AU = 21el, 2 ,B B BU U kx= = where 0.25 mBx = other F BW W Fx= = Thus 2 21 12 2 .B B BFx mv kx= + (The work done by F goes partly to the potential energy of the stretched spring and partly to the kinetic energy of the block.) (20.0 N)(0.25 m) 5.0 JBFx = = and 2 21 12 2 (40.0 N/m)(0.25 m) 1.25 JBkx = = Thus 2125.0 J 1.25 JBmv= + and 2(3.75 J) 3.87 m/s0.500 kgBv = = (b) IDENTIFY: Apply Eq.(7.15) to the motion of the block. Let point C be where the block is closest to the wall. When the block is at point C the spring is compressed an amount ,Cx so the block is 0.60 m Cx? from the wall, and the distance between B and C is .B Cx x+ SET UP: The motion from A to B to C is described in Figure 7.75. otherB B C CK U W K U+ + = + EXECUTE: other 0W = 21 2 5.0 J 1.25 J 3.75 JB BK mv= = ? = (from part (a)) 21 2 1.25 JB BU kx= = 0CK = (instantaneously at rest at point closest to wall) 21 2C CU k x= Figure 7.75 Thus 2123.75 J 1.25 J Ck x+ = 2(5.0 J) 0.50 m 40.0 N/mC x = = The distance of the block from the wall is 0.60 m 0.50 m 0.10 m.? = EVALUATE: The work (20.0 N)(0.25 m) 5.0 J= done by F puts 5.0 J of mechanical energy into the system. No mechanical energy is taken away by friction, so the total energy at points B and C is 5.0 J. 7.76. IDENTIFY: Apply Eq.(7.14) to the motion of the student. SET UP: Let 0 0.18 mx = , 1 0.71 mx = . The spring constants (assumed identical) are then known in terms of the unknown weight w, 04 kx w= . Let 0y = at the initial position of the student. EXECUTE: (a) The speed of the brother at a given height h above the point of maximum compression is then found from 2 21 1 1 (4 ) , 2 2 w k x v mgh g ? ?= +? ?? ? or 2 2 2 1 1 0 (4 ) 2 2 k g x v x gh g h w x ? ?= ? = ?? ?? ? . Therefore, 2 2(9.80 m/s )((0.71 m) (0.18 m) 2(0.90 m)) 3.13 m/sv = ? = , or 3.1 m/s to two figures. (b) Setting 0v = and solving for h, 2 2 1 1 0 2 1.40 m, 2 kx x h mg x = = = or 1.4 m to two figures. (c) No; the distance 0x will be different, and the ratio 22 2 1 1 1 0 1 1 ( 0.53 m) 0.53 m 1 x x x x x x ? ?+= = +? ?? ? will be different. Note that on a planet with lower g , 1x will be smaller and h will be larger. EVALUATE: We are able to solve the problem without knowing either the mass of the student or the force constant of the spring. 7-26 Chapter 7 7.77. IDENTIFY: 2 2/xa d x dt= , 2 2/ya d y dt= . x xF ma= , y yF ma= . x yU F dx F dy= +? ? . SET UP: 0 0 0(cos ) sind t tdt ? ? ?= ? . 0 0 0(sin ) cos d t t dt ? ? ?= . 0 0 0 1 cos sint dt t? ??=? , 0 00 1 sin cost dt t? ??= ?? . /xv dx dt= , /yv dy dt= . E K U= + . EXECUTE: (a) 2 2 2 20 0/ , .x x xa d x dt x F ma m x? ?= = ? = = ? 2 2 2 20 0/ , y y ya d y dt y F ma m y? ?= = ? = = ? (b) 2 2 2 20 0 1 ( ) 2x y U F dx F dy m xdx ydy m x y? ?? ? ? ?= ? + = + = +? ? ? ?? ? ? ? (c) 0 0 0 0 0 0/ sin ( / ).xv dx dt x t x y y? ? ?= = ? = ? 0 0 0 0 0 0/ cos ( / ).yv dy dt y t y x x? ? ?= = + = + (i) When 0x x= and 0, 0xy v= = and 0 0yv y ?= , 2 2 2 2 2 2 2 2 2 0 0 0 0 0 0 0 1 1 1 1 ( ) , and ( ) 2 2 2 2x y K m v v my U mx E K U m x y? ? ?= + = = = + = + (ii) When 0x = and 0 0 0, xy y v x ?= = ? and 0yv = , 2 2 2 2 2 2 2 0 0 0 0 0 0 0 1 1 1 , and ( ) 2 2 2 K mx U m y E K U m x y? ? ?= = = + = + EVALUATE: The total energy is the same at the two points in part (c); the total energy of the system is constant. 7.78. IDENTIFY: Calculate the increase in kinetic energy for the car. SET UP: The car gets 8(0.15)(1.3 10 J)× of energy from one gallon of gasoline. EXECUTE: (a) The mechanical energy increase of the car is 2 612 1 2 (1500 kg)(37 m/s) 1.027 10 J.K K? = = × Let ? be the number of gallons of gasoline consumed. 8 6(1.3 10 J)(0.15) 1.027 10 J? × = × and 0.053gallons? = . (b) (1.00 gallons) 19 accelerations? = EVALUATE: The time over which the increase in velocity occurs doesn't enter into the calculation. 7.79. IDENTIFY: U mgh= . Use 150 mh = for all the water that passes through the dam. SET UP: m V?= and V A h= ? is the volume of water in a height h? of water in the lake. EXECUTE: (a) Stored energy ( ) (1 m)mgh V gh A gh? ?= = = . 3 6 2 2 12stored energy (1000 kg/m )(3.0 10 m )(1 m)(9.8 m/s )(150 m) 4.4 10 J.= × = × (b) 90% of the stored energy is converted to electrical energy, so (0.90)( ) 1000 kWhmgh = . (0.90) 1000 kWhVgh? = . 3 33 2(1000 kWh)((3600 s) (1 h)) 2 7 10 m(0.90)(1000 kg/m )(150 m)(9.8 m/s )V .= = × . Change in level of the lake: waterA h V? = . 3 3 4 6 2 2.7 10 m 9.0 10 m 3.0 10 m V h A ?×? = = = ×× . EVALUATE: h? is much less than 150 m, so using 150 mh = for all the water that passed through the dam was a very good approximation. 7.80. IDENTIFY and SET UP: The potential energy of a horizontal layer of thickness dy, area A, and height y is ( ) .dU dm gy= Let ? be the density of water. EXECUTE: ,dm dV A dy? ?= = so .dU Agy dy?= The total potential energy U is 21 20 0 . h h U dU Ag y dy Agh? ?= = =? ? 6 23.0 10 mA = × and 150 m,h = so 14 73.3 10 J 9.2 10 kWhU = × = × EVALUATE: The volume is Ah and the mass of water is .V Ah? ?= The average depth is av /2,h h= so av.U mgh= 7.81. IDENTIFY: Apply x UF x ?= ? ? , y U F y ?= ? ? and z U F z ?= ? ? . SET UP: 2 2 2 1/ 2( )r x y z= + + . 2 2 3/ 2(1/ ) ( ) r x x x y ? = ?? + , 2 2 3/ 2 (1/ ) ( ) r y y x y ? = ?? + and 2 2 3/ 2 (1/ ) ( ) r z z x y ? = ?? + . Potential Energy and Energy Conservation 7-27 EXECUTE: (a) 1 2( ) Gm mU r r = ? . 1 21 2 2 2 2 3/ 2(1/ ) ( )x U r Gm m x F Gm m x x x y z ? ?? ?= ? = + = ?? ?? ? + +? ? . Similarly, 1 2 2 2 2 3/ 2( )y Gm m y F x y z = ? + + and 1 2 2 2 2 3/ 2( )z Gm m z F x y z = ? + + . (b) 2 2 2 3/2 3( )x y z r+ + = so 1 23x Gm m xF r= ? , 1 2 3y Gm m y F r = ? and 1 2 3z Gm m z F r = ? . 2 2 2 2 2 21 2 1 2 3 2x y z Gm m Gm m F F F F x y z r r = + + = + + = . (c) xF , yF and zF are negative. xF x?= , yF y?= and zF z?= , where ? is a constant, so F ! and the vector r ! from 1m to 2m are in the same direction. Therefore, F ! is directed toward 1m at the origin and F ! is attractive. EVALUATE: When 2m moves to larger r, the work done on it by the attractive gravity force is negative. Since W U= ?? , negative work done by gravity means the gravitational potential energy increases. 1 2( ) Gm m U r r = ? does increase (becomes less negative) as r increases. For an object near the surface of the earth, 1 2( ) Gm m U r r = ? will be shown in Chapter 12 to be equivalent to gravU mgy= . 7.82. IDENTIFY: Calculate the work W done by this force. If the force is conservative, the work is path independent. SET UP: 2 1 P P W d= ?? F l!! . EXECUTE: (a) 2 2 1 1 2P P yP P W F dy C y dy= =? ? . W doesn't depend on x, so it is the same for all paths between 1P and 2P . The force is conservative. (b) 2 2 1 1 2P P xP P W F dx C y dx= =? ? . W will be different for paths between points 1P and 2P for which y has different values. For example, if y has the constant value 0y along the path, then 0 2 1( )W Cy x x= ? . W depends on the value of 0y . The force is not conservative. EVALUATE: 2 ?CyF = j! has the potential energy function 3 ( ) 3 Cy U y = ? . We cannot find a potential energy function for 2 ?CyF = i ! . 7.83. 2 ?,xy?= ?F j! 32.50 N/m? = IDENTIFY: F! is not constant so use Eq.(6.14) to calculate W. F! must be evaluated along the path. (a) SET UP: The path is sketched in Figure 7.83a. ? ?d dx dy= +l i j! 2d xy dy?? = ?F l!! On the path, x y= so 3d y dy?? = ?F l!! Figure 7.83a EXECUTE: ( )2 2 11 2 3 4 4 4 2 11 ( ) ( / 4) ( / 4)( )| y y yy W d y dy y y y? ? ?= ? = ? = ? = ? ?? ?F l!! 1 0,y = 2 3.00 m,y = so 3 414 (2.50 N/m )(3.00 m) 50.6 JW = ? = ? (b) SET UP: The path is sketched in Figure 7.83b. Figure 7.83b For the displacement from point 1 to point 2, ?,d dx=l i! so 0d? =F l!! and 0.W = (The force is perpendicular to the displacement at each point along the path, so 0.)W = 7-28 Chapter 7 For the displacement from point 2 to point 3, ?,d dy=l j! so 2 .d xy dy?? = ?F l!! On this path, 3.00 m,x = so 3 2 2 2(2.50 N/m )(3.00 m) (7.50 N/m ) .d y dy y dy? = ? = ?F l!! EXECUTE: 3 2 3 2 2 2 3 31 3 232 (7.50 N/m ) (7.50 N/m ) ( ) y y W d y dy y y= ? = ? = ? ?? ?F l!! ( )2 313(7.50 N/m ) (3.00 m) 67.5 JW = ? = ? (c) EVALUATE: For these two paths between the same starting and ending points the work is different, so the force is nonconservative. 7.84. IDENTIFY: Use 2 1 P P W d= ?? F l!! to calculate W for each segment of the path. SET UP: xd F dx xy dx?? =F l = !! EXECUTE: (a) The path is sketched in Figure 7.84. (b) (1): 0x = along this leg, so 0=F! and 0W = . (2): Along this leg, 1.50 my = , so (3.00 N m)d xdx? =F l!! , and 2(1.50 N m)((1.50 m) 0) 3.38 JW = ? = (3) 0d? =F l!! , so 0W = (4) 0y = , so 0=F! and 0W = . The work done in moving around the closed path is 3.38 J. (c) The work done in moving around a closed path is not zero, and the force is not conservative. EVALUATE: There is no potential energy function for this force. Figure 7.84 7.85. IDENTIFY: Use Eq.(7.16) to relate xF and ( )U x . The equilibrium is stable where ( )U x is a local minimum and the equilibrium is unstable where ( )U x is a local maximum. SET UP: The maximum and minimum values of x are those for which ( )U x E= . K E U= ? , so the maximum speed is where U is a minimum. EXECUTE: (a) For the given proposed potential ( ), dUU x kx F dx ? = ? + , so this is a possible potential function. For this potential, 2(0) 2U F k= ? , not zero. Setting the zero of potential is equivalent to adding a constant to the potential; any additive constant will not change the derivative, and will correspond to the same force. (b) At equilibrium, the force is zero; solving 0kx F? + = for x gives 0 /x F k= . 20( ) /U x F k= ? , and this is a minimum of U , and hence a stable point. (c) The graph is given in Figure 7.85. (d) No; tot 0F = at only one point, and this is a stable point. (e) The extreme values of x correspond to zero velocity, hence zero kinetic energy, so ( )U x E± = , where x± are the extreme points of the motion. Rather than solve a quadratic, note that 2 212 ( / ) /k x F k F k? ? , so ( )U x E± = becomes 2 2 21 / 2 F F k x F k k k± ? ?? ? =? ?? ? . 2 , F F x k k± ? = ± so 3 .F Fx x k k+ ? = = ? (f) The maximum kinetic energy occurs when ( )U x is a minimum, the point 0 /x F k= found in part (b). At this point 2 2 2( / ) ( / ) 2 /K E U F k F k F k= ? = ? ? = , so 2v F mk= . Potential Energy and Energy Conservation 7-29 EVALUATE: As E increases, the magnitudes of x+ and x? increase. The particle cannot reach values of x for which ( )E U x< because K cannot be negative. Figure 7.85 7.86. IDENTIFY: Use Eq.(7.16) to relate xF and ( )U x . The equilibrium is stable where ( )U x is a local minimum and the equilibrium is unstable where ( )U x is a local maximum. SET UP: /dU dx is the slope of the graph of U versus x. K E U= ? , so K is a maximum when U is a minimum. The maximum x is where E U= . EXECUTE: (a) The slope of the U vs. x curve is negative at point A, so xF is positive (Eq. (7.16)). (b) The slope of the curve at point B is positive, so the force is negative. (c) The kinetic energy is a maximum when the potential energy is a minimum, and that figures to be at around 0.75 m. (d) The curve at point C looks pretty close to flat, so the force is zero. (e) The object had zero kinetic energy at point A, and in order to reach a point with more potential energy than ( )U A , the kinetic energy would need to be negative. Kinetic energy is never negative, so the object can never be at any point where the potential energy is larger than ( )U A . On the graph, that looks to be at about 2.2 m. (f) The point of minimum potential (found in part (c)) is a stable point, as is the relative minimum near 1.9 m. (g) The only potential maximum, and hence the only point of unstable equilibrium, is at point C. EVALUATE: If E is less than U at point C, the particle is trapped in one or the other of the potential "wells" and cannot move from one allowed region of x to the other. 7.87. IDENTIFY: K E U= ? determines ( )v x . SET UP: v is a maximum when U is a minimum and v is a minimum when U is a maximum. /xF dU dx= ? . The extreme values of x are where ( )E U x= . EXECUTE: (a) Eliminating ? in favor of ? and 0 0( / )x ? x?= , 22 0 0 0 2 2 2 2 0 0 0 ( ) . x x x U x x x x x x x x x x ? ? ? ? ? ? ?? ? ? ?= ? = ? = ?? ?? ? ? ?? ? ? ?? ?? ? 0 2 0 ( ) (1 1) 0U x x ?? ?= ? =? ?? ? . ( )U x is positive for 0x x< and negative for 0x x> (? and ? must be taken as positive). The graph of ( )U x is sketched in Figure 7.87a. (b) 2 0 0 2 0 2 2 ( ) x x v x U m mx x x ? ? ?? ? ? ? ? ?= ? = ?? ?? ? ? ? ? ?? ?? ? ? ?? ? ? ? . The proton moves in the positive x-direction, speeding up until it reaches a maximum speed (see part (c)), and then slows down, although it never stops. The minus sign in the square root in the expression for ( )v x indicates that the particle will be found only in the region where 0U < , that is, 0x x> . The graph of ( )v x is sketched in Figure 7.87b. (c) The maximum speed corresponds to the maximum kinetic energy, and hence the minimum potential energy. This minimum occurs when 0dU dx = , or 3 2 0 0 0 3 2 0, dU x x dx x x x ? ? ?? ? ? ?= ? + =? ?? ? ? ?? ? ? ?? ?? ? which has the solution 02x x= . 0 2 0 (2 ) 4 U x x ?= ? , so 2 02 v mx ?= . (d) The maximum speed occurs at a point where 0dU dx = , and from Eq. (7.15), the force at this point is zero. 7-30 Chapter 7 (e) 1 03x x= , and 0 2 0 2 (3 ) 9 U x x ?= ? . 2 2 0 0 0 0 1 22 2 00 0 2 2 2 2( ) ( ( ) ( )) 2 9 9 x x x x v x U x U x x xmxm m x x x x ? ? ?? ?? ? ? ?? ?? ? ? ? ? ? ?= ? = ? ? = ? ?? ?? ? ? ?? ? ? ? ? ?? ? ? ?? ? ? ? ? ?? ?? ?? ? ? ?? ?? ? . The particle is confined to the region where 1( ) ( )U x U x< . The maximum speed still occurs at 02x x= , but now the particle will oscillate between 1x and some minimum value (see part (f)). (f) Note that 1( ) ( )U x U x? can be written as 2 0 0 0 0 2 2 0 0 2 1 2 , 9 3 3 x x x x x x x x x x ? ?? ? ? ? ? ?? ? ? ? ? ? ? ? ? ?? + = ? ?? ?? ? ? ? ? ? ? ? ? ?? ? ? ?? ? ? ? ? ? ? ? ? ?? ? ? ?? ?? ? which is zero (and hence the kinetic energy is zero) at 0 13x x x= = and 3 02x x= . Thus, when the particle is released from 0x , it goes on to infinity, and doesn?t reach any maximum distance. When released from 1x , it oscillates between 3 02 x and 03x . EVALUATE: In each case the proton is released from rest and ( )iE U x= , where ix is the point where it is released. When 0ix x= the total energy is zero. When 1ix x= the total energy is negative. ( ) 0U x ? as x ?? , so for this case the proton can't reach x ?? and the maximum x it can have is limited. Figure 7.87 8-1 M OMENTUM , IMPULSE , AND COLLISIONS 8.1. IDENTIFY and SET UP: .p mv= 212 .K mv= EXECUTE: (a) 5(10,000 kg)(12.0 m/s) 1.20 10 kg m/sp = = × ? (b) (i) 51.20 10 kg m/s 60.0 m/s 2000 kg p v m × ?= = = . (ii) 2 21 1T T SUV SUV2 2m v m v= , so T SUV T SUV 10,000 kg (12.0 m/s) 26.8 m/s 2000 kg m v v m = = = EVALUATE: The SUV must have less speed to have the same kinetic energy as the truck than to have the same momentum as the truck. 8.2. IDENTIFY: Example 8.1 shows that the two iceboats have the same kinetic energy at the finish line. 212K mv= . p mv= . SET UP: Let A be the iceboat with mass m and let B be the iceboat with mass 2m , so 2B Am m= . EXECUTE: A BK K= gives 2 21 12 2A Bmv mv= . 2BA B B A m v v v m = = . A A Ap m v= . ( )(2 ) / 2 2 2B B B A A A A Ap m v m v m v p= = = = . EVALUATE: The more massive boat must have less speed but greater momentum than the other boat in order to have the same kinetic energy. 8.3. IDENTIFY and SET UP: p mv= . 212K mv= . EXECUTE: (a) p v m = and 2 2 1 2 2 p p K m m m ? ?= =? ?? ? . (b) c bK K= and the result from part (a) gives 2 2 c b c b2 2 p p m m = . bb c c c c 0.145 kg 1.90 0.040 kg m p p p p m = = = . The baseball has the greater magnitude of momentum. c b/ 0.526p p = . (c) 2 2p mK= so m wp p= gives m m w w2 2m K m K= . w mg= , so m m w ww K w K= . m w m m m w 700 N 1.56 450 N w K K K K w ? ? ? ?= = =? ? ? ?? ?? ? . The woman has greater kinetic energy. m w/ 0.641K K = . EVALUATE: For equal kinetic energy, the more massive object has the greater momentum. For equal momenta, the less massive object has the greater kinetic energy. 8.4. IDENTIFY: Each momentum component is the mass times the corresponding velocity component. SET UP: Let + x be along the horizontal motion of the shotput. Let + y be vertically upward. cosxv v ?= , sinyv v ?= . EXECUTE: The horizontal component of the initial momentum is cos (7.30 kg)(15.0 m/s)cos40.0 83.9 kg m/sx xp mv mv ?= = = = ?° . The vertical component of the initial momentum is sin (7.30 kg)(15.0 m/s)sin40.0 70.4 kg m/sy yp mv mv ?= = = = ?° EVALUATE: The initial momentum is directed at 40.0° above the horizontal. 8 8-2 Chapter 8 8.5. IDENTIFY: For each object, m ! ! p = v and 212K mv= . The total momentum is the vector sum of the momenta of each object. The total kinetic energy is the scalar sum of the kinetic energies of each object. SET UP: Let object A be the 110 kg lineman and object B the 125 kg lineman. Let + x be the object to the right, so 2.75 m/sAxv = + and 2.60 m/sBxv = ? . EXECUTE: (a) (110 kg)(2.75 m/s) (125 kg)( 2.60 m/s) 22.5 kg m/sx A Ax B BxP m v m v= + = + ? = ? ? . The net momentum has magnitude 22.5 kg m/s? and is directed to the left. (b) 2 2 2 21 1 1 12 2 2 2(110 kg)(2.75 m/s) (125 kg)(2.60 m/s) 838 JA A B BK m v m v= + = + = EVALUATE: The kinetic energy of an object is a scalar and is never negative. It depends only on the magnitude of the velocity of the object, not on its direction. The momentum of an object is a vector and has both magnitude and direction. When two objects are in motion, their total kinetic energy is greater than the kinetic energy of either one. But if they are moving in opposite directions, the net momentum of the system has a smaller magnitude than the magnitude of the momentum of either object. 8.6. IDENTIFY: For each object m ! ! p = v and the net momentum of the system is A B ! ! ! P = p + p . The momentum vectors are added by adding components. The magnitude and direction of the net momentum is calculated from its x and y components. SET UP: Let object A be the pickup and object B be the sedan. 14.0 m/sAxv = ? , 0Ayv = . 0Bxv = , 23.0 m/sByv = + . EXECUTE: (a) 4(2500 kg)( 14.0 m/s) 0 3.50 10 kg m/sx Ax Bx A Ax B BxP p p m v m v= + = + = ? + = ? × ? 4(1500 kg)( 23.0 m/s) 3.45 10 kg m/sy Ay By A Ay B ByP p p m v m v= + = + = + = + × ? (b) 2 2 44.91 10 kg m/sx yP P P= + = × ? . From Figure 8.6, 4 4 3.50 10 kg m/s tan 3.45 10 kg m/s x y P P ? × ?= = × ? and 45.4? = ° . The net momentum has magnitude 44.91 10 kg m/s× ? and is directed at 45.4° west of north. EVALUATE: The momenta of the two objects must be added as vectors. The momentum of one object is west and the other is north. The momenta of the two objects are nearly equal in magnitude, so the net momentum is directed approximately midway between west and north. Figure 8.6 8.7. IDENTIFY: The average force on an object and the object?s change in momentum are related by Eq. 8.9. The weight of the ball is w mg= . SET UP: Let + x be in the direction of the final velocity of the ball, so 1 0xv = and 2 25.0 m/sxv = . EXECUTE: av 2 1 2 1( ) ( )x x xF t t mv mv? = ? gives 2 1av 3 2 1 (0.0450 kg)(25.0 m/s) ( ) 562 N 2.00 10 s x x x mv mv F t t ? ?= = =? × . 2(0.0450 kg)(9.80 m/s ) 0.441 Nw = = . The force exerted by the club is much greater than the weight of the ball, so the effect of the weight of the ball during the time of contact is not significant. EVALUATE: Forces exerted during collisions typically are very large but act for a short time. 8.8. IDENTIFY: The change in momentum, the impulse and the average force are related by Eq. 8.9. SET UP: Let the direction in which the batted ball is traveling be the + x direction, so 1 45.0 m/sxv = ? and 2 55.0 m/sxv = . EXECUTE: (a) 2 1 2 1( ) (0.145 kg)(55.0 m/s [ 45.0 m/s]) 14.5 kg m/sx x x x xp p p m v v? = ? = ? = ? ? = ? . x xJ p= ? , so 14.5 kg m/sxJ = ? . Both the change in momentum and the impulse have magnitude 14.5 kg m/s? . (b) av 3 14.5 kg m/s ( ) 7250 N 2.00 10 s x x J F t ? ?= = =? × . EVALUATE: The force is in the direction of the momentum change. 8.9. IDENTIFY: Use Eq. 8.9. We know the intial momentum and the impluse so can solve for the final momentum and then the final velocity. Momentum, Impulse, and Collisions 8-3 SET UP: Take the x-axis to be toward the right, so 1 3.00 m /s.xv = + Use Eq. 8.5 to calculate the impulse, since the force is constant. EXECUTE: (a) 2 1x x xJ p p= ? 2 1( ) ( 25.0 N)(0.050 s) 1.25 kg m/sx xJ F t t= ? = + = + ? Thus 2 1 1.25 kg m/s (0.160 kg)( 3.00 m/s)x x xp J p= + = + ? + + = 1.73 kg m/s+ ? 2 2 1.73 kg m/s 10.8 kg m/s (to the right) 0.160 kg x x p v m ?= = = + ? (b) 2 1( ) ( 12.0 N)(0.050 s) 0.600 kg m/sx xJ F t t= ? = ? = ? ? (negative since force is to left) 2 1 0.600 kg m/s (0.160 kg)( 3.00 m/s) 0.120 kg m/sx x xp J p= + = ? ? + + = ? ? 2 2 0.120 kg m/s 0.75 m/s (to the left) 0.160 kg x x p v m ? ?= = = ? EVALUATE: In part (a) the impulse and initial momentum are in the same direction and xv increases. In part (b) the impulse and initial momentum are in opposite directions and the velocity decreases. 8.10. IDENTIFY: The impulse, change in momentum and change in velocity are related by Eq. 8.9. SET UP: 26,700 NyF = and 0xF = . The force is constant, so av( ) y yF F= . EXECUTE: (a) 5(26,700 N)(3.90 s) 1.04 10 N sy yJ F t= ? = = × ? . (b) 51.04 10 kg m/sy yp J? = = × ? . (c) y yp m v? = ? . 51.04 10 kg m/s 1.09 m/s 95,000 kg y y p v m ? × ?? = = = . (d) The initial velocity of the shuttle isn?t known. The change in kinetic energy is 2 212 1 2 12 ( )K K K m v v? = ? = ? . It depends on the initial and final speeds and isn?t determined solely by the change in speed. EVALUATE: The force in the + y direction produces an increase of the velocity in the + y direction. 8.11. IDENTIFY: The force is not constant so 2 1 t t dt?J = F! ! . The impulse is related to the change in velocity by Eq. 8.9. SET UP: Only the x component of the force is nonzero, so 2 1 t x xt J F dt= ? is the only nonzero component of !J . 2 1( )x x xJ m v v= ? . 1 2.00 st = , 2 3.50 st = . EXECUTE: (a) 22 2 781.25 N 500 N/s (1.25 s) xFA t = = = . (b) 2 1 2 3 3 2 3 3 31 1 2 13 3( ) (500 N/s )([3.50 s] [2.00 s] ) 5.81 10 N s t x t J At dt A t t= = ? = ? = × ?? . (c) 3 2 1 5.81 10 N s 2.70 m/s 2150 kg x x x x J v v v m × ?? = ? = = = . The x component of the velocity of the rocket increases by 2.70 m/s. EVALUATE: The change in velocity is in the same direction as the impulse, which in turn is in the direction of the net force. In this problem the net force equals the force applied by the engine, since that is the only force on the rocket. 8.12. IDENTIFY: Apply Eq. 8.9 to relate the change in momentum of the momentum to the components of the average force on it. SET UP: Let + x be to the right and + y be upward. EXECUTE: (a) 2 1 (0.145 kg)( [65.0 m/s]cos30 50.0 m/s) 15.4 kg m/sx x x xJ p mv mv= ? = ? = ? ? = ? ?° . 2 1 (0.145 kg)([65.0 m/s]sin30 0) 4.71 kg m/sy y y yJ p mv mv= ? = ? = ? = ?° The horizontal component is 15.4 kg m/s? , to the left and the vertical component is 4.71 kg m/s? , upward. (b) av- 3 15.4 kg m/s 8800 N 1.75 10 s x x J F t ? ? ?= = = ?? × . av- 3 4.71 kg m/s 2690 N 1.75 10 s y y J F t ? ?= = =? × . The horizontal component is 8800 N, to the left, and the vertical component is 2690 N, upward. EVALUATE: The ball gains momentum to the left and upward and the force components are in these directions. 8.13. IDENTIFY: The force is constant during the 1.0 ms interval that it acts, so t?! !J = F . 2 1 ( )m 2 1J p p v v ! ! ! ! !5 2 5 2 . SET UP: Let + x be to the right, so 1 5.00 m/sxv = + . Only the x component of ! J is nonzero, and 2 1( )x x xJ m v v= ? . 8-4 Chapter 8 EXECUTE: (a) The magnitude of the impulse is 3 3(2.50 10 N)(1.00 10 s) 2.50 N sJ F t ?= ? = × × = ? . The direction of the impulse is the direction of the force. (b) (i) 2 1 x x x J v v m = + . 2.50 N sxJ = + ? . 2 2.50 N s 5.00 m/s 6.25 m/s2.00 kgxv + ?= + = . The stone?s velocity has magnitude 6.25 m/s and is directed to the right. (ii) Now 2.50 N sxJ = ? ? and 2 2.50 N s 5.00 m/s 3.75 m/s2.00 kgxv ? ?= + = . The stone?s velocity has magnitude 3.75 m/s and is directed to the right. EVALUATE: When the force and initial velocity are in the same direction the speed increases and when they are in opposite directions the speed decreases. 8.14. IDENTIFY: Apply conservation of momentum to the system of the astronaut and tool. SET UP: Let A be the astronaut and B be the tool. Let + x be the direction in which she throws the tool, so 2 3.20 m/sB xv = + . Assume she is initially at rest, so 1 1 0A x B xv v= = . Solve for 2A xv . EXECUTE: 1 2x xP P= . 1 1 1 0x A A x B B xP m v m v= + = . 2 2 2 0x A A x B B xP m v m v= + = and 2 2 (2.25 kg)(3.20 m/s) 0.105 m/s 68.5 kg B A x A x A m v v m = ? = ? = ? . Her speed is 0.105 m/s and she moves opposite to the direction in which she throws the tool. EVALUATE: Her mass is much larger than that of the tool so to have the same magnitude of momentum as the tool her speed is much less. 8.15. IDENTIFY: Since drag effects are neglected there is no net external force on the system of squid plus expelled water and the total momentum of the system is conserved. Since the squid is initially at rest, with the water in its cavity, the initial momentum of the system is zero. For each object, 212K mv= . SET UP: Let A be the squid and B be the water it expels, so 6.50 kg 1.75 kg 4.75 kgAm = ? = . Let + x be the direction in which the water is expelled. 2 2.50 m/sA xv = ? . Solve for 2B xv . EXECUTE: (a) 1 0xP = . 2 1x xP P= , so 2 20 A A x B B xm v m v= + . 22 (4.75 kg)( 2.50 m/s) 6.79 m/s1.75 kg A A x B x B m v v m ?= ? = ? = + . (b) 2 2 2 21 1 1 12 2 2 2 22 2 2 2(4.75 kg)(2.50 m/s) (1.75 kg)(6.79 m/s) 55.2 JA B A A B BK K K m v m v= + = + = + = The initial kinetic energy is zero, so the kinetic energy produced is 2 55.2 JK = . EVALUATE: The two objects end up with momenta that are equal in magnitude and opposite in direction, so the total momentum of the system remains zero. The kinetic energy is created by the work done by the squid as it expels the water. 8.16. IDENTIFY: Apply conservation of momentum to the system of you and the ball. In part (a) both objects have the same final velocity. SET UP: Let + x be in the direction the ball is traveling initially. 0.400 kgAm = (ball). 70.0 kgBm = (you). EXECUTE: (a) 1 2x xP P= gives 2(0.400 kg)(10.0 m/s) (0.400 kg 70.0 kg)v= + and 2 0.0568 m/sv = . (b) 1 2x xP P= gives 2(0.400 kg)(10.0 m/s) (0.400 kg)( 8.00 m/s) (70.0 kg) Bv= ? + and 2 0.103 m/sBv = . EVALUATE: When the ball bounces off it has a greater change in momentum and you acquire a greater final speed. 8.17. IDENTIFY: Apply conservation of momentum to the system of the two pucks. SET UP: Let + x be to the right. EXECUTE: (a) 1 2x xP P= says 1(0.250) (0.250 kg)( 0.120 m/s) (0.350 kg)(0.650 m/s)Av = ? + and 1 0.790 m/sAv = . (b) 211 2 (0.250 kg)(0.790 m/s) 0.0780 JK = = . 2 21 1 2 2 2(0.250 kg)(0.120 m/s) (0.350 kg)(0.650 m/s) 0.0757 JK = + = and 2 1 0.0023 JK K K? = ? = ? . EVALUATE: The total momentum of the system is conserved but the total kinetic energy decreases. 8.18. IDENTIFY: Since road friction is neglected, there is no net external force on the system of the two cars and the total momentum of the system is conserved. For each object, 212K mv= . SET UP: Let A be the 1750 kg car and B be the 1450 kg car. Let + x be to the right, so 1 1.50 m/sA xv = + , 1 1.10 m/sB xv = ? , and 2 0.250 m/sA xv = + . Solve for 2B xv . EXECUTE: (a) 1 2x xP P= . 1 1 2 2A A x B B x A A x B B xm v m v m v m v+ = + . 1 1 22 A A x B B x A A xB x B m v m v m v v m + ?= . 2 (1750 kg)(1.50 m/s) (1450 kg)( 1.10 m/s) (1750 kg)(0.250 m/s) 0.409 m/s 1450 kgB x v + ? ?= = . After the collision the lighter car is moving to the right with a speed of 0.409 m/s. Momentum, Impulse, and Collisions 8-5 (b) 2 2 2 21 1 1 11 1 12 2 2 2(1750 kg)(1.50 m/s) (1450 kg)(1.10 m/s) 2846 JA A B BK m v m v= + = + = . 2 2 2 21 1 1 1 2 2 22 2 2 2(1750 kg)(0.250 m/s) (1450 kg)(0.409 m/s) 176 JA A B BK m v m v= + = + = . The change in kinetic energy is 2 1 176 J 2846 J 2670 JK K K? = ? = ? = ? . EVALUATE: The total momentum of the system is constant because there is no net external force during the collision. The kinetic energy of the system decreases because of negative work done by the forces the cars exert on each other during the collision. 8.19. IDENTIFY: Since the rifle is loosely held there is no net external force on the system consisting of the rifle, bullet and propellant gases and the momentum of this system is conserved. Before the rifle is fired everything in the system is at rest and the initial momentum of the system is zero. SET UP: Let + x be in the direction of the bullet?s motion. The bullet has speed 601 m/s 1.85 m/s 599 m/s? = relative to the earth. 2 r b gx x x xP p p p= + + , the momenta of the rifle, bullet and gases. r 1.85 m/sxv = ? and b 599 m/sxv = + . EXECUTE: 2 1 0x xP P= = . r b g 0x x xp p p+ + = . g r b (2.80 kg)( 1.85 m/s) (0.00720 kg)(599 m/s)x x xp p p= ? ? = ? ? ? and g 5.18 kg m/s 4.31 kg m/s 0.87 kg m/sxp = + ? ? ? = ? . The propellant gases have momentum 0.87 kg m/s? , in the same direction as the bullet is traveling. EVALUATE: The magnitude of the momentum of the recoiling rifle equals the magnitude of the momentum of the bullet plus that of the gases as both exit the muzzle. 8.20. IDENTIFY: In part (a) no horizontal force implies xP is constant. In part (b) use the energy expression, Eq. 7.14, to find the potential energy intially in the spring. SET UP: Initially both blocks are at rest. Figure 8.20 EXECUTE: (a) 1 1 2 2A A x B B x A A x B B xm v m v m v m v+ = + 2 20 A A x B B xm v m v= + 2 2 3.00 kg ( 1.20 m/s) 3.60 m/s 1.00 kg B A x B x A m v v m ? ? ? ?= ? = ? + = ?? ? ? ?? ?? ? Block A has a final speed of 3.60 m/s, and moves off in the opposite direction to B. (b) Use energy conservation: 1 1 other 2 2K U W K U+ + = + . Only the spring force does work so other el0 and .W U U= = 1 0K = (the blocks initially are at rest) 2 0U = (no potential energy is left in the spring) 2 2 2 21 1 1 1 2 2 22 2 2 2(1.00 kg)(3.60 m/s) (3.00 kg)(1.20 m/s) 8.64 JA A B BK m v m v= + = + = 1 1,elU U= the potential energy stored in the compressed spring. Thus 1,el 2 8.64 JU K= = EVALUATE: The blocks have equal and opposite momenta as they move apart, since the total momentum is zero. The kinetic energy of each block is positive and doesn?t depend on the direction of the block?s velocity, just on its magnitude. 8.21. IDENTIFY: Since friction at the pond surface is neglected, there is no net external horizontal force and the horizontal component of the momentum of the system of hunter plus bullet is conserved. Both objects are initially at rest, so the initial momentum of the system is zero. Gravity and the normal force exerted by the ice together produce a net vertical force while the rifle is firing, so the vertical component of momentum is not conserved. SET UP: Let object A be the hunter and object B be the bullet. Let + x be the direction of the horizontal component of velocity of the bullet. Solve for 2A xv . 8-6 Chapter 8 EXECUTE: (a) 2 965 m/sB xv = + . 1 2 0x xP P= = . 2 20 A A x B B xm v m v= + and 3 2 2 4.20 10 kg (965 m/s) 0.0559 m/s 72.5 kg B A x B x A m v v m ?? ?×= ? = ? = ?? ?? ? . (b) 2 2 cos (965 m/s)cos56.0 540 m/sB x Bv v ?= = =° . 3 2 4.20 10 kg (540 m/s) 0.0313 m/s 72.5 kgA x v ?? ?×= ? = ?? ?? ? . EVALUATE: The mass of the bullet is much less than the mass of the hunter, so the final mass of the hunter plus gun is still 72.5 kg, to three significant figures. Since the hunter has much larger mass, her final speed is much less than the speed of the bullet. 8.22. IDENTIFY: Assume the nucleus is initially at rest. 212K mv= . SET UP: Let + x be to the right. 2A x Av v= ? and 2B x Bv v= + . EXECUTE: (a) 2 1 0x xP P= = gives 2 2 0A A x B B xm v m v+ = . AB A B m v v m ? ?= ? ?? ? . (b) ( ) 2 21 2 221 2 / A AA A A B B B B AB A A B m vK m v m K m v mm m v m = = = . EVALUATE: The lighter fragment has the greater kinetic energy. 8.23. IDENTIFY: Apply conservation of momentum to the nucleus and its fragments. The initial momentum is zero. The 214 Po nucleus has mass 27 25214(1.67 10 kg) 3.57 10 kg? ?× = × , where 271.67 10 kg?× is the mass of a nucleon (proton or neutron). 212K mv= . SET UP: Let + x be the direction in which the alpha particle is emitted. The nucleus that is left after the decay has mass 25 25 27 25n 3.75 10 kg 3.57 10 kg 6.65 10 kg 3.50 10 kgm m? ? ? ? ?= × ? = × ? × = × . EXECUTE: 2 1 0x xP P= = gives n n 0m v m v? ? + = . n n m v v m ? ?= . 12 7 27 2 2(1.23 10 J) 1.92 10 m/s 6.65 10 kg K v m ? ? ? ? ? ×= = = ×× . 27 7 5 n 25 6.65 10 kg (1.92 10 m/s) 3.65 10 m/s 3.50 10 kg v ? ? ? ?×= × = ×? ?×? ? . EVALUATE: The recoil velocity of the more massive nucleus is much less than the speed of the emitted alpha particle. 8.24. IDENTIFY and SET UP: Let the + x-direction be horizontal, along the direction the rock is thrown. There is no net horizontal force, so xP is constant. Let object A be you and object B be the rock. EXECUTE: 0 cos35.0A A B Bm v m v= ? + ° cos35.0 2.11 m/sB BA A m v v m °= = EVALUATE: yP is not conserved because there is a net external force in the vertical direction; as you throw the rock the normal force exerted on you by the ice is larger than the total weight of the system. 8.25. IDENTIFY: Each horizontal component of momentum is conserved. 212K mv= . SET UP: Let + x be the direction of Rebecca?s initial velocity and let the + y axis make an angle of 36.9° with respect to the direction of her final velocity. D1 D1 0x yv v= = . R1 13.0 m/sxv = ; R1 0yv = . R 2 (8.00 m/s)cos53.1 4.80 m/sxv = =° ; R 2 (8.00 m/s)sin53.1 6.40 m/syv = =° . Solve for D2xv and D2 yv . EXECUTE: (a) 1 2x xP P= gives R R1 R R 2 D D2x x xm v m v m v= + . R R1 R 2 D2 D ( ) (45.0 kg)(13.0 m/s 4.80 m/s) 5.68 m/s 65.0 kg x x x m v v v m ? ?= = = . 1 2y yP P= gives R R 2 D D20 y ym v m v= + . RD2 R 2 D 45.0 kg (6.40 m/s) 4.43 m/s 65.0 kgy y m v v m ? ?= ? = ? = ?? ?? ? . The directions of R1 ! v , R2 ! v and D2 ! v are sketched in Figure 8.25. D2 D2 4.43 m/s tan 5.68 m/s y x v v ? = = and 38.0? = ° . 2 2 D D2 D2 7.20 m/sx yv v v= + = . Momentum, Impulse, and Collisions 8-7 (b) 2 2 31 11 R R12 2 (45.0 kg)(13.0 m/s) 3.80 10 JK m v= = = × . 2 2 2 2 31 1 1 1 2 R R2 D D22 2 2 2(45.0 kg)(8.00 m/s) (65.0 kg)(7.20 m/s) 3.12 10 JK m v m v= + = + = × . 2 1 680 JK K K? = ? = ? . EVALUATE: Each component of momentum is separately conserved. The kinetic energy of the system increases. vR1 vR2 vD2 y x u Figure 8.25 8.26. IDENTIFY: There is no net external force on the system of astronaut plus canister, so the momentum of the system is conserved. SET UP: Let object A be the astronaut and object B be the canister. Assume the astronaut is initially at rest. After the collision she must be moving in the same direction as the canister. Let + x be the direction in which the canister is traveling initially, so 1 0A xv = , 2 2.40 m/sA xv = + , 1 3.50 m/sB xv = + , and 2 1.20 m/sB xv = + . Solve for Bm . EXECUTE: 1 2x xP P= . 1 1 2 2A A x B B x A A x B B xm v m v m v m v+ = + . 2 1 1 2 ( ) (78.4 kg)(2.40 m/s 0) 81.8 kg 3.50 m/s 1.20 m/s A A x A x B B x B x m v v m v v ? ?= = =? ? . EVALUATE: She must exert a force on the canister in the x? direction to reduce its velocity component in the + x direction. By Newton?s third law, the canister exerts a force on her that is in the + x direction and she gains velocity in that direction. 8.27. IDENTIFY: The horizontal component of the momentum of the system of the rain and freight car is conserved. SET UP: Let + x be the direction the car is moving initially. Before it lands in the car the rain has no momentum along the x axis. EXECUTE: (a) 1 2x xP P= says 2(24,000 kg)(4.00 m/s) (27,000 kg) xv= and 2 3.56 m/sxv = . (b) After it lands in the car the water must gain horizontal momentum, so the car loses horizontal momentum. EVALUATE: The vertical component of the momentum is not conserved, because of the vertical external force exerted by the track. 8.28. IDENTIFY: The x and y components of the momentum of the system of the two asteroids are separately conserved. SET UP: The before and after diagrams are given in Figure 8.28 and the choice of coordinates is indicated. Each asteroid has mass m. EXECUTE: (a) 1 2x xP P= gives 1 2 2cos30.0 cos45.0A A Bmv mv mv= +° ° . 2 240.0 m/s 0.866 0.707A Bv v= + and 2 20.707 40.0 m/s 0.866B Av v= ? . 2 2y yP P= gives 2 20 sin30.0 sin 45.0A Bmv mv= ?° ° and 2 20.500 0.707A Bv v= . Combining these two equations gives 2 20.500 40.0 m/s 0.866A Av v= ? and 2 29.3 m/sAv = . Then 2 0.500 (29.3 m/s) 20.7 m/s 0.707B v ? ?= =? ?? ? . (b) 211 12 AK mv= . 2 21 12 2 22 2A BK mv mv= + . 2 2 2 2 2 2 2 2 2 1 1 (29.3 m/s) (20.7 m/s) 0.804 (40.0 m/s) A B A K v v K v + += = = . 2 1 2 1 1 1 1 0.196 K K K K K K K ? ?= = ? = ? . 19.6% of the original kinetic energy is dissipated during the collision. EVALUATE: We could use any directions we wish for the x and y coordinate directions, but the particular choice we have made is especially convenient. 8-8 Chapter 8 Figure 8.28 8.29. IDENTIFY: Since drag effects are neglected there is no net external force on the system of two fish and the momentum of the system is conserved. The mechanical energy equals the kinetic energy, which is 212K mv= for each object. SET UP: Let object A be the 15.0 kg fish and B be the 4.50 kg fish. Let + x be the direction the large fish is moving initially, so 1 1.10 m/sA xv = and 1 0B xv = . After the collision the two objects are combined and move with velocity 2 ! v . Solve for 2xv . EXECUTE: (a) 1 2x xP P= . 1 1 2( )A A x B B x A B xm v m v m m v+ = + . 1 1 2 (15.0 kg)(1.10 m/s) 0 0.846 m/s 15.0 kg 4.50 kg A A x B B x x A B m v m v v m m + += = =+ + . (b) 2 2 21 1 11 1 12 2 2 (15.0 kg)(1.10 m/s) 9.08 JA A B BK m v m v= + = = . 2 21 12 22 2( ) (19.5 kg)(0.846 m/s) 6.98 JA BK m m v= + = = . 2 1 2.10 JK K K? = ? = ? . 2.10 J of mechanical energy is dissipated. EVALUATE: The total kinetic energy always decreases in a collision where the two objects become combined. 8.30. IDENTIFY: There is no net external force on the system of the two otters and the momentum of the system is conserved. The mechanical energy equals the kinetic energy, which is 212K mv= for each object. SET UP: Let A be the 7.50 kg otter and B be the 5.75 kg otter. After the collision their combined velocity is 2 ! v . Let + x be to the right, so 1 5.00 m/sA xv = ? and 1 6.00 m/sB xv = + . Solve for 2xv . EXECUTE: (a) 1 2x xP P= . 1 1 2( )A A x B B x A B xm v m v m m v+ = + . 1 1 2 (7.50 kg)( 5.00 m/s) (5.75)( 6.00 m/s) 0.226 m/s 7.50 kg 5.75 kg A A x B B x x A B m v m v v m m + ? + += = = ?+ + . (b) 2 2 2 21 1 1 11 1 12 2 2 2(7.50 kg)(5.00 m/s) (5.75 kg)(6.00 m/s) 197.2 JA A B BK m v m v= + = + = . 2 21 1 2 22 2( ) (13.25 kg)(0.226 m/s) 0.338 JA BK m m v= + = = . 2 1 197 JK K K? = ? = ? . 197 J of mechanical energy is dissipated. EVALUATE: The total kinetic energy always decreases in a collision where the two objects become combined. 8.31. IDENTIFY: Treat the comet and probe as an isolated system for which momentum is conserved. SET UP: In part (a) let object A be the probe and object B be the comet. Let x? be the direction the probe is traveling just before the collision. After the collision the combined object moves with speed 2v . The change in velocity is 2 1x B xv v v? = ? . In part (a) the impact speed of 37,000 km/h is the speed of the probe relative to the comet just before impact: 1 1 37,000 km/hA x B xv v? = ? . In part (b) let object A be the comet and object B be the earth. Let x? be the direction the comet is traveling just before the collision. The impact speed is 40,000 km/h, so 1 1 40,000 km/hA x B xv v? = ? . EXECUTE: (a) 1 2x xP P= . 1 12 A A x B B xx A B m v m v v m m += + . ( )2 1 1 1 1 1A B A B Ax B x A x B x A x B x A B A B A B m m m m m v v v v v v v m m m m m m ? ? ? ? ? ?? ?? = ? = + = ?? ? ? ? ? ?+ + +? ? ? ? ? ? . 6 14 372 kg ( 37,000 km/h) 1.4 10 km/h 372 kg 0.10 10 kg v ? ? ?? = ? = ? ×? ?+ ×? ? . The speed of the comet decreased by 61.4 10 km/h?× . This change is not noticeable. Momentum, Impulse, and Collisions 8-9 (b) 14 8 14 24 0.10 10 kg ( 40,000 km/h) 6.7 10 km/h 0.10 10 kg 5.97 10 kg v ? ? ?×? = ? = ? ×? ?× + ×? ? . The speed of the earth would change by 86.7 10 km/h?× . This change is not noticeable. EVALUATE: 1 1A x B xv v? is the velocity of the projectile (probe or comet) relative to the target (comet or earth). The expression for v? can be derived directly by applying momentum conservation in coordinates in which the target is initially at rest. 8.32. IDENTIFY: The forces the two vehicles exert on each other during the collision are much larger than the horizontal forces exerted by the road, and it is a good approximation to assume momentum conservation. SET UP: Let + x be eastward. After the collision two vehicles move with a common velocity 2 ! v . EXECUTE: (a) 1 2x xP P= gives SC SC T T SC T 2( )x x xm v m v m m v+ = + . SC SC T T 2 SC T (1050 kg)( 15.0 m/s) (6320 kg)( 10.0 m/s) 6.44 m/s 1050 kg 6320 kg x x x m v m v v m m + ? + += = =+ + . The final velocity is 6.44 m/s, eastward. (b) 1 2 0x xP P= = gives SC SC T T 0x xm v m v+ = . SCT SC T 1050 kg ( 15.0 m/s) 2.50 m/s 6320 kgx x m v v m ? ? ? ?= ? = ? ? =? ? ? ?? ?? ? . The truck would need to have initial speed 2.50 m/s. (c) part (a): 2 2 2 51 1 12 2 2(7370 kg)(6.44 m/s) (1050 kg)(15.0 m/s) (6320 kg)(10.0 m/s) 2.81 10 JK? = ? ? = ? × part (b): 2 2 51 12 20 (1050 kg)(15.0 m/s) (6320 kg)(2.50 m/s) 1.38 10 JK? = ? ? = ? × . The change in kinetic energy has the greater magnitude in part (a). EVALUATE: In part (a) the eastward momentum of the truck has a greater magnitude than the westward momentum of the car and the wreckage moves eastward after the collision. In part (b) the two vehicles have equal magnitudes of momentum, the total momentum of the system is zero, and the wreckage is at rest after the collision. 8.33. IDENTIFY: The forces the two players exert on each other during the collision are much larger than the horizontal forces exerted by the slippery ground and it is a good approximation to assume momentum conservation. Each component of momentum is separately conserved. SET UP: Let + x be east and + y be north. After the collision the two players have velocity 2 ! v . Let the linebacker be object A and the halfback be object B, so 1 0A xv = , 1 8.8 m/sA yv = , 1 7.2 m/sB xv = and 1 0B yv = . Solve for 2xv and 2 yv . EXECUTE: 1 2x xP P= gives 1 1 2( )A A x B B x A B xm v m v m m v+ = + . 1 1 2 (85 kg)(7.2 m/s) 3.14 m/s 110 kg 85 kg A A x B B x x A B m v m v v m m += = =+ + . 1 2y yP P= gives 1 1 2( )A A y B B y A B ym v m v m m v+ = + . 1 1 2 (110 kg)(8.8 m/s) 4.96 m/s 110 kg 85 kg A A y B B y y A B m v m v v m m += = =+ + . 2 2 2 2 5.9 m/sx yv v v= + = . 2 2 4.96 m/s tan 3.14 m/s y x v v ? = = and 58? = ° . The players move with a speed of 5.9 m/s and in a direction 58° north of east. EVALUATE: Each component of momentum is separately conserved. 8.34. IDENTIFY: There is no net external force on the system of the two skaters and the momentum of the system is conserved. SET UP: Let object A be the skater with mass 70.0 kg and object B be the skater with mass 65.0 kg. Let + x be to the right, so 1 2.00 m/sA xv = + and 1 2.50 m/sB xv = ? . After the collision the two objects are combined and move with velocity 2 ! v . Solve for 2xv . EXECUTE: 1 2x xP P= . 1 1 2( )A A x B B x A B xm v m v m m v+ = + . 1 1 2 (70.0 kg)(2.00 m/s) (65.0)( 2.50 m/s) 0.167 m/s 70.0 kg 65.0 kg A A x B B x x A B m v m v v m m + + ?= = = ?+ + . The two skaters move to the left at 0.167 m/s. EVALUATE: There is a large decrease in kinetic energy. 8-10 Chapter 8 8.35. IDENTIFY: Neglect external forces during the collision. Then the momentum of the system of the two cars is conserved. SET UP: S 1200 kgm = , L 3000 kgm = . The small car has velocity Sv and the large car has velocity Lv . EXECUTE: (a) The total momentum of the system is conserved, so the momentum lost by one car equals the momentum gained by the other car. They have the same magnitude of change in momentum. Since m=! !p v and ?!p is the same, the car with the smaller mass has a greater change in velocity. S S L Lm v m v? = ? and LS L S 3000 kg 2.50 1200 kg m v v v v m ? ? ? ?? = ? = ? = ?? ? ? ?? ?? ? . (b) The acceleration of the small car is greater, since it has a greater change in velocity during the collision. The large acceleration means a large force on the occupants of the small car and they would sustain greater injuries. EVALUATE: Each car exerts the same magnitude of force on the other car but the force on the compact has a greater effect on its velocity since its mass is less. 8.36. IDENTIFY: The collision forces are large so gravity can be neglected during the collision. Therefore, the horizontal and vertical components of the momentum of the system of the two birds are conserved. SET UP: The system before and after the collision is sketched in Figure 8.36. Use the coordinates shown. Figure 8.36 EXECUTE: There is no external force on the system so 1 2x xP P= and 1 2y yP P= . 1 2x xP P= gives raven-2(1.5 kg)(9.0 m/s) (1.5 kg) cosv ?= and raven-2 cos 9.0 m/sv ? = . 1 2y yP P= gives raven-2(0.600 kg)(20.0 m/s) (0.600 kg)( 5.0 m/s) (1.5 kg) sinv ?= ? + and raven-2 sin 10.0 m/sv ? = . Combining these two equations gives 10.0 m/s tan 9.0 m/s ? = and 48? = ° . EVALUATE: Due to its large initial speed the lighter falcon was able to produce a large change in the raven?s direction of motion. 8.37. IDENTIFY: Since friction forces from the road are ignored, the x and y components of momentum are conserved. SET UP: Let object A be the subcompact and object B be the truck. After the collision the two objects move together with velocity 2 ! v . Use the x and y coordinates given in the problem. 1 1 0A y B yv v= = . 2 (16.0 m/s)sin 24.0 6.5 m/sxv = =° ; 2 (16.0 m/s)cos24.0 14.6 m/syv = =° . EXECUTE: 1 2x xP P= gives 1 2( )A A x A B xm v m m v= + . 1 2 950 kg 1900 kg (6.5 m/s) 19.5 m/s 950 kg A B A x x A m m v v m ? ? ? ?+ += = =? ? ? ?? ?? ? . 1 2y yP P= gives 1 2( )A B y A B ym v m m v= + . 1 2 950 kg 1900 kg (14.6 m/s) 21.9 m/s 1900 kg A B B y y A m m v v m ? ? ? ?+ += = =? ? ? ?? ?? ? . Before the collision the subcompact car has speed 19.5 m/s and the truck has speed 21.9 m/s. EVALUATE: Each component of momentum is independently conserved. 8.38. IDENTIFY: Apply conservation of momentum to the collision. Apply conservation of energy to the motion of the block after the collision. Momentum, Impulse, and Collisions 8-11 SET UP: Conservation of momentum applied to the collision between the bullet and the block: Let object A be the bullet and object B be the block. Let Av be the speed of the bullet before the collision and let V be the speed of the block with the bullet inside just after the collision. Figure 8.38a xP is constant gives ( )A A A Bm v m m V= + . Conservation of energy applied to the motion of the block after the collision: V y A1B x #1 #2 v 5 0 0.230 m Figure 8.38b 1 1 other 2 2K U W K U+ + = + EXECUTE: Work is done by friction so other k k k( cos )fW W f s f s mgs? ?= = = ? = ? 1 2 0U U= = (no work done by gravity) 21 1 2 ;K mV= 2 0K = (block has come to rest) Thus 21 k2 0mV mgs?? = 2 k2 2(0.20)(9.80 m/s )(0.230 m) 0.9495 m/sV gs?= = = Use this in the conservation of momentum equation 3 3 5.00 10 kg 1.20 kg (0.9495 m/s) 229 m/s 5.00 10 kg A B A A m m v V m ? ? ? ? ? ?+ × += = =? ? ? ?×? ?? ? EVALUATE: When we apply conservation of momentum to the collision we are ignoring the impulse of the friction force exerted by the surface during the collision. This is reasonable since this force is much smaller than the forces the bullet and block exert on each other during the collision. This force does work as the block moves after the collision, and takes away all the kinetic energy. 8.39. IDENTIFY: Apply conservation of momentum to the collision and conservation of energy to the motion after the collision. After the collision the kinetic energy of the combined object is converted to gravitational potential energy. SET UP: Immediately after the collision the combined object has speed V. Let h be the vertical height through which the pendulum rises. EXECUTE: (a) Conservation of momentum applied to the collision gives 3 3(12.0 10 kg)(380 m/s) (6.00 kg 12.0 10 kg)V? ?× = + × and 0.758 m/sV = . Conservation of energy applied to the motion after the collision gives 21 tot tot2 m V m gh= and 2 2 2 (0.758 m/s) 0.0293 m = 2.93 cm 2 2(9.80 m/s ) V h g = = = . (b) 2 3 21 1b b2 2 (12.0 10 kg)(380 m/s) 866 JK m v ?= = × = . (c) 2 3 21 1tot2 2 (6.00 kg 12.0 10 kg)(0.758 m/s) 1.73 JK m V ?= = + × = . EVALUATE: Most of the initial kinetic energy of the bullet is dissipated in the collision. 8.40. IDENTIFY: Each component of horizontal momentum is conserved. SET UP: Let + x be east and + y be north. S1 A1 0y xv v= = . S2 (6.00 m/s)cos37.0 4.79 m/sxv = =° , S2 (6.00 m/s)sin37.0 3.61 m/syv = =° , 2 (9.00 m/s)cos23.0 8.28 m/sA xv = =° and 2 (9.00 m/s)sin 23.0 3.52 m/sA yv = ? = ?° . EXECUTE: 1 2x xP P= gives S S1 S S2 A A2x x xm v m v m v= + . S S2 A A2 S1 S (80.0 kg)(4.79 m/s) (50.0 kg)(8.28 m/s) 9.97 m/s 80.0 kg x x x m v m v v m + += = = . Sam?s speed before the collision was 9.97 m/s. 8-12 Chapter 8 1 2y yP P= gives A A1 S S2y A A2y ym v m v m v= + . S S2y A A2 A1 S (80.0 kg)(3.61 m/s) (50.0 kg)( 3.52 m/s) 2.26 m/s 50.0 kg y y m v m v v m + + ?= = = . Abigail?s speed before the collision was 2.26 m/s. (b) 2 2 2 21 1 1 12 2 2 2(80.0 kg)(6.00 m/s) (50.0 kg)(9.00 m/s) (80.0 kg)(9.97 m/s) (50.0 kg)(2.26 m/s) .K? = + ? ? 639 JK? =? . EVALUATE: The total momentum is conserved because there is no net external horizontal force. The kinetic energy decreases because the forces between the objects do negative work during the collision. 8.41. IDENTIFY: When the spring is compressed the maximum amount the two blocks aren?t moving relative to each other and have the same velocity ! V relative to the surface. Apply conservation of momentum to find V and conservation of energy to find the energy stored in the spring. Since the collision is elastic, Eqs. 8.24 and 8.25 give the final velocity of each block after the collision. SET UP: Let + x be the direction of the initial motion of A. EXECUTE: (a) Momentum conservation gives (2.00 kg)(2.00 m/s) (12.0 kg)V= and 0.333 m/sV = . Both blocks are moving at 0.333 m/s, in the direction of the initial motion of block A. Conservation of energy says the initial kinetic energy of A equals the total kinetic energy at maximum compression plus the potential energy bU stored in the bumpers: 2 21 1b2 2(2.00 kg)(2.00 m/s) (12.0 kg)(0.333 m/s)U= + and b 3.33 JU = . (b) 2 1 2.00 kg 10.0 kg (2.00 m/s) 1.33 m/s 12.0 kg A B A x A x A B m m v v m m ? ? ? ?? ?= = = ?? ? ? ?+ ? ?? ? . Block A is moving in the x? direction at 1.33 m/s. 2 1 2 2(2.00 kg) (2.00 m/s) 0.667 m/s 12.0 kg A B x A x A B m v v m m ? ?= = = +? ?+? ? . Block B is moving in the + x direction at 0.667 m/s. EVALUATE: When the spring is compressed the maximum amount the system must still be moving in order to conserve momentum. 8.42. IDENTIFY: No net external horizontal force so xP is conserved. Elastic collision so 1 2K K= and can use Eq. 8.27. SET UP: Figure 8.42 EXECUTE: From conservation of x- component of momentum: 1 1 2 2A A x B B x A A x B B xm v m v m v m v+ = + 1 1 2 2A A B B A A x B B xm v m v m v m v? = + 2 2(0.150 kg)(0.80 m/s) (0.300 kg)(2.20 m/s) (0.150 kg) (0.300 kg)A x B xv v? = + A2 23.60 m/s 2x B xv v? = + From the relative velocity equation for an elastic collision Eq. 8.27: 2 2 1 1( ) ( 2.20 m/s 0.80 m/s) 3.00 m/sB x A x B x A xv v v v? = ? ? = ? ? ? = + A2 23.00 m/s x B xv v= ? + Adding the two equations gives 20.60 m/s 3 B xv? = and 2 0.20 m/s.B xv = ? Then 2 2 3.00 m/s 3.20 m/s.A x B xv v= ? = ? The 0.150 kg glider (A) is moving to the left at 3.20 m/s and the 0.300 kg glider (B) is moving to the left at 0.20 m/s. EVALUATE: We can use our 2A xv and 2B xv to show that xP is constant and 1 2K K= 8.43. IDENTIFY: Since the collision is elastic, both momentum conservation and Eq. 8.27 apply. SET UP: Let object A be the 30.0 kg marble and let object B be the 10.0 g marble. Let + x be to the right. EXECUTE: (a) Conservation of momentum gives 2 2(0.0300 kg)(0.200 m/s) (0.0100 kg)( 0.400 m/s) (0.0300 kg) (0.0100 kg)A x B xv v+ ? = + . 2 23 0.200 m/sA x B xv v+ = . Eq. 8.27 says 2 2 ( 0.400 m/s 0.200 m/s) 0.600 m/sB x A xv v? = ? ? ? = + . Solving this pair of equations gives 2 0.100 m/sA xv = ? and 2 0.500 m/sB xv = + . The 30.0 g marble is moving to the left at 0.100 m/s and the 10.0 g marble is moving to the right at 0.500 m/s. Momentum, Impulse, and Collisions 8-13 (b) For marble A, 2 1 (0.0300 kg)( 0.100 m/s 0.200 m/s) 0.00900 kg m/sAx A A x A A xP m v m v? = ? = ? ? = ? ? . For marble B, 2 1 (0.0100 kg)(0.500 m/s [ 0.400 m/s]) 0.00900 kg m/sBx B B x B B xP m v m v? = ? = ? ? = + ? . The changes in momentum have the same magnitude and opposite sign. (c) For marble A, 2 2 2 2 41 1 12 12 2 2 (0.0300 kg)([0.100 m/s] [0.200 m/s] ) 4.5 10 JA A A A AK m v m v ?? = ? = ? = ? × . For marble B, 2 2 2 2 41 1 12 12 2 2 (0.0100 kg)([0.500 m/s] [0.400 m/s] ) 4.5 10 JB B B B BK m v m v ?? = ? = ? = + × . The changes in kinetic energy have the same magnitude and opposite sign. EVALUATE: The results of parts (b) and (c) show that momentum and kinetic energy are conserved in the collision. 8.44. IDENTIFY and SET UP: Without rounding, the calculation in Example 8.12 gives 2 20 m/sBv = . EXECUTE: The two equations in Example 8.12 for ? and ? are (0.500 kg)(4.00 m/s) (0.500 kg)(2.00 m/s)(cos ) (0.300 kg)( 20 m/s)(cos )? ?= + Eq. 1 and 0 (0.500 kg)(2.00 m/s)(sin ) (0.300 kg)( 20 m/s)sin? ?= ? Eq. 2. Dividing each equation by (0.500 kg)(1.00 m/s) gives 4.00 2.00cos 0.6 20 cos? ?= + Eq. 3 and 0 2.00sin 0.6 20 sin? ?= ? Eq. 4. Eq. 3 gives 4.00 2.00cos cos 0.6 20 ?? ?= and 2 2cos 2.222 2.222cos 0.5556cos? ? ?= ? + . Eq. 4 gives sin 0.7454sin? ?= and 2 2 2sin 0.5556sin 0.5556 0.5556cos? ? ?= = ? . Adding the two equations and using 2 2sin cos 1? ?+ = gives 1 2.778 2.222cos?= ? and cos 0.8002? = . 36.9? = ° . Then sin 0.7454sin? ?= gives 26.6? = ° . EVALUATE: For these values of ? and ? , the x component of momentum, the y component of momentum and the kinetic energy are all conserved in the collision. 8.45. IDENTIFY: Eqs. 8.24 and 8.25 apply, with object A being the neutron. SET UP: Let + x be the direction of the initial momentum of the neutron. The mass of a neutron is n 1.0 um = . EXECUTE: (a) 2 1 1 1 1.0 u 2.0 u /3.0 1.0 u 2.0 u A B A x A x A x A x A B m m v v v v m m ? ?? ?= = = ?? ?+ +? ? . The speed of the neutron after the collision is one-third its initial speed. (b) 2 21 12 n n n 1 12 2 1 ( /3.0) 9.0A K m v m v K= = = . (c) After n collisions, 2 1 1 3.0 n A Av v ? ?= ? ?? ? . 1 1 3.0 59,000 n? ? =? ?? ? , so 3.0 59,000 n = . log3.0 log59,000n = and 10n = . EVALUATE: Since the collision is elastic, in each collision the kinetic energy lost by the neutron equals the kinetic energy gained by the deuteron. 8.46. IDENTIFY: Elastic collision. Solve for mass and speed of target nucleus. SET UP: (a) Let A be the proton and B be the target nucleus. The collision is elastic, all velocities lie along a line, and B is at rest before the collision. Hence the results of Eqs. 8.24 and 8.25 apply. EXECUTE: Eq. 8.24: ( ) ( ),B x Ax A x Axm v v m v v+ = ? where xv is the velocity component of A before the collision and Axv is the velocity component of A after the collision. Here, 71.50 10 m/sxv = × (take direction of incident beam to be positive) and 71.20 10 m/sAxv = ? × (negative since traveling in direction opposite to incident beam). 7 7 7 7 1.50 10 m/s 1.20 10 m/s 2.70 9.00 . 1.50 10 m/s 1.20 10 m/s 0.30 x Ax B A x Ax v v m m m m m v v ? ? ? ?? × + × ? ?= = = =? ? ? ? ? ?+ × ? × ? ?? ?? ? (b) Eq. 8.25: 7 6 2 2 (1.50 10 m/s) 3.00 10 m/s. 9.00 A Bx A B m m v v m m m m ? ? ? ?= = × = ×? ? ? ?+ +? ?? ? EVALUATE: Can use our calculated Bxv and Bm to show that xP is constant and that 1 2.K K= 8-14 Chapter 8 8.47. IDENTIFY: Apply Eq. 8.28. SET UP: 0.300 kgAm = , 0.400 kgBm = , 0.200 kgCm = . EXECUTE: cm A A B B C C A B C m x m x m x x m m m + += + + . cm (0.300 kg)(0.200 m) (0.400 kg)(0.100 m) (0.200 kg)( 0.300 m) 0.0444 m 0.300 kg 0.400 kg 0.200 kg x + + ?= =+ + . cm A A B B C C A B C m y m y m y y m m m + += + + . cm (0.300 kg)(0.300 m) (0.400 kg)( 0.400 m) (0.200 kg)(0.600 m) 0.0556 m 0.300 kg 0.400 kg 0.200 kg y + ? += =+ + . EVALUATE: There is mass at both positive and negative x and at positive and negative y and therefore the center of mass is close to the origin. 8.48. IDENTIFY: Calculate cm.x SET UP: Apply Eq. 8.28 with the sun as mass 1 and Jupiter as mass 2. Take the origin at the sun and let Jupiter lie on the positive x-axis. Figure 8.48 1 1 2 2 cm 1 2 m x m x x m m += + EXECUTE: 1 0x = and 112 7.78 10 mx = × ( )( )27 11 8 cm 30 27 1.90 10 kg 7.78 10 m 7.42 10 m 1.99 10 kg 1.90 10 kg x × ×= = ×× + × The center of mass is 87.42 10 m× from the center of the sun and is on the line connecting the centers of the sun and Jupiter. The sun?s radius is 86.96 10 m× so the center of mass lies just outside the sun. EVALUATE: The mass of the sun is much greater than the mass of Jupiter so the center of mass is much closer to the sun. For each object we have considered all the mass as being at the center of mass (geometrical center) of the object. 8.49. IDENTIFY: The location of the center of mass is given by Eq. 8.48. The mass can be expressed in terms of the diameter. Each object can be replaced by a point mass at its center. SET UP: Use coordinates with the origin at the center of Pluto and the + x direction toward Charon, so P 0x = C 19,700 kmx = . 3 34 13 6m V r d? ? ? ??= = = . EXECUTE: 3 31 CP P C C C C6 cm C C C3 3 3 31 1 P C P C P C P C6 6 dm x m x m d x x x x m m m m d d d d ?? ?? ?? ? ?? ? ? ?+= = = =? ?? ? ? ?+ + + +? ? ? ?? ? . 3 3 cm 3 3 [1250 km] (19,700 km) 2.52 10 km [2370 km] [1250 km] x ? ?= = ×? ?+? ? . The center of mass of the system is 32.52 10 km× from the center of Pluto. EVALUATE: The center of mass is closer to Pluto because Pluto has more mass than Charon. 8.50. IDENTIFY: Apply Eqs. 8.28, 8.30 and 8.32. There is only one component of position and velocity. SET UP: 1200 kgAm = , 1800 kgBm = . 3000 kgA BM m m= + = . Let + x be to the right and let the origin be at the center of mass of the station wagon. EXECUTE: (a) cm 0 (1800 kg)(40.0 m) 24.0 m. 1200 kg 1800 kg A A B B A B m x m x x m m + += = =+ + The center of mass is between the two cars, 24.0 m to the right of the station wagon and 16.0 m behind the lead car. Momentum, Impulse, and Collisions 8-15 (b) 41 1 (1200 kg)(12.0 m/s) (1800 kg)(20.0 m/s) 5.04 10 kg m/s.x A A B BP m v m v= + = + = × ? (c) , ,cm, (1200 kg)(12.0 m/s) (1800 kg)(20.0 m/s) 16.8 m/s. 1200 kg 1800 kg A A x B B x x A B m v m v v m m + += = =+ + (d) 4cm- (3000 kg)(16.8 m/s) 5.04 10 kg m/sx xP Mv= = = × ? , the same as in part (b). EVALUATE: The total momentum can be calculated either as the vector sum of the momenta of the individual objects in the system, or as the total mass of the system times the velocity of the center of mass. 8.51. IDENTIFY: Use Eq. 8.28 to find the x and y coordinates of the center of mass of the machine part for each configuration of the part. In calculating the center of mass of the machine part, each uniform bar can be represented by a point mass at its geometrical center. SET UP: Use coordinates with the axis at the hinge and the + x and + y axes along the horizontal and vertical bars in the figure in the problem. Let i i( , )x y and f f( , )x y be the coordinates of the bar before and after the vertical bar is pivoted. Let object 1 be the horizontal bar, object 2 be the vertical bar and 3 be the ball. EXECUTE: 1 1 2 2 3 3i 1 2 3 (4.00 kg)(0.750 m) 0 0 0.333 m 4.00 kg 3.00 kg 2.00 kg m x m x m x x m m m + + + += = =+ + + + . 1 1 2 2 3 3 i 1 2 3 0 (3.00 kg)(0.900 m) (2.00 kg)(1.80 m) 0.700 m 9.00 kg m y m y m y y m m m + + + += = =+ + . f (4.00 kg)(0.750 m) (3.00 kg)( 0.900 m) (2.00 kg)( 1.80 m) 0.366 m 9.00 kg x + ? + ?= = ? . f 0y = . f i 0.700 mx x? = ? and f i 0.700 my y? = ? . The center of mass moves 0.700 m to the right and 0.700 m upward. EVALUATE: The vertical bar moves upward and to the right so it is sensible for the center of mass of the machine part to move in these directions. 8.52. (a) IDENTIFY: Use Eq. 8.28. SET UP: The target variable is 1.m EXECUTE: cm 2.0 m,x = cm 0y = ( ) ( )( ) ( ) 11 1 2 2 cm 1 2 1 1 0 0.10 kg 8.0 m 0.80 kg m 0.10 kg 0.10 kg mm x m x x m m m m ++ ?= = =+ + + . cm 2.0 mx = gives 1 0.80 kg m 2.0 m 0.10 kgm ?= + . 1 0.80 kg m 0.10 kg 0.40 kg. 2.0 m m ?+ = = 1 0.30 kg.m = EVALUATE: The cm is closer to 1m so its mass is larger then 2.m (b) IDENTIFY: Use Eq. 8.32 to calculate .P ! SET UP: ( )cm ?5.0 m/s . v j! 5 ( )( ) ( )cm ? ?0.10 kg 0.30 kg 5.0 m/s 2.0 kg m/s .M + ?P v i i! !5 5 5 (c) IDENTIFY: Use Eq. 8.31. SET UP: 1 1 2 2cm 1 2 . m m m m + + v v v ! !! 5 The target variable is 1.v! Particle 2 at rest says 2 0.v = EXECUTE: ( ) ( )1 21 cm 1 0.30 kg 0.10 kg ? ?5.00 m/s 6.7 m/s . 0.30 kg m m m ? ? ? ?+ + ? ? ? ?? ?? ? v v i i ! !5 5 5 EVALUATE: Using the result of part (c) we can calculate 1p ! and 2p ! and show that P ! as calculated in part (b) does equal 1 2. p p! !1 8.53. IDENTIFY: There is no net external force on the system of James, Ramon and the rope and the momentum of the system is conserved and the velocity of its center of mass is constant. Initially there is no motion, and the velocity of the center of mass remains zero after Ramon has started to move. 8-16 Chapter 8 SET UP: Let + x be in the direction of Ramon?s motion. Ramon has mass R 60.0 kgm = and James has mass J 90.0 kgm = . EXECUTE: R R J Jcm- R J 0x xx m v m v v m m += =+ . R J R J 60.0 kg (0.700 m/s) 0.47 m/s 90.0 kgx x m v v m ? ? ? ?= ? = ? = ?? ? ? ?? ?? ? . James? speed is 0.47 m/s. EVALUATE: As they move, the two men have momenta that are equal in magnitude and opposite in direction, and the total momentum of the system is zero. Also, Example 8.14 shows that Ramon moves farther than James in the same time interval. This is consistent with Ramon having a greater speed. 8.54. (a) IDENTIFY and SET UP: Apply Eq. 8.28 and solve for 1m and 2.m EXECUTE: 1 1 2 2cm 1 2 m y m y y m m += + 1 1 2 2 1 1 2 cm (0) (0.50 kg)(6.0 m) 1.25 kg 2.4 m m y m y m m m y + ++ = = = and 1 0.75 kg.m = EVALUATE: cmy is closer to 1m since 1 2.m m> (b) IDENTIFY and SET UP: Apply /d dt a v! !5 for the cm motion. EXECUTE: ( )3cmcm ?1.5 m/s .d tdt va i !! 5 5 (c) IDENTIFY and SET UP: Apply Eq. 8.34. EXECUTE: ( )( )3ext cm ?1.25 kg 1.5 m/s .M t ?F a i! !5 5 At 3.0 s,t = ( )( )( ) ( )3ext ? ?1.25 kg 1.5 m/s 3.0 s 5.6 N . ?F i i! 5 5 EVALUATE: cm-xv is positive and increasing so cm xa ? is positive and extF ! is in the -direction.x+ There is no motion and no force component in the y-direction. 8.55. IDENTIFY: Apply d dt? PF = !! to the airplane. SET UP: ( ) 1n nd t nt dt ?= . 21 N 1 kg m/s= ? . EXECUTE: 3 2[ (1.50 kg m/s ) ] (0.25 kg m/s ) d t dt ? ? ?P i j ! ! ! 5 1 . (1.50 N/s)xF t= ? , 0.25 NyF = , 0zF = . EVALUATE: There is no momentum or change in momentum in the z direction and there is no force component in this direction. 8.56. IDENTIFY: Use Eq. 8.38, applied to a finite time interval. SET UP: ex 1600 m/sv = EXECUTE: (a) ex 0.0500 kg (1600 m/s) 80.0 N 1.00 s m F v t ? ?= ? = ? = +? . (b) The absence of atmosphere would not prevent the rocket from operating. The rocket could be steered by ejecting the gas in a direction with a component perpendicular to the rocket?s velocity and braked by ejecting it in a direction parallel (as opposed to antiparallel) to the rocket?s velocity. EVALUATE: The thrust depends on the speed of the ejected gas relative to the rocket and on the mass of gas ejected per second. 8.57. IDENTIFY: ex v dm a m dt = ? . Assume that /dm dt is constant over the 5.0 s interval, since m doesn?t change much during that interval. The thrust is ex dm F v dt = ? . SET UP: Take m to have the constant value 110 kg 70 kg 180 kg+ = . /dm dt is negative since the mass of the MMU decreases as gas is ejected. EXECUTE: (a) 2 ex 180 kg (0.029 m/s ) 0.0106 kg/s 490 m/s dm m a dt v ? ?= ? = ? = ?? ?? ? . In 5.0 s the mass that is ejected is (0.0106 kg/s)(5.0 s) 0.053 kg= . (b) ex (490 m/s)( 0.0106 kg/s) 5.19 N dm F v dt = ? = ? ? = . Momentum, Impulse, and Collisions 8-17 EVALUATE: The mass change in the 5.0 s is a very small fraction of the total mass m, so it is accurate to take m to be constant. 8.58. IDENTIFY and SET UP: Apply Eq. 8.39: ex . v dm a m dt = ? Solve for / .dm dt EXECUTE: ( )( )2 ex 6000 kg 25.0 m/s 75.0 kg/s 2000 m/s dm ma dt v = ? = ? = ? . So in 1 s the rocket must eject 75.0 kg of gas. EVALUATE: We have approximated /dm dt by / .m t? ? We have assumed that 225.0 m/s is the average acceleration for the first second. 8.59. IDENTIFY: Use Eq. 8.39, applied to a finite time interval. Solve for exv . SET UP: 160 m m t ? = ?? . EXECUTE: ex v m a m t ?= ? ? . 2 3 ex 15.0 m/s 2.40 10 m/s 2.40 km/s / / 160 a v m m m m t ?= ? = = × =?? ? ? ??? ? ? ??? ? ? ? EVALUATE: The acceleration is proportional to the speed of the exhaust gas and to the rate at which mass is ejected. 8.60. IDENTIFY and SET UP: av( )F t J? = relates the impulse J to the average thrust avF . Eq. 8.38 applied to a finite time interval gives av ex m F v t ?= ? ? . 0 0 ex ln m v v v m ? ?? = ? ?? ? . The remaining mass m after 1.70 s is 0.0133 kg. EXECUTE: (a) 10.0 N s 5.88 N 1.70 s J F t ?= = =? . av max/ 0.442F F = . (b) avex 800 m/s0.0125 kg F t v ?= ? =? . (c) 0 0v = and 0ex 0.0258 kgln (800 m/s)ln 530 m/s0.0133 kg m v v m ? ?? ?= = =? ?? ?? ? ? ? . EVALUATE: The acceleration of the rocket is not constant. It increases as the mass remaining decreases. 8.61. IDENTIFY: 00 ex ln m v v v m ? ?? = ? ?? ? . SET UP: 0 0v = . EXECUTE: 3 0 ex 8.00 10 m/s ln 3.81 2100 m/s m v m v ×? ? = = =? ?? ? and 3.810 45.2 m e m = = . EVALUATE: Note that the final speed of the rocket is greater than the relative speed of the exhaust gas. 8.62. IDENTIFY and SET UP: Use Eq. 8.40: ( )0 ex 0ln /v v v m m? = . 0 0v = (?fired from rest?), so ( )ex 0/ ln /v v m m= . Thus ex/0 / , v vm m e= or ex/0/ v vm m e ?= . If v is the final speed then m is the mass left when all the fuel has been expended; 0/m m is the fraction of the initial mass that is not fuel. (a) EXECUTE: 3 51.00 10 3.00 10 m/sv c?= × = × gives 5(3.00 10 m/s) /(2000 m/s) 66 0/ 7.2 10m m e ? × ?= = × . EVALUATE: This is clearly not feasible, for so little of the initial mass to not be fuel. (b) EXECUTE: 3000 m/sv = gives (3000 m/s)/(2000 m/s)0/ 0.223m m e ?= = . EVALUATE: 22.3% of the total initial mass not fuel, so 77.7% is fuel; this is possible. 8.63. IDENTIFY: Use the heights to find 1yv and 2 yv , the velocity of the ball just before and just after it strikes the slab. Then apply y y yJ F t p= ? = ? . SET UP: Let + y be downward. 8-18 Chapter 8 EXECUTE: (a) 212 mv mgh= so 2v gh= ± . 2 1 2(9.80 m/s )(2.00 m) 6.26 m/syv = + = . 22 2(9.80 m/s )(1.60 m) 5.60 m/syv = ? = ? . 3 2 1( ) (40.0 10 kg)( 5.60 m/s 6.26 m/s) 0.474 kg m/sy y y yJ p m v v ?= ? = ? = × ? ? = ? ? . The impulse is 0.474 kg m/s? , upward. (b) 3 0.474 kg m/s 237 N 2.00 10 s y y J F t ? ? ?= = = ?? × . The average force on the ball is 237 N, upward. EVALUATE: The upward force on the ball changes the direction of its momentum. 8.64. IDENTIFY: Momentum is conserved in the explosion. At the highest point the velocity of the boulder is zero. Since one fragment moves horizontally the other fragment also moves horizontally. Use projectile motion to relate the initial horizontal velocity of each fragment to its horizontal displacement. SET UP: Use coordinates where + x is north. Since both fragments start at the same height with zero vertical component of velocity, the time in the air, t, is the same for both. Call the fragments A and B, with A being the one that lands to the north. Therefore, 3B Am m= . EXECUTE: Apply 1 2x xP P= to the collision: 0 A Ax B Bxm v m v= + . /3ABx Ax Ax B m v v v m = ? = ? . Apply projectile motion to the motion after the collision: 0 0xx x v t? = . Since t is the same, 0 0( ) ( )A B Ax Bx x x x x v v ? ?= and 0 0 0 /3 ( ) ( ) ( ) (274 m) /3 91.3 mBx AxB A A Ax Ax v v x x x x x x v v ? ? ? ??? = ? = ? = ? = ?? ? ? ?? ? ? ? . The other fragment lands 91.3 m directly south of the point of explosion. EVALUATE: The fragment that has three times the mass travels one-third as far. 8.65. IDENTIFY: The impulse, force and change in velocity are related by Eq. 8.9 SET UP: / 0.0571 kgm w g= = . Since the force is constant, avF = F ! ! . EXECUTE: (a) 3( 380 N)(3.00 10 s) 1.14 N sx xJ F t ?= ? = ? × = ? ? . 3(110 N)(3.00 10 s) 0.330 N sy yJ F t ?= ? = × = ? . (b) 2 1 1.14 N s 20.0 m/s 0.04 m/s 0.0571 kg x x x J v v m ? ?= + = + = . 2 1 0.330 N s ( 4.0 m/s) 1.8 m/s0.0571 kg y y y J v v m ?= ? = + ? = + . EVALUATE: The change in velocity ?v! is in the same direction as the force, so ?v! has a negative x component and a positive y component. 8.66. IDENTIFY: The horizontal component of the momentum of the system of cars is conserved. SET UP: Let + x be the direction the cars are traveling. Each car has mass m. Let 1v be the initial speed of the three cars. 12 15v v= . Let N be the number of cars in the final collection. EXECUTE: 1 2x xP P= . 1 2(3 ) ( )m v Nm v= . 1 1 2 1 3 3 15 /5 v v N v v = = = . EVALUATE: In the complete absence of friction or other external horizontal forces this process of adding cars and slowing down continues forever. 8.67. IDENTIFY: x Ax BxP p p= + and y Ay ByP p p= + . SET UP: Let object A be the convertible and object B be the SUV. Let + x be west and + y be south, 0Axp = and 0Byp = . EXECUTE: (8000 kg m/s)sin 60.0 6928 kg m/sxP = ? = ?° , so 6928 kg m/sBxp = ? and 6928 kg m/s 3.46 m/s 2000 kgBx v ?= = . (8000 kg m/s)cos60.0 4000 kg m/syP = ? = ?° , so 4000 kg m/sBxp = ? and 4000 kg m/s 2.67 m/s1500 kgAyv ?= = . The convertible has speed 2.67 m/s and the SUV has speed 3.46 m/s. EVALUATE: Each component of the total momentum arises from a single vehicle. 8.68. IDENTIFY: The total momentum of the system is conserved and is equal to zero, since the pucks are released from rest. SET UP: Each puck has the same mass m. Let + x be east and + y be north. Let object A be the puck that moves west. All three pucks have the same speed v. Momentum, Impulse, and Collisions 8-19 EXECUTE: 1 2x xP P= gives 0 Bx Cxmv mv mv= ? + + and Bx Cxv v v= + . 1 2y yP P= gives 0 By Cymv mv= + and By Cyv v= ? . Since B Cv v= and the y components are equal in magnitude, the x components must also be equal: Bx Cxv v= and Bx Cxv v v= + says / 2Bx Cxv v v= = . If Byv is positive then Cyv is negative. The angle ? that puck B makes with the x axis is given by / 2 cos v v ? = and 60? = ° . One puck moves in a direction 60° north of east and the other puck moves in a direction 60° south of east. EVALUATE: Each component of momentum is separately conserved. 8.69. IDENTIFY: The x and y components of the momentum of the system are conserved. Set Up: After the collision the combined object with mass tot 0.100 kgm = moves with velocity 2v! . Solve for Cxv and Cyv . EXECUTE: (a) 1 2x xP P= gives tot 2A Ax B Bx C Cx xm v m v m v m v+ + = . tot 2A Ax B Bx x Cx C m v m v m v v m + ?= ? (0.020 kg)( 1.50 m/s) (0.030 kg)( 0.50 m/s)cos60 (0.100 kg)(0.50 m/s) 0.050 kgCx v ? + ? ?= ? ° . 1.75 m/sCxv = . 1 2y yP P= gives tot 2A Ay B By C Cy ym v m v m v m v+ + = . tot 2 (0.030 kg)( 0.50 m/s)sin 60 0.260 m/s 0.050 kg A Ay B By y Cy C m v m v m v v m + ? ?= ? = ? = +° . (b) 2 2 1.77 m/sC Cx Cyv v v= + = . 2 1K K K? = ? . 2 2 2 21 1 1 1 2 2 2 2(0.100 kg)(0.50 m/s) [ (0.020 kg)(1.50 m/s) (0.030)(0.50 m/s) (0.050 kg)(1.77 m/s) ]K? = ? + + 0.092 JK? = ? . EVALUATE: Since there is no horizontal external force the vector momentum of the system is conserved. The forces the spheres exert on each other do negative work during the collision and this reduces the kinetic energy of the system. 8.70. IDENTIFY: Use a coordinate system attached to the ground. Take the x-axis to be east (along the tracks) and the y-axis to be north (parallel to the ground and perpendicular to the tracks). Then xP is conserved and yP is not conserved, due to the sideways force exerted by the tracks, the force that keeps the handcar on the tracks. (a) SET UP: Let A be the 25.0 kg mass and B be the car (mass 175 kg). After the mass is thrown sideways relative to the car it still has the same eastward component of velocity, 5.00 m/s, as it had before it was thrown. Figure 8.70a xP is conserved so ( ) 1 2 2A B A A x B B xm m v m v m v+ = + EXECUTE: ( )( ) ( )( ) ( ) 2200 kg 5.00 m/s 25.0 kg 5.00 m/s 175 kg B xv= + . 2 1000 kg m/s 125 kg m/s 5.00 m/s. 175 kgB x v ? ? ?= = The final velocity of the car is 5.00 m/s, east (unchanged). EVALUATE: The thrower exerts a force on the mass in the y-direction and by Newton?s 3rd law the mass exerts an equal and opposite force in the -directiony? on the thrower and car. (b) SET UP: We are applying P x = constant in coordinates attached to the ground, so we need the final velocity of A relative to the ground. Use the relative velocity addition equation. Then use P x = constant to find the final velocity of the car. 8-20 Chapter 8 EXECUTE: / / /A E A B B E= +v v v ! ! ! / 5.00 m/sB Ev = + / 5.00 m/sA Bv = ? (minus since the mass is moving west relative to the car). This gives / 0;A Ev = the mass is at rest relative to the earth after it is thrown backwards from the car. As in part (a), ( ) 1 2 2 .A B A A x B B xm m v m v m v+ = + Now 2 0,A xv = so ( ) 1 2 .A B B B xm m v m v+ = ( )2 1 200 kg 5.00 m/s 5.71 m/s.175 kgA BB x B m m v v m ? ? ? ?+= = =? ? ? ?? ?? ? The final velocity of the car is 5.71 m/s, east. EVALUATE: The thrower exerts a force in the -directionx? so the mass exerts a force on him in the -directionx+ and he and the car speed up. (c) SET UP: Let A be the 25.0 kg mass and B be the car (mass 200 kg).Bm = Figure 8.70b xP is conserved so ( )1 1 2A A x B B x A B xm v m v m m v+ = + . EXECUTE: ( )1 1 2A A B B A B xm v m v m m v? + = + . ( )( ) ( )( )1 1 2 200 kg 5.00 m/s 25.0 kg 6.00 m/s 3.78 m/s. 200 kg 25.0 kg B B A A x A B m v m v v m m ??= = =+ + The final velocity of the car is 3.78 m/s, east. EVALUATE: The mass has negative xp so reduces the total xP of the system and the car slows down. 8.71. IDENTIFY: The horizontal component of the momentum of the sand plus railroad system is conserved. SET UP: As the sand leaks out it retains its horizontal velocity of 15.0 m/s. EXECUTE: The horizontal component of the momentum of the sand doesn?t change when it leaks out so the speed of the railroad car doesn?t change; it remains 15.0 m/s. In Exercise 8.27 the rain is falling vertically and initially has no horizontal component of momentum. Its momentum changes as it lands in the freight car. Therefore, in order to conserve the horizontal momentum of the system the freight car must slow down. EVALUATE: The horizontal momentum of the sand does change when it strikes the ground, due to the force that is external to the system of sand plus railroad car. 8.72. IDENTIFY: Kinetic energy is 212K mv= and the magnitude of the momentum is p mv= . The force and the time t it acts are related to the change in momentum whereas the force and distance d it acts are related to the change in kinetic energy. SET UP: Assume the net forces are constant and let the forces and the motion be along the x axis. The impulse- momentum theorem then says Ft p= ? and the work-energy theorem says Fd K= ? . EXECUTE: (a) 2 41N 2 (840 kg)(9.0 m/s) 3.40 10 JK = = × . 2 41P 2 (1620 kg)(5.0 m/s) 2.02 10 JK = = × . The Nash has the greater kinetic energy and N P 1.68 K K = . (b) 3N (840 kg)(9.0 m/s) 7.56 10 kg m/sp = = × ? . 3P (1620 kg)(5.0 m/s) 8.10 10 kg m/sp = = × ? . The Packard has the greater magnitude of momentum and N P 0.933 p p = . (c) Since the cars stop the magnitude of the change in momentum equals the initial momentum. Since P Np p> , P NF F> and N N P P 0.933 F p F p = = . (d) Since the cars stop the magnitude of the change in kinetic energy equals the initial kinetic energy. Since N PK K> , N PF F> and N N P P 1.68 F K F K = = . EVALUATE: If the stopping forces were the same, the Packard would have a larger stopping time but would travel a shorter distance while stopping. This consistent with it having a smaller initial speed. Momentum, Impulse, and Collisions 8-21 8.73. IDENTIFY: Use the impulse-momentum theorem to relate the average force on the bullets to their rate of change in momentum. By Newton?s third law, the average force the weapon exerts on the bullets is equal in magnitude and opposite in direction to the recoil force the bullets exert on the weapon. SET UP: Consider a time interval of 1.00 minute. Let + x be the direction of motion of the bullets and use coordinated fixed to the ground. The bullets start from rest. EXECUTE: avF t p? = ? gives 3 av (1000)(7.45 10 kg)(293 m/s) 36.4 N 60.0 s F ?×= = . The recoil force is 36.4 N. EVALUATE: The change in momentum for each bullet is small since the mass is small, but over 16 bullets are fired per second. 8.74. IDENTIFY: Find k for the spring from the forces when the frame hangs at rest, use constant acceleration equations to find the speed of the putty just before it strikes the frame, apply conservation of momentum to the collision between the putty and the frame and then apply conservation of energy to the motion of the frame after the collision. SET UP: Use the free-body diagram for the frame when it hangs at rest on the end of the spring to find the force constant k of the spring. Let s be the amount the spring is stretched. Figure 8.74a EXECUTE: y yF ma=? . 0mg ks? + = . ( )( )20.150 kg 9.80 m/s 29.4 N/m 0.050 m mg k s = = = . SET UP: Next find the speed of the putty when it reaches the frame. The putty falls with acceleration ,a g= downward. Figure 8.74b 0 0v = 0 0.300 my y? = 29.80 m/sa = + ?v = 2 2 0 02 ( )v v a y y= + ? EXECUTE: ( ) ( )( )202 2 9.80 m/s 0.300 m 2.425 m/sv a y y= ? = = . SET UP: Apply conservation of momentum to the collision between the putty (A) and the frame (B): Figure 8.74c yP is conserved, so ( )1 2A A A Bm v m m v? = ? + . EXECUTE: ( )2 1 0.200 kg 2.425 m/s 1.386 m/s0.350 kgA AA B m v v m m ? ? ? ?= = =? ? ? ?+ ? ?? ? . 8-22 Chapter 8 SET UP: Apply conservation of energy to the motion of the frame on the end of the spring after the collision. Let point 1 be just after the putty strikes and point 2 be when the frame has its maximum downward displacement. Let d be the amount the frame moves downward. Figure 8.74d When the frame is at position 1 the spring is stretched a distance 1 0.050 m.x = When the frame is at position 2 the spring is stretched a distance 2 0.050 m .x d= + Use coordinates with the y-direction upward and 0y = at the lowest point reached by the frame, so that 1y d= and 2 0.y = Work is done on the frame by gravity and by the spring force, so other 0,W = and el gravity.U U U= + EXECUTE: 1 1 other 2 2K U W K U+ + = + . other 0W = . ( )( )221 11 12 2 0.350 kg 1.386 m/s 0.3362 JK mv= = = . ( )( ) ( )( )22 21 11 1,el 1, grav 1 12 2 29.4 N/m 0.050 m 0.350 kg 9.80 m/sU U U kx mgy d= + = + = + . ( )1 0.03675 J 3.43 NU d= + . ( )( )221 12 2,el 2,grav 2 22 2 29.4 N/m 0.050 mU U U kx mgy d= + = + = + . ( ) ( ) 22 0.03675 J 1.47 N 14.7 N/mU d d= + + . Thus ( ) ( ) ( ) 20.3362 J 0.03675 J 3.43 N 0.03675 J 1.47 N 14.7 N/md d d+ + = + + . ( ) ( )214.7 N/m 1.96 N 0.3362 J 0d d? ? = . ( ) ( ) ( )( )21/ 29.4 1.96 1.96 4 14.7 0.3362 m 0.0667 m 0.1653 m.d ? ?= ± ? ? = ±? ?? ? The solution we want is a positive (downward) distance, so 0.0667 m 0.1653 m 0.232 m.d = + = EVALUATE: The collision is inelastic and mechanical energy is lost. Thus the decrease in gravitational potential energy is not equal to the increase in potential energy stored in the spring. 8.75. IDENTIFY: Apply conservation of momentum to the collision and conservation of energy to the motion after the collision. SET UP: Let + x be to the right. The total mass is bullet block 1.00 kgm m m= + = . The spring has force constant 2 0.750 N 300 N/m 0.250 10 m F k x ? = = =× . Let V be the velocity of the block just after impact. EXECUTE: (a) Conservation of energy for the motion after the collision gives 1 el2K U= . 2 21 12 2mV kx= and 300 N/m (0.150 m) 2.60 m/s 1.00 kg k V x m = = = . (b) Conservation of momentum applied to the collision gives bullet 1m v mV= . 1 3 bullet (1.00 kg)(2.60 m/s) 325 m/s 8.00 10 kg mV v m ? = = =× . EVALUATE: The initial kinetic energy of the bullet is 422 J. The energy stored in the spring at maximum compression is 3.38 J. Most of the initial mechanical energy of the bullet is dissipated in the collision. 8.76. IDENTIFY: The horizontal components of momentum of the system of bullet plus stone are conserved. The collision is elastic if 1 2.K K= Momentum, Impulse, and Collisions 8-23 SET UP: Let A be the bullet and B be the stone. (a) Figure 8.76 EXECUTE: xP is conserved so 1 1 2 2A A x B B x A A x B B xm v m v m v m v+ = + . 1 2A A B B xm v m v= . ( )32 1 6.00 10 kg 350 m/s 21.0 m/s0.100 kgAB x AB m v v m ?? ? ? ?×= = =? ? ? ?? ?? ? yP is conserved so 1 1 2 2A A y B B y A A y B B ym v m v m v m v+ = + . 2 20 A A B B ym v m v= ? + . ( )32 2 6.00 10 kg 250 m/s 15.0 m/s0.100 kgAB y AB m v v m ?? ? ? ?×= = =? ? ? ?? ?? ? . ( ) ( )2 22 22 2 2 21.0 m/s 15.0 m/s 25.8 m/sB B x B yv v v= + = + = . 2 2 15.0 m/s tan 0.7143; 21.0 m/s B y B x v v ? = = = 35.5? = ° (defined in the sketch). (b) To answer this question compare 1K and 2K for the system: ( )( )22 2 31 1 11 1 12 2 2 6.00 10 kg 350 m/s 368 JA A B BK m v m v ?= + = × = . ( )( ) ( )( )2 22 2 31 1 1 12 2 22 2 2 26.00 10 kg 250 m/s 0.100 kg 25.8 m/s 221 JA A B BK m v m v ?= + = × + = . 2 1 221 J 368 J 147 JK K K? = ? = ? = ? . EVALUATE: The kinetic energy of the system decreases by 147 J as a result of the collision; the collision is not elastic. Momentum is conserved because ext, 0xF =? and ext, 0.yF =? But there are internal forces between the bullet and the stone. These forces do negative work that reduces K . 8.77. IDENTIFY: Apply conservation of momentum to the collision between the two people. Apply conservation of energy to the motion of the stuntman before the collision and to the entwined people after the collision. SET UP: For the motion of the stuntman, 1 2 5.0 my y? = . Let Sv be the magnitude of his horizontal velocity just before the collision. Let V be the speed of the entwined people just after the collision. Let d be the distance they slide along the floor. EXECUTE: (a) Motion before the collision: 1 1 2 2K U K U+ = + . 1 0K = and 21 S 1 22 ( )mv mg y y= ? . 2 S 1 22 ( ) 2(9.80 m/s )(5.0 m) 9.90 m/sv g y y= ? = = . Collision: S S totm v m V= . S S tot 80.0 kg (9.90 m/s) 5.28 m/s 150.0 kg m V v m ? ?= = =? ?? ? . (b) Motion after the collision: 1 1 other 2 2K U W K U+ + = + gives 21 tot k tot2 0m V m gd?? = . 2 2 2 k (5.28 m/s) 5.7 m 2 2(0.250)(9.80 m/s ) V d g?= = = . EVALUATE: Mechanical energy is dissipated in the inelastic collision, so the kinetic energy just after the collision is less than the initial potential energy of the stuntman. 8.78. IDENTIFY: Apply conservation of energy to the motion before and after the collision and apply conservation of momentum to the collision. SET UP: Let v be the speed of the mass released at the rim just before it strikes the second mass. Let each object have mass m. EXECUTE: Conservation of energy says 212 ;mv mgR= 2v gR= . 8-24 Chapter 8 SET UP: This is speed 1v for the collision. Let 2v be the speed of the combined object just after the collision. EXECUTE: Conservation of momentum applied to the collision gives 1 22mv mv= so 2 1 / 2 / 2v v gR= = SET UP: Apply conservation of energy to the motion of the combined object after the collision. Let 3y be the final height above the bottom of the bowl. EXECUTE: ( ) ( )21 2 32 2 2m v m gy= . 2 2 3 1 / 4 2 2 2 v gR y R g g ? ?= = =? ?? ? . EVALUATE: Mechanical energy is lost in the collision, so the final gravitational potential energy is less than the initial gravitational potential energy. 8.79. IDENTIFY: Eqs. 8.24 and 8.25 give the outcome of the elastic collision. Apply conservation of energy to the motion of the block after the collision. SET UP: Object B is the block, initially at rest. If L is the length of the wire and ? is the angle it makes with the vertical, the height of the block is (1 cos )y L ?= ? . Initially, 1 0y = . EXECUTE: Eq. 8.25 gives 2 2 (5.00 m/s) 2.50 m/s 3 A B A A B m M v v m m M M ? ? ? ?= = =? ? ? ?+ +? ?? ? . Conservation of energy gives 21 2 (1 cos )B B Bm v m gL ?= ? . 2 2 2 (2.50 m/s) cos 1 1 0.362 2 2(9.80 m/s )(0.500 m) Bv gL ? = ? = ? = and 68.8? = ° . EVALUATE: Only a portion of the initial kinetic energy of the ball is transferred to the block in the collision. 8.80. IDENTIFY: Apply conservation of energy to the motion before and after the collision. Apply conservation of momentum to the collision. SET UP: First consider the motion after the collision. The combined object has mass tot 25.0 kg.m = Apply m ?F a! !5 to the object at the top of the circular loop, where the object has speed 3.v The acceleration is 2 rad 3 / ,a v R= downward. EXECUTE: 2 3vT mg m R + = . The minimum speed 3v for the object not to fall out of the circle is given by setting 0.T = This gives 3 ,v Rg= where 3.50 m.R = SET UP: Next, use conservation of energy with point 2 at the bottom of the loop and point 3 at the top of the loop. Take 0y = at point 2. Only gravity does work, so 2 2 3 3K U K U+ = + EXECUTE: ( )2 21 1tot 2 tot 3 tot2 2 2m v m v m g R= + . Use 3v Rg= and solve for 2:v 2 5 13.1 m/sv gR= = . SET UP: Now apply conservation of momentum to the collision between the dart and the sphere. Let 1v be the speed of the dart before the collision. EXECUTE: ( ) ( )( )15.00 kg 25.0 kg 13.1 m/sv = . 1 65.5 m/sv = . EVALUATE: The collision is inelastic and mechanical energy is removed from the system by the negative work done by the forces between the dart and the sphere. 8.81. IDENTIFY: Use Eq. 8.25 to find the speed of the hanging ball just after the collision. Apply m?F = a! ! to find the tension in the wire. After the collision the hanging ball moves in an arc of a circle with radius 1.35 mR = and acceleration 2rad /a v R= . SET UP: Let A be the 2.00 kg ball and B be the 8.00 kg ball. For applying m?F = a! ! to the hanging ball, let + y be upward, since rada ! is upward. The free-body force diagram for the 8.00 kg ball is given in Figure 8.81. EXECUTE: 2 1 2 2[2.00kg] (5.00 m/s) 2.00 m/s 2.00 kg 8.00 kg A B x A x A B m v v m m ? ? ? ?= = =? ? ? ?+ +? ?? ? . Just after the collision the 8.00 kg ball has speed 2.00 m/sv = . Using the free-body diagram, y yF ma=? gives radT mg ma? = . 2 2 2 [2.00 m/s](8.00 kg) 9.80 m/s 102 N 1.35 m v T m g R ? ? ? ?= + = + =? ? ? ?? ? ? ? . Momentum, Impulse, and Collisions 8-25 EVALUATE: The tension before the collision is the weight of the ball, 78.4 N. Just after the collision, when the ball has started to move, the tension is greater than this. arad mg y x T Figure 8.81 8.82. IDENTIFY: The impulse applied to the ball equals its change in momentum. The height of the ball and its speed are related by conservation of energy. SET UP: Let + y be upward. EXECUTE: Applying conservation of energy to the motion of the ball from its height h to the floor gives 21 12 mv mgh= , where 1v is its speed just before it hits the floor. Just before it hits, it is traveling downward, so the velocity of the ball just before it hits the floor is 1 2yv gh= ? . Applying conservation of energy to the motion of the ball from just after it bounces off the floor with speed 2v to its maximum height of 0.90h gives 21 22 (0.90 )mv mg h= . It is moving upward, so 2 2 (0.90 )yv g h= + . The impulse applied to the ball is 2 1 2 1( )y y y y yJ p p m v v= ? = ? = 2 (0.90 ) 2 2.76m g h m gh m gh+ = . The floor exerts an upward impulse of 2.76m gh to the ball. EVALUATE: The impulse increases when m increases and when h increases. The ball does not return to its initial height because some mechanical energy is dissipated during the collision with the floor. 8.83. IDENTIFY: Apply conservation of momentum to the collision between the bullet and the block and apply conservation of energy to the motion of the block after the collision. (a) SET UP: Collision between the bullet and the block: Let object A be the bullet and object B be the block. Apply momentum conservation to find the speed 2Bv of the block just after the collision. Figure 8.83a EXECUTE: xP is conserved so 1 1 2 2A A x B B x A A x B B xm v m v m v m v+ = + . 1 2 2A A A A B B xm v m v m v= + . ( ) ( )31 2 2 4.00 10 kg 400 m/s 120 m/s 1.40 m/s 0.800 kg A A A B x B m v v v m ?? × ?= = = . SET UP: Motion of the block after the collision. Let point 1 in the motion be just after the collision, where the block has the speed 1.40 m/s calculated above, and let point 2 be where the block has come to rest. Figure 8.83b 1 1 other 2 2K U W K U+ + = + . EXECUTE: Work is done on the block by friction, so other .fW W= ( )other k kcos ,f kW W f s f s mgs? ?= = = ? = ? where 0.450 ms = 1 20, 0U U= = 21 1 1 22 , 0K mv K= = (block has come to rest) Thus 21 1 k2 0.mv mgs?? = ( ) ( )( ) 22 1 k 2 1.40 m/s 0.222 2 2 9.80 m/s 0.450 m v gs ? = = = . 8-26 Chapter 8 (b) For the bullet, ( )( )22 31 11 12 2 4.00 10 kg 400 m/s 320 JK mv ?= = × = . ( )( )22 31 12 22 2 4.00 10 kg 120 m/s 28.8 JK mv ?= = × = . 2 1 28.8 J 320 J 291 JK K K? = ? = ? = ? . The kinetic energy of the bullet decreases by 291 J. (c) Immediately after the collision the speed of the block is 1.40 m/s so its kinetic energy is ( )( )221 12 2 0.800 kg 1.40 m/s 0.784 J.K mv= = = EVALUATE: The collision is highly inelastic. The bullet loses 291 J of kinetic energy but only 0.784 J is gained by the block. But momentum is conserved in the collision. All the momentum lost by the bullet is gained by the block. 8.84. IDENTIFY: Apply conservation of momentum to the collision and conservation of energy to the motion of the block after the collision. SET UP: Let + x be to the right. Let the bullet be A and the block be B. Let V be the velocity of the block just after the collision. EXECUTE: Motion of block after the collision: 1 grav2K U= . 212 B Bm V m gh= . 2 22 2(9.80 m/s )(0.450 10 m) 0.297 m/sV gh ?= = × = . Collision: 2 0.297 m/sBv = . 1 2x xP P= gives 1 2 2A A A A B Bm v m v m v= + . 3 1 2 2 3 (5.00 10 kg)(450 m/s) (1.00 kg)(0.297 m/s) 391 m/s 5.00 10 kg A A B B A A m v m v v m ? ? ? × ?= = =× . EVALUATE: We assume the block moves very little during the time it takes the bullet to pass through it. 8.85. IDENTIFY: Eqs. 8.24 and 8.25 give the outcome of the elastic collision. The value of M where the kinetic energy loss lossK of the neutron is a maximum satisfies loss / 0dK dM = . SET UP: Let object A be the neutron and object B be the nucleus. Let the initial speed of the neutron be 1Av . All motion is along the x-axis. 210 12 AK mv= . EXECUTE: (a) 2 1A A m M v v m M ?= + . 2 2 2 2 2 2 01 1 1 loss 1 2 1 12 2 2 2 2 2 4 1 ( ) ( )A A A A m M m M K mM K mv mv m v v m M M m M m ? ??? ?= ? = ? = =? ?? ?? ?+ + +? ?? ? , as was to be shown. (b) loss 0 2 3 1 2 4 0 ( ) ( ) dK M K m dM M m M m ? ?= ? =? ?+ +? ? . 2 1 M M m =+ and M m= . The incident neutron loses the most kinetic energy when the target has the same mass as the neutron. (c) When A Bm m= , Eq. 8.24 says 2 0Av = . The final speed of the neutron is zero and the neutron loses all of its kinetic energy. EVALUATE: When M m>> , 2 1A x A xv v? ? and the neutron rebounds with speed almost equal to its initial speed. In this case very little kinetic energy is lost; loss 04 /K K m M= , which is very small. 8.86. IDENTIFY: Eqs. 8.24 and 8.25 give the outcome of the elastic collision. SET UP: Let all the motion be along the x axis. 1 0A xv v= . EXECUTE: (a) 2 0 A B A x A B m m v v m m ? ??= ? ?+? ? and 2 0 2 A B x A B m v v m m ? ?= ? ?+? ? . 211 02 AK m v= . 2 2 2 21 1 2 2 0 12 2 A B A B A A A x A A B A B m m m m K m v m v K m m m m ? ? ? ?? ?= = =? ? ? ?+ +? ? ? ? and 2 2 1 A A B A B K m m K m m ? ??= ? ?+? ? . ( ) 2 2 21 1 2 2 0 12 2 2 2 4A A B B B B x B A B A B m m m K m v m v K m m m m ? ?= = =? ?+ +? ? and ( ) 2 2 1 4B A B A B K m m K m m = + . (b) (i) For A Bm m= , 2 1 0A K K = and 2 1 1B K K = . (ii) For 5A Bm m= , 2 1 4 9 AK K = and 2 1 5 9 BK K = . Momentum, Impulse, and Collisions 8-27 (c) Equal sharing of the kinetic energy means 2 2 1 1 1 2 A BK K K K = = . 2 1 2 A B A B m m m m ? ?? =? ?+? ? . 2 2 2 22 2 4 2A B A B A A B Bm m m m m m m m+ ? = + + . 2 26 0A A B Bm m m m? + = . The quadratic formula gives 5.83A B m m = or 0.172A B m m = . We can also verify that these values give 2 1 1 2 BK K = . EVALUATE: When A Bm m<< or when A Bm m>> , object A retains almost all of the original kinetic energy. 8.87. IDENTIFY: Apply conservation of energy to the motion of the package before the collision and apply conservation of the horizontal component of momentum to the collision. (a) SET UP: Apply conservation of energy to the motion of the package from point 1 as it leaves the chute to point 2 just before it lands in the cart. Take 0y = at point 2, so 1 4.00 m.y = Only gravity does work, so 1 1 2 2K U K U+ = + . EXECUTE: 2 21 11 1 22 2mv mgy mv+ = . 2 2 1 12 9.35 m/sv v gy= + = . (b) SET UP: In the collision between the package and the cart momentum is conserved in the horizontal direction. (But not in the vertical direction, due to the vertical force the floor exerts on the cart.) Take x+ to be to the right. Let A be the package and B be the cart. EXECUTE: xP is constant gives ( )1 1 2A A x B B x A B xm v m v m m v+ = + . 1 5.00 m/sB xv = ? . ( )1 3.00 m/s cos37.0A xv = ° . (The horizontal velocity of the package is constant during its free-fall.) Solving for 2xv gives 2 3.29 m/s.xv = ? The cart is moving to the left at 3.29 m/s after the package lands in it. EVALUATE: The cart is slowed by its collision with the package, whose horizontal component of momentum is in the opposite direction to the motion of the cart. 8.88. IDENTIFY: Eqs. 8.24, 8.25, and 8.27 give the outcome of the elastic collision. SET UP: The blue puck is object A and the red puck is object B. Let + x be the direction of the initial motion of A. 1 0.200 m/sA xv = , 2 0.050 m/sA xv = and 1 0B xv = EXECUTE: (a) Eq. 8.27 gives 2 2 1 1 0.250 m/sB x A x B x A xv v v v= ? + = . (b) Eq. 8.25 gives 1 2 0.200 m/s 2 1 (0.0400 kg) 2 1 0.024 kg 0.250 m/s A x B A B x v m m v ? ? ? ?? ?= ? = ? =? ? ? ?? ?? ?? ?? ? . EVALUATE: We can verify that our results give 1 2K K= and 1 2x xP P= , as required in an elastic collision. 8.89. (a) IDENTIFY and SET UP: 2 21 12 2 .A A B BK m v m v= + Use cmA A?= +v v v! ! and cmB B?= +v v v! ! to replace Av and Bv in this equation. Note A?v! and B?v! as defined in the problem are the velocities of A and B in coordinates moving with the center of mass. Note also that cmA A B Bm m M? ? ?+ =v v v! ! ! where cm?v! is the velocity of the car in these coordinates. But that?s zero, so 0;A A B Bm m? ? v v! !1 5 we can use this in the proof. In part (b), use that P ! is conserved in a collision. EXECUTE: cm ,A A?= +v v v! ! so 2 2 2cm cm2A A Av v v? ?= + + ?v v! ! . cm ,B B?= +v v v! ! so 2 2 2cm cm2B B Bv v v? ?= + + ?v v! ! . (We have used that for a vector ,A ! 2 .)A = ?A A! ! Thus 2 2 2 21 1 1 1cm cm cm cm2 2 2 2A A A A A B B B B BK m v m v m m v m v m? ? ? ?= + + ? + + + ?v v v v! ! ! ! . ( ) ( ) ( )2 2 21 1cm cm2 2A B A A B B A A B BK m m v m v m v m m? ? ? ?= + + + + ?+v v v! ! ! . But A Bm m M+ = and as noted earlier 0,A A B Bm m? ? =+v v! ! so ( )2 2 21 1cm2 2 .A A B BK Mv m v m v? ?= + + This is the result the problem asked us to derive. (b) EVALUATE: In the collision cmM=P v ! ! is constant, so 21 cm2 Mv stays constant. The asteroids can lose all their relative kinetic energy but the 21 cm2 Mv must remain. 8-28 Chapter 8 8.90. IDENTIFY: Eq. 8.27 describes the elastic collision, with x replaced by y. Speed and height are related by conservation of energy. SET UP: Let + y be upward. Let A be the large ball and B be the small ball, so 1B yv v= ? and 1A yv v= + . If the large ball has much greater mass than the small ball its speed is changed very little in the collision and 2A yv v= + . EXECUTE: (a) 2 2 1 1( )B y A y B y A yv v v v? = ? ? gives 2 2 1 1 ( ) 3B y A y B y A yv v v v v v v v= + ? + = ? ? + = + . The small ball moves upward with speed 3v after the collision. (b) Let 1h be the height the small ball fell before the collision. Conservation of energy applied to the motion from the release point to the floor gives 1 2U K= and 211 2mgh mv= . 2 1 2 v h g = . Conservation of energy applied to the motion of the small ball from immediately after the collision to its maximum height 2h (rebound distance) gives 1 2K U= and 21 22 (3 )m v mgh= . 2 2 1 9 9 2 v h h g = = . The ball?s rebound distance is nine times the distance it fell. EVALUATE: The mechanical energy gained by the small ball comes from the energy of the large ball. But since the large ball?s mass is much larger it can give up this energy with very little decrease in speed. 8.91. IDENTIFY: Apply conservation of momentum to the system consisting of Jack, Jill and the crate. The speed of Jack or Jill relative to the ground will be different from 4.00 m/s. SET UP: Use an inertial coordinate system attached to the ground. Let + x be the direction in which the people jump. Let Jack be object A, Jill be B, and the crate be C. EXECUTE: (a) If the final speed of the crate is v, 2C xv v= ? , and 2 2 4.00 m/sA x B xv v v= = ? . 2 1x xP P= gives 2 2 2 0A A x B Bx C Cxm v m v m v+ + = . (75.0 kg)(4.00 m/s ) (45.0 kg)(4.00 m/s ) (15.0 kg)( ) 0v v v? + ? + ? = and (75.0 kg 45.0 kg)(4.00 m/s) 3.56 m/s 75.0 kg 45.0 kg 15.0 kg v += =+ + . (b) Let v? be the speed of the crate after Jack jumps. Apply momentum conservation to Jack jumping: (75.0 kg)(4.00 m/s ) (60.0 kg)( ) 0v v? ?? + ? = and (75.0 kg)(4.00 m/s) 2.22 m/s 135.0 kg v? = = . Then apply momentum conservation to Jill jumping, with v being the final speed of the crate: 1 2x xP P= gives (60.0 kg)( ) (45.0 kg)(4.00 m/s ) (15.0 kg)( )v v v?? = ? + ? . (45.0 kg)(4.00 m/s) (60.0 kg)(2.22 m/s) 5.22 m/s 60.0 kg v += = . (c) Repeat the calculation in (b), but now with Jill jumping first. Jill jumps: (45.0 kg)(4.00 m/s ) (90.0 kg)( ) 0v v? ?? + ? = and 1.33 m/sv? = . Jack jumps: (90.0 kg)( ) (75.0 kg)(4.00 m/s ) (15.0 kg)( )v v v?? = ? + ? . (75.0 kg)(4.00 m/s) (90.0 kg)(1.33 m/s) 4.66 m/s 90.0 kg v += = . EVALUATE: The final speed of the crate is greater when Jack jumps first, then Jill. In this case Jack leaves with a speed of 1.78 m/s relative to the ground, whereas when they both jump simultaneously Jack and Jill each leave with a speed of only 0.44 m/s relative to the ground. 8.92. IDENTIFY: Momentum is conserved in the explosion. The total kinetic energy of the two fragments is Q . SET UP: Let the final speed of the two fragments be Av and Bv . They must move in opposite directions after the explosion. EXECUTE: (a) Since the initial momentum of the system is zero, conservation of momentum says A A B Bm v m v= and AB A B m v v m ? ?= ? ?? ? . A BK K Q+ = gives 2 2 21 1 2 2 A A A B A B m m v m v Q m ? ?+ =? ?? ? . 212 1 A A A B m m v Q m ? ?+ =? ?? ? . 1 / B A A B A B Q m K Q m m m m ? ?= = ? ?+ +? ? . 1 B AB A A B A B m m K Q K Q Q m m m m ? ? ? ?= ? = ? =? ? ? ?+ +? ? ? ? . (b) If 4B Am m= , then 45AK Q= and 1 5B K Q= . The lighter fragment gets 80% of the energy that is released. EVALUATE: If A Bm m= the fragments share the energy equally. In the limit that B Am m>> , the lighter fragment gets almost all of the released energy. Momentum, Impulse, and Collisions 8-29 8.93. IDENTIFY: Apply conservation of momentum to the system of the neutron and its decay products. SET UP: Let the proton be moving in the + x direction with speed pv after the decay. The initial momentum of the neutron is zero, so to conserve momentum the electron must be moving in the x? direction after the decay. Let the speed of the electron be ev . EXECUTE: 1 2x xP P= gives p p e e0 m v m v= ? and pe p e m v v m ? ?= ? ?? ? . The total kinetic energy after the decay is 2 p p2 2 2 2 21 1 1 1 1 tot e e p p e p p p p p2 2 2 2 2 e e 1 m m K m v m v m v m v m v m m ? ? ? ?= + = + = +? ? ? ?? ? ? ? . Thus, p 4 tot p e 1 1 5.44 10 0.0544% 1 / 1 1836 K K m m ?= = = × =+ + . EVALUATE: Most of the released energy goes to the electron, since it is much lighter than the proton. 8.94. IDENTIFY: Momentum is conserved in the decay. The results of Problem 8.92 give the kinetic energy of each fragment. SET UP: Let A be the alpha particle and let B be the radium nucleus, so / 0.0176A Bm m = . 136.54 10 JQ ?= × . EXECUTE: 13 136.54 10 J 6.43 10 J 1 / 1 0.0176A A B Q K m m ? ?×= = = ×+ + and 130.11 10 JBK ?= × . EVALUATE: The lighter particle receives most of the released energy. 8.95. IDENTIFY: The momentum of the system is conserved. SET UP: Let + x be to the right. 1 0xP = . exp , nxp and anxp are the momenta of the electron, polonium nucleus and antineutrino, respectively. EXECUTE: 1 2x xP P= gives e n an 0x x xp p p+ + = . an e n( )x x xp p p= ? + . 22 25 3 22 an (5.60 10 kg m/s [3.50 10 kg][ 1.14 10 m/s]) 1.66 10 kg m/sxp ? ? ?= ? × ? + × ? × = ? × ? . The antineutrino has momentum to the left with magnitude 221.66 10 kg m/s?× ? . EVALUATE: The antineutrino interacts very weakly with matter and most easily shows its presence by the momentum it carries away. 8.96. IDENTIFY: Momentum components in the x and y directions are separately conserved. For an elastic collision 1 2K K= . SET UP: 1 1A x Av v= + , 1 0B xv = . 2 2 cosA x Av v ?= , 2 2 sinA y Av v ?= . 2 2 cosB x Bv v ?= , 2 2 sinB y Bv v ?= ? . 2 2sin cos 1? ?+ = , for any angle ? . cos( ) cos cos sin sin? ? ? ? ? ?+ = ? . EXECUTE: (a) 1 2x xP P= gives 1 2 2cos cosA A A A B Bm v m v m v? ?= + . 1 2y yP P= gives 2 20 sin sinA A B Bm v m v? ?= ? . (b) 2 2 2 2 2 2 2 21 2 2 2 2cos cos 2 cos cosA A A A B B A B A Bm v m v m v m m v v? ? ? ?= + + and 2 2 2 2 2 2 2 2 2 20 sin sin 2 sin sinA A B B A B A Bm v m v m m v v? ? ? ?= + ? . Adding these two equations and using the trig identities in the SET UP step gives 2 2 2 2 2 21 2 2 2 22 cos( )A A A A B B A B A Bm v m v m v m m v v ? ?= + + + . (c) 1 2K K= says 2 2 21 1 11 2 22 2 2A A A A B Bm v m v m v= + . The result in part (b) agrees with this expression only if cos( ) 0? ?+ = . This requires that 90 rad 2 ?? ?+ = =° . EVALUATE: The result of part (c) says that the two protons move in perpendicular directions after the collision. 8.97. IDENTIFY and SET UP: Figure 8.97 xP and yP are conserved in the collision since there is no external horizontal force. 8-30 Chapter 8 The result of Problem 8.96 part (d) applies here since the collision is elastic This says that 25.0 90 ,B?° + = ° so that 65.0 .B? = ° (A and B move off in perpendicular directions.) EXECUTE: xP is conserved so 1 1 2 2A A x B B x A A x B B xm v m v m v m v+ = + . But A Bm m= so 1 2 2cos25.0 cos65.0A A Bv v v= ° + ° . yP is conserved so 1 1 2 2A A y B B y A A y B B ym v m v m v m v+ = + . 2 20 A y B yv v= + . 2 20 sin 25.0 sin 65.0A Bv v= ° ? ° . ( )2 2sin 25.0 / sin65.0B Av v= ° ° . This result in the first equation gives 1 2 2 sin 25.0 cos65.0 cos25.0 sin65.0A A A v v v ° °? ?= ° + ? ?°? ? . 1 21.103A Av v= . 2 1 /1.103 (15.0 m/s)/1.103 13.6 m/sA Av v= = = . And then ( )( )2 sin 25.0 / sin65.0 13.6 m/s 6.34 m/s.Bv = ° ° = EVALUATE: We can use our numerical results to show that 1 2K K= and that 1 2x xP P= and 1 2 .y yP P= 8.98. IDENTIFY: Since there is no friction, the horizontal component of momentum of the system of Jonathan, Jane and the sleigh is conserved. SET UP: Let + x be to the right. 800 NAw = , 600 NBw = and 1000 NCw = . EXECUTE: 1 2x xP P= gives 2 2 20 A A x B B x C C xm v m v m v= + + . 2 2 2 22 A A x B B x A A x B B xC x C C m v m v w v w v v m w + += = . 2 (800 N)( [5.00 m/s]cos30.0 (600 N)( [7.00 m/s]cos36.9 ) 0.105 m/s 1000 NC x v ? + += = ?°) ° . The sleigh?s velocity is 0.105 m/s, to the left. EVALUATE: The vertical component of the momentum of the system consisting of the two people and the sleigh is not conserved, because of the net force exerted on the sleigh by the ice while they jump. 8.99. IDENTIFY: In Eq. 8.28 treat each straight piece as an object in the system. SET UP: The center of mass of each piece of length L is at its center. EXECUTE: (a) From symmetry, the center of mass is on the vertical axis, a distance ( / 2)cos( / 2)L ? below the apex. (b) The center of mass is on the vertical axis of symmetry, a distance 2( / 2) /3 /3L L= above the center of the horizontal segment. (c) Using the wire frame as a coordinate system, the coordinates of the center of mass are equal and each is equal to ( / 2) / 2 / 4L L= . The center of mass is along the bisector of the angle, a distance / 8L from the corner. (d) By symmetry, the center of mass is at the center of the equilateral triangle, a distance ( /3)sin60 / 12L L=° above the center of the horizontal segment. EVALUATE: The center of mass need not lie on any point of the object, it can be in empty space. 8.100. IDENTIFY: There is no net horizontal external force so cmv is constant. SET UP: Let + x be to the right, with the origin at the initial position of the left-hand end of the canoe. A 45.0 kgm = , 60.0 kgBm = . The center of mass of the canoe is at its center. EXECUTE: Initially, cm 0v = , so the center of mass doesn?t move. Initially, 1 1cm1 A A B B A B m x m x x m m += + . After she walks, 2 2cm2 A A B B A B m x m x x m m += + . cm1 cm2x x= gives 1 1 2 2A A B B A A B Bm x m x m x m x+ = + . She walks to a point 1.00 m from the right-hand end of the canoe, so she is 1.50 m to the right of the center of mass of the canoe and 2 2 1.50 mA Bx x= + . 2 2(45.0 kg)(1.00 m) (60.0 kg)(2.50 m) (45.0 kg)( 1.50 m) (60.0 kg)B Bx x+ = + + . 2(105.0 kg) 127.5 kg mBx = ? and 2 1.21 mBx = . 2 1 1.21 m 2.50 m 1.29 mB Bx x? = ? = ? . The canoe moves 1.29 m to the left. Momentum, Impulse, and Collisions 8-31 EVALUATE: When the woman walks to the right, the canoe moves to the left. The woman walks 3.00 m to the right relative to the canoe and the canoe moves 1.29 m to the left, so she moves 3.00 m 1.29 m 1.71 m? = to the right relative to the water. Note that this distance is (60.0 kg / 45.0 kg)(1.29 m) . 8.101. IDENTIFY: Take as the system you and the slab. There is no horizontal force, so horizontal momentum is conserved. By Eq. 8.32, P ! is constant cmv ! is constant (for a system of constant mass). Use coordinates fixed to the ice, with the direction you walk as the x-direction. cmv ! is constant and initially cm 0.=v! Figure 8.101 p p s s cm p s m m m m + 0 v v v ! !! 15 5 . p cm s sm m = 0v v! !1 . p p s s 0x xm v m v+ = . ( ) ( )s p s p p p/ /5 2.00 m/s 0.400 m/sx xv m m v m m= ? = ? = ? . The slab moves at 0.400 m/s, in the direction opposite to the direction you are walking. EVALUATE: The initial momentum of the system is zero. You gain momentum in the -directionx+ so the slab gains momentum in the -direction.x? The slab exerts a force on you in the -directionx+ so you exert a force on the slab in the -direction.x? 8.102. IDENTIFY: Conservation of x and y components of momentum applies to the collision. At the highest point of the trajectory the vertical component of the velocity of the projectile is zero. SET UP: Let + y be upward and + x be horizontal and to the right. Let the two fragments be A and B, each with mass m. For the projectile before the explosion and the fragments after the explosion. 0xa = , 29.80 m/sya = ? . EXECUTE: (a) 2 20 02 ( )y y yv v a y y= + ? with 0yv = gives that the maximum height of the projectile is 2 2 0 2 ([80.0 m/s]sin 60.0 ) 244.9 m 2 2( 9.80 m/s ) y y v h a = ? = ? =? ° . Just before the explosion the projectile is moving to the right with horizontal velocity 0 0 cos60.0 40.0 m/sx xv v v= = =° . After the explosion 0Axv = since fragment A falls vertically. Conservation of momentum applied to the explosion gives (2 )(40.0 m/s) Bxm mv= and 80.0 m/sBxv = . Fragment B has zero initial vertical velocity so 210 0 2y yy y v t a t? = + gives a time of fall of 2 2 2(244.9 m) 7.07 s 9.80 m/sy h t a = ? = ? =? . During this time the fragment travels horizontally a distance (80.0 m/s)(7.07 s) 566 m= . It also took the projectile 7.07 s to travel from launch to maximum height and during this time it travels a horizontal distance of ([80.0 m/s]cos60.0 )(7.07 s) 283 m=° . The second fragment lands 283 m 566 m 849 m+ = from the firing point. (b) For the explosion, 2 411 2 (20.0 kg)(40.0 m/s) 1.60 10 JK = = × . 2 412 2 (10.0 kg)(80.0 m/s) 3.20 10 JK = = × . The energy released in the explosion is 41.60 10 J× . EVALUATE: The kinetic energy of the projectile just after it is launched is 46.40 10 J× . We can calculate the speed of each fragment just before it strikes the ground and verify that the total kinetic energy of the fragments just before they strike the ground is 4 4 46.40 10 J 1.60 10 J 8.00 10 J× + × = × . Fragment A has speed 69.3 m/s just before it strikes the ground, and hence has kinetic energy 42.40 10 J× . Fragment B has speed 2 2(80.0 m/s) (69.3 m/s) 105.8 m/s+ = just before it strikes the ground, and hence has kinetic energy 45.60 10 J× . Also, the center of mass of the system has the same horizontal range 2 0 0sin(2 ) 565 m v R g ?= = that the projectile would have had if no explosion had occurred. One fragment lands at / 2R so the other, equal mass fragment lands at a distance 3 / 2R from the launch point. 8.103. IDENTIFY: The rocket moves in projectile motion before the explosion and its fragments move in projectile motion after the explosion. Apply conservation of energy and conservation of momentum to the explosion. 8-32 Chapter 8 SET UP: Apply conservation of energy to the explosion. Just before the explosion the shell is at its maximum height and has zero kinetic energy. Let A be the piece with mass 1.40 kg and B be the piece with mass 0.28 kg. Let Av and Bv be the speeds of the two pieces immediately after the collision. EXECUTE: 2 21 12 2 860 JA A B Bm v m v+ = SET UP: Since the two fragments reach the ground at the same time, their velocities just after the explosion must be horizontal. The initial momentum of the shell before the explosion is zero, so after the explosion the pieces must be moving in opposite horizontal directions and have equal magnitude of momentum: .A A B Bm v m v= EXECUTE: Use this to eliminate Av in the first equation and solve for :Bv ( )212 1 / 860 JB B B Am v m m+ = and 71.6 m/s.Bv = Then ( )/ 14.3 m/s.A B A Bv m m v= = (b) SET UP: Use the vertical motion from the maximum height to the ground to find the time it takes the pieces to fall to the ground after the explosion. Take y+ downward. 0 0,yv = 29.80 m/s ,ya = + 0 80.0 m,y y? = ?t = EXECUTE: 210 0 2y yy y v t a t? = + gives 4.04 s.t = During this time the horizontal distance each piece moves is 57.8 mA Ax v t= = and 289.1 m.B Bx v t= = They move in opposite directions, so they are 347 mA Bx x+ = apart when they land. EVALUATE: Fragment A has more mass so it is moving slower right after the collision, and it travels horizontally a smaller distance as it falls to the ground. 8.104. IDENTIFY: Apply conservation of momentum to the collision. At the highest point of its trajectory the shell is moving horizontally. If one fragment received some upward momentum in the collision, the other fragment would have had to receive a downward component. Since they each the ground at the same time, each must have zero vertical velocity immediately after the explosion. SET UP: Let + x be horizontal, along the initial direction of motion of the projectile and let + y be upward. At its maximum height the projectile has 0 cos55.0 86.0 m/sxv v= =° . Let the heavier fragment be A and the lighter fragment be B. 9.00 kgAm = and 3.00 kgBm = . EXECUTE: Since fragment A returns to the launch point, immediately after the explosion it has 86.0 m/sAxv = ? . Conservation of momentum applied to the explosion gives (12.0 kg)(86.0 m/s) (9.00 kg)( 86.0 m/s) (3.00 kg) Bxv= ? + and 602 m/sBxv = . The horizontal range of the projectile, if no explosion occurred, would be 2 0 0sin(2 ) 2157 m v R g ?= = . The horizontal distance each fragment travels is proportional to its initial speed and the heavier fragment travels a horizontal distance / 2 1078 mR = after the explosion, so the lighter fragment travels a horizontal distance 602 m (1078 m) 7546 m 86 m ? ? =? ?? ? from the point of explosion and 1078 m 7546 m 8624 m+ = from the launch point. The energy released in the explosion is 2 2 2 51 1 1 2 1 2 2 2(9.00 kg)(86.0 m/s) (3.00 kg)(602 m/s) (12.0 kg)(86.0 m/s) 5.33 10 JK K? = + ? = × . EVALUATE: The center of mass of the system has the same horizontal range 2157 mR = as if the explosion didn?t occur. This gives (12.0 kg)(2157 m) (9.00 kg)(0) (3.00 kg)d= + and 8630 md = , where d is the distance from the launch point to where the lighter fragment lands. This agrees with our calculation. 8.105. IDENTIFY: No external force, so P ! is conserved in the collision. SET UP: Apply momentum conservation in the x and y directions: Figure 8.105 Solve for 1v and 2.v Momentum, Impulse, and Collisions 8-33 EXECUTE: xP is conserved so ( )0 1 f 2cos45 cos10 cos30mv m v v v= ° + ° + ° . 0 f 1 2cos10 cos45 cos30v v v v? ° = ° + ° . 1 21030.4 m/s cos45 cos30v v= ° + ° . xP is conserved so ( )1 2 f0 sin 45 sin30 sin10m v v v= ° ? ° + ° . 1 2sin 45 sin30 347.3 m/sv v° = ° ? . sin 45 cos45° = ° so 2 21030.4 m/s sin30 347.3 m/s cos30v v= ° ? + ° . 2 1030.4 m/s 347.3 m/s 1010 m/s sin30 cos30.0 v += =° + ° . And then 21 sin30 347.3 m/s 223 m/s. sin 45 v v ° ?= =° Then two emitted neutrons have speeds of 223 m/s and 1010 m/s. The speeds of the Ba and Kr nuclei are related by zP conservation. zP is constant implies that Ba Ba Kr Kr0 m v m v= ? 25 Ba Kr Ba Ba Ba25 Kr 2.3 10 kg 1.5 . 1.5 10 kg m v v v v m ? ? ? ? ? ?×= = =? ? ? ?×? ?? ? We can?t say what these speeds are but they must satisfy this relation. The value of Bav depends on energy considerations. EVALUATE: ( ) ( )23 611 n n2 3.0 10 m/s 4.5 10 J/kg .K m m= × = × ( ) ( ) ( )2 2 231 1 12 n n n Ba Kr2 2 22.0 10 m/s 223 m/s 1010 m/sK m m m K K= × + + + + ( )6 n Ba Kr2.5 10 J/kg .m K K= × + + We don?t know what BaK and KrK are, but they are positive. We will study such nuclear reactions further in Chapter 43 and will find that energy is released in this process; 2 1.K K> Some of the potential energy stored in the 235 U nucleus is released as kinetic energy and shared by the collision fragments. 8.106. IDENTIFY: The velocity of the center of mass of the system of the two blocks is given by Eq. 8.30. Conservation of momentum says the center of mass moves at constant speed. SET UP: 1 1A x Av v= , 1 0B xv = . The velocity u! in the center of mass frame is related to the velocity v! in the stationary frame by cm?u = v v! ! ! . We can express kinetic energy as 2 2 p K m = . EXECUTE: (a) 1cm- A A x A B m v v m m = + . (b) The center of mass moves with constant speed so this coordinate system is an inertial frame. (c) 11 1 cm- B A A x A x x A B m v u v v m m = ? = + . 1 1 1 cm- A A B x B x x A B m v u v v m m = ? = ? + . In this frame 1 1 1 0x A A x B B xP m u m u= + = . (d) 2 1 0x xP P= = gives 1 1 0A x B xp p+ = and 2 2 0A x B xp p+ = , so 1 1B x A xp p= ? and 2 2B x A xp p= ? . Conservation of kinetic energy gives 2 2 2 2 2 2 1 1 2 2 2 2 A x B x A x B x A B A B p p p p m m m m + = + . Using 2 2B x A xp p= ? and 1 1B x A xp p= ? gives 2 22 1A x A xp p= and 2 1A x A xp p= ± . If a collision occurs Axp changes and 2 1A x A xp p= ? . But 2 2B x A xp p= ? and 1 1B x A xp p= ? , so 2 1B x B xp p= ? . In the center of mass frame the momentum and hence the velocity of each puck keeps the same magnitude and reverses direction. (e) cm- 0.400 kg (6.00 m/s) 4.00 m/s 0.600 kgx v ? ?= =? ?? ? . 1 6.00 m/s 4.00 m/s 2.00 m/sA xu = ? = . 1 0 4.00 m/s 4.00 m/sB xu = ? = ? . 2 2.00 m/sA xu = ? and 2 4.00 m/sB xu = + . 2 2 cm- 2.00 m/s 4.00 m/s 2.00 m/sA x A x xv u v= + = ? + = . 2 2 cm- 4.00 m/s 4.00 m/s 8.00 m/sB x B x xv u v= + = + = . Eq. 8.24 says 2 0.400 kg 0.200 kg (6.00 m/s) 2.00 m/s 0.400 kg 0.200 kgA x v ? ??= =? ?+? ? . Eq. 8.25 says 2 2[0.400 kg] (6.00 m/s) 8.00 m/s 0.400 kg 0.200 kgA x v ? ?= =? ?+? ? . Our result agrees with Eqs. 8.24 and 8.25. 8-34 Chapter 8 EVALUATE: Eqs. 8.24 and 8.25 apply only when 1 0Bv = . The result that the velocity of each puck in the center of mass frame reverses direction and retains the same magnitude applies to all elastic collisions, even when both are moving initially. 8.107. IDENTIFY and SET UP: Apply conservation of energy to find the total energy before and after the collision with the floor from the initial and final maximum heights. EXECUTE: (a) Objects stick together says that the relative speed after the collision is zero, so 0.=P (b) In an elastic collision the relative velocity of the two bodies has the same magnitude before and after the collision, so 1.=P (c) Speed of ball just before collision: 21 12mgh mv= . 1 2v gh= Speed of ball just after collision: 211 22mgH mv= . 2 12v gH= The second object (the surface) is stationary, so 2 1 1/ / .v v H h= =P (d) 1 /H h=P implies ( )( )221 1.2 m 0.85 0.87 mH h= = =P . (e) 21H h= P . 2 4 2 1H H h= =P P . ( )2 4 2 63 2H H h h= = =P P P P . Generalize to 2 2( 1) 2 21 n n n nH H h h ? ?= = =P P P P . (f) 8th bounce implies 8n = . ( )16168 1.2 m 0.85 0.089 mH h= = =P . EVALUATE: P is a measure of the kinetic energy lost in the collision. The collision here is between a ball and the earth. Momentum lost by the ball is gained by the earth, but the velocity gained by the earth is very small and can be taken to be zero. 8.108. IDENTIFY: Momentum is conserved in the collision. Conservation of energy says 2 1K K= + ? . SET UP: For part (b) let 0v be the common speed of each atom before the collision and let V ! and 3v ! be the velocities after the collision of the molecule and the atom that remains. 271.67 10 kgm ?= × is the mass of one hydrogen atom. EXECUTE: (a) In the center of mass frame 1 0xP = so 2 0xP = and cm2 0v = . But in this frame the potential energy decreases and the kinetic energy increases. This is inconsistent with 212cm tot cm22 0K m v= = . (b) Before the collision cm 0v = . After the collision the molecule and remaining atom move in opposite directions and 3(2 )m V mv= ; 3 2v V= . Conservation of energy gives 2 2 2 21 1 13 02 2 2(2 ) 3( )m V mv mv+ = + ? . With 3 2v V= this becomes 2 21 02 3 V v m ?= + . 19 3 2 41 2 27 7.23 10 J (1.00 10 m/s) 1.20 10 m/s 3(1.67 10 ) V ? ? ×= × + = ×× and 4 3 2 2.40 10 m/sv V= = × . EVALUATE: ( )2 211 023 2.50 10 JK mv ?= = × , which is much less than the binding energy of the molecule. Other initial conditions also lead to molecule formation; the one of zero initial momentum is just particularly simple to analyze. 8.109. IDENTIFY: Apply conservation of energy to the motion of the wagon before the collision. After the collision the combined object moves with constant speed on the level ground. In the collision the horizontal component of momentum is conserved. SET UP: Let the wagon be object A and treat the two people together as object B. Let + x be horizontal and to the right. Let V be the speed of the combined object after the collision. EXECUTE: (a) The speed 1Av of the wagon just before the collision is given by conservation of energy applied to the motion of the wagon prior to the collision. 1 2U K= says 21 12([50 m][sin6.0 ])A A Am g m v=° . 1 10.12 m/sAv = . 1 2x xP P= for the collision says 1 ( )A A A Bm v m m V= + and 300 kg (10.12 m/s) 6.98 m/s300 kg 75.0 kg 60.0 kgV ? ?= =? ?+ +? ? . In 5.0 s the wagon travels (6.98 m/s)(5.0 s) 34.9 m= , and the people will have time to jump out of the wagon before it reaches the edge of the cliff. Momentum, Impulse, and Collisions 8-35 (b) For the wagon, 2 411 2 (300 kg)(10.12 m/s) 1.54 10 JK = = × . Assume that the two heroes drop from a small height, so their kinetic energy just before the wagon can be neglected compared to 1K of the wagon. 2 41 2 2 (435 kg)(6.98 m/s) 1.06 10 JK = = × . The kinetic energy of the system decreases by 31 2 4.8 10 JK K? = × . EVALUATE: The wagon slows down when the two heroes drop into it. The mass that is moving horizontally increases, so the speed decreases to maintain the same horizontal momentum. In the collision the vertical momentum is not conserved, because of the net external force due to the ground. 8.110. IDENTIFY: Gravity gives a downward external force of magnitude mg . The impulse of this force equals the change in momentum of the rocket. SET UP: Let + y be upward. Consider an infinitesimal time interval dt. In Example 8.15, ex 2400 m/sv = and 0 120 s dm m dt = ? . In Example 8.16, 0 / 4m m= after 90 st = . EXECUTE: (a) The impulse-momentum theorem gives ex( )( ) ( )( )mgdt m dm v dv dm v v mv? = + + + ? ? . This simplifies to exmgdt mdv v dm? = + and exdv dmm v mgdt dt= ? ? . (b) ex dv v dm a g dt m dt = = ? ? . (c) At 0t = , 2 2ex 0 1 (2400 m/s) 9.80 m/s 10.2 m/s 120 s v dm a g m dt ? ?= ? ? = ? ? ? =? ?? ? . (d) ex v dv dm gdt m = ? ? . Integrating gives 00 ex ln mv v v gtm? = + ? . 0 0v = and 2(2400 m/s)ln 4 (9.80 m/s )(90 s) 2445 m/sv = + ? = . EVALUATE: Both the initial acceleration in Example 8.15 and the final speed of the rocket in Example 8.16 are reduced by the presence of gravity. 8.111. IDENTIFY and SET UP: Apply Eq. 8.40 to the single-stage rocket and to each stage of the two-stage rocket. (a) EXECUTE: ( )0 ex 0ln / ;v v v m m? = 0 0v = so ( )ex 0ln /v v m m= The total initial mass of the rocket is 0 12,000 kg 1000 kg 13,000 kg.m = + = Of this, 9000 kg 700 kg 9700 kg+ = is fuel, so the mass m left after all the fuel is burned is 13,000 kg 9700 kg 3300 kg.? = ( )ex exln 13,000 kg/3300 kg 1.37v v v= = . (b) First stage: ( )ex 0ln /v v m m= 0 13,000 kgm = The first stage has 9000 kg of fuel, so the mass left after the first stage fuel has burned is 13,000 kg 9000 kg 4000 kg.? = ( )ex exln 13,000 kg/4000 kg 1.18v v v= = . (c) Second stage: 0 1000 kg,m = 1000 kg 700 kg 300 kgm = ? = . ( ) ( )0 ex 0 ex ex exln / 1.18 ln 1000 kg/300 kg 2.38v v v m m v v v= + = + = . (d) 7.00 km/sv = ( )ex / 2.38 7.00 km/s / 2.38 2.94 km/sv v= = = . q EVALUATE: The two-stage rocket achieves a greater final speed because it jetisons the left-over mass of the first stage before the second-state fires and this reduces the final m and increases 0 / .m m 8.112. IDENTIFY: During an interval where the mass is constant the speed of the rocket is constant. During an interval where the mass is changing at a constant rate, the equations of Section 8.6 apply. SET UP: For 0 90 st? ? , 0 120 s dm m dt = ? . From Example 8.15, ex 2400 m/sv = . EXECUTE: (a) For 0t ? , 0v = . For 0 90 st? ? , Eq. 8.40 says (2400 m/s)ln 4 3327 m/sv = = . For 90 st > , v has the constant value 3327 m/s. The graph of ( )v t is given in Fig. 8.112a. (b) For 0 90 st? ? , Eq. 8.39 gives 2 ex 0 0 2400 m/s 20 m/s (1 /[120 s]) 120 s 1 /[120 s] v dm m a m dt m t t ? ?= ? = ? ? =? ?? ?? ? . 220 m/sa = at 0t = (as in Example 8.15) and 280 m/sa = at 90 st = . For 90 st > , 0a = . The graph of ( )a t is given in Fig. 8.112b. (c) The astronaut has the same acceleration as the rocket. This is maximum at 90 st = and 2 3 max astronaut max (75 kg)(80 m/s ) 6.0 10 NF m a= = = × . This is 8.2 times her weight on earth, since maxa is 8.2 times g . 8-36 Chapter 8 EVALUATE: The acceleration increases because the mass decreases while the thrust ex dm F v dt = ? remains constant. 20 40 60 80 100 0 200 40 60 12010080 a (m/s2) t (s) 1000 2000 3000 4000 0 (a) (b) 200 40 60 12010080 v (m/s) t (s) Figure 8.112 8.113. IDENTIFY and SET UP: dm dV?= . dV Adx= . Since the thin rod lies along the x axis, cm 0y = . The mass of the rod is given by M dm= ? . EXECUTE: (a) 2 cm 0 0 1 2 L L A L x xdm A xdx M M M ? ?= = =? ? . The volume of the rod is AL and M AL?= . 2 cm 2 2 AL L x AL ? ?= = . The center of mass of the uniform rod is at its geometrical center, midway between its ends. (b) 3 2 cm 0 0 0 1 1 . 3 L L LA A L x xdm x Adx x dx M M M M ? ??= = = =? ? ? 20 0 .2 L L AL M dm Adx A xdx ?? ?= = = =? ? ? Therefore, 3 cm 2 2 2 . 3 3 A L L x AL ? ? ? ?? ?= =? ?? ?? ?? ? EVALUATE: When the density increases with x, the center of mass is to the right of the center of the rod. 8.114. IDENTIFY: cm 1 x xdm M = ? and cm 1 .y ydmM= ? At the upper surface of the plate, 2 2 2.y x a+ = SET UP: To find cmx , divide the plate into thin strips parallel to the y-axis, as shown in Fig. 8.114a. To find cmy , divide the plate into thin strips parallel to the x-axis as shown in Fig. 8.114b. The plate has volume one-half that of a circular disk, so 212V a t?= and 212 .M a t??= EXECUTE: In Fig.114a each strip has length 2 2 .y a x= ? cm 1 ,x xdmM= ? where 2 2 .dm tydx t a x dx? ?= = ? 2 2 cm 0, a a t x x a x dx M ? ?= ? =? since the integrand is an odd function of x. cm 0x = because of symmetry. In Fig.114b each strip has length 2 22 2 .x a y= ? cm 1 ,y ydmM= ? where 2 22 2 .dm txdy t a y dy? ?= = ? 2 2 cm 0 2 at y y a y dy M ?= ?? . The integral can be evaluated using 2 2u a y= ? , 2du ydy= ? . This substitution gives 2 3 30 1/ 2 cm 2 2 1 2 2 2 4 2 3 3 3a t ta ta a y u du M M a t ? ? ? ?? ? ? ?? ?? ?= ? = = =? ?? ?? ?? ? ? ?? ?? . EVALUATE: 4 0.424. 3? = cmy is less than /2,a as expected, since the plate becomes wider as y decreases. dx (a) y y x 2x x dy (b) y y x Figure 8.114 Momentum, Impulse, and Collisions 8-37 8.115. IDENTIFY: The work is related to the force by 2 1 x x W Fdx= ? . The force the person must apply equals the weight of the hanging portion. Since the rope is uniform, the center of mass of the hanging portion is at its geometrical center. SET UP: Let y be the length of the rope hanging over the edge and use coordinates where the origin is at the edge of the table and + y is downward. When the rope is pulled onto the table, y goes from / 4l to zero. A length y of the rope has mass y? . EXECUTE: (a) When a length y hangs over the edge, the person must apply an upward force ( )yF m y g yg?= ? = ? . 20 0 / 4 / 4 ( ) 32yl l g l W F y dy g ydy ??= = ? =? ? . (b) Initially, cm /8y l= . The work done to raise an object of mass M a distance cmy is cmW Mgy= . 2 4 8 32 l l gl W g ? ?? ? ? ?= =? ? ? ?? ? ? ? . EVALUATE: The answers from methods (a) and (b) agree. The change in gravitational potential energy of the rope can be calculated by considering all its mass acting at its center of mass, and the work done by the person equals the increase in gravitational potential energy of the rope. 8.116. IDENTIFY: From our analysis of motion with constant acceleration, if v at= and a is constant, then 21 0 0 2x x v t at? = + . SET UP: Take 0 0v = , 0 0x = and let + x downward. EXECUTE: (a) dv a dt = , v at= and 212x at= . Substituting into 2dvxg x vdt= + gives 2 2 2 2 2 231 1 2 2 2at g at a a t a t= + = . The nonzero solution is /3a g= . (b) 2 2 2 21 1 12 6 6 (9.80 m/s )(3.00 s) 14.7 mx at gt= = = = . (c) (2.00 g/m)(14.7 m) 29.4 gm kx= = = . EVALUATE: The acceleration is less than g because the small water droplets are initially at rest, before they adhere to the falling drop. The small droplets are suspended by buoyant forces that we ignore for the raindrops. 9-1 ROTATION OF RIGID B ODIES 9.1. IDENTIFY: s r?= , with ? in radians. SET UP: rad 180? = ° . EXECUTE: (a) 1.50 m 0.600 rad 34.4 2.50 m s r ? = = = = ° (b) 14.0 cm 6.27 cm (128 )( rad /180 ) s r ? ?= = =° ° (c) (1.50 m)(0.700 rad) 1.05 ms r ?= = = EVALUATE: An angle is the ratio of two lengths and is dimensionless. But, when s r?= is used, ? must be in radians. Or, if /s r? = is used to calculate ? , the calculation gives ? in radians. 9.2. IDENTIFY: 0 t? ? ?? = , since the angular velocity is constant. SET UP: 1 rpm (2 / 60) rad/s?= . EXECUTE: (a) (1900)(2 rad / 60 s) 199 rad/s? ?= = (b) 35 (35 )( /180 ) 0.611 rad?= =° ° ° . 30 0.611 rad 3.1 10 s 199 rad/s t ? ? ? ??= = = × EVALUATE: In 0t ? ? ? ?= we must use the same angular measure (radians, degrees or revolutions) for both 0? ?? and ? . 9.3. IDENTIFY: ( ) zz d t dt ?? = . Writing Eq.(2.16) in terms of angular quantities gives 2 1 t zt dt? ? ?? = ? . SET UP: 1n nd t nt dt ?= and 11 1 n nt dt t n += +? EXECUTE: (a) A must have units of rad/s and B must have units of 3rad/s . (b) 3( ) 2 (3.00 rad/s )z t Bt t? = = . (i) For 0t = , 0z? = . (ii) For 5.00 st = , 215.0 rad/sz? = . (c) 2 1 2 3 31 2 1 2 1 2 13( ) ( ) ( ) t t A Bt dt A t t B t t? ?? = + = ? + ?? . For 1 0t = and 2 2.00 st = , 3 31 2 1 3(2.75 rad/s)(2.00 s) (1.50 rad/s )(2.00 s) 9.50 rad? ?? = + = . EVALUATE: Both z? and z? are positive and the angular speed is increasing. 9.4. IDENTIFY: /z zd dt? ?= . av- zz t ?? ?= ? . SET UP: 2( ) 2d t t dt = EXECUTE: (a) 3( ) 2 ( 1.60 rad s ) .zz d?? t ?t tdt= = ? = ? (b) 3 2(3.0 s) ( 1.60 rad s )(3.0 s) 4.80 rad s .z? = ? = ? 2. av- (3.0 s) (0) 2.20 rad s 5.00 rad s 2.40 rad s , 3.0 s 3.0 s z z z ? ? ? ? ? ?= = = ? which is half as large (in magnitude) as the acceleration at 3.0 s.t = EVALUATE: ( )z t? increases linearly with time, so av- (0) (3.0 s)2 z z z ? ?? += . (0) 0z? = . 9 9-2 Chapter 9 9.5. IDENTIFY and SET UP: Use Eq.(9.3) to calculate the angular velocity and Eq.(9.2) to calculate the average angular velocity for the specified time interval. EXECUTE: 3;t t? ? ?= + 0.400 rad/s,? = 30.0120 rad/s? = (a) 23z d t dt ?? ? ?= = + (b) At 0,t = 0.400 rad/sz? ?= = (c) At 5.00 s,t = 3 20.400 rad/s 3(0.0120 rad/s )(5.00 s) 1.30 rad/sz? = + = 2 1 av- 2 1 z t t t ? ? ?? ? ?= =? ? For 1 0,t = 1 0.? = For 2 5.00 s,? = 3 32 (0.400 rad/s)(5.00 s) (0.012 rad/s )(5.00 s) 3.50 rad? = + = So av- 3.50 rad 0 0.700 rad/s. 5.00 s 0z ? ?= =? EVALUATE: The average of the instantaneous angular velocities at the beginning and end of the time interval is 1 2 (0.400 rad/s 1.30 rad/s) 0.850 rad/s.+ = This is larger than av- ,z? because ( )z t? is increasing faster than linearly. 9.6. IDENTIFY: ( )z dt dt ?? = . ( ) zz dt dt ?? = . av-z t ?? ?= ? . SET UP: 2 3 2(250 rad s) (40.0 rad s ) (4.50 rad s )z? t t= ? ? . 2 3z (40.0 rad s ) (9.00 rad s )? t= ? ? . EXECUTE: (a) Setting 0z? = results in a quadratic in t. The only positive root is 4.23 st = . (b) At 4.23 st = , 278.1 rad s .z? = ? (c) At 4.23 st = , 586 rad 93.3 rev? = = . (d) At 0t = , 250 rad/sz? = . (e) av- 586 rad 138 rad s. 4.23 sz ? = = EVALUATE: Between 0t = and 4.23 st = , z? decreases from 250 rad/s to zero. z? is not linear in t, so av-z? is not midway between the values of z? at the beginning and end of the interval. 9.7. IDENTIFY: ( )z dt dt ?? = . ( ) zz dt dt ?? = . Use the values of ? and z? at 0t = and z? at 1.50 s to calculate a, b, and c. SET UP: 1n nd t nt dt ?= EXECUTE: (a) 2( ) 3z t b ct? = ? . ( ) 6z t ct? = ? . At 0t = , / 4 rada? ?= = and 2.00 rad/sz b? = = . At 1.50 st = , 26 (1.50 s) 1.25 rad/sz c? = ? = and 30.139 rad/sc = ? . (b) / 4 rad? ?= and 0z? = at 0t = . (c) 23.50 rad/sz? = at 2 3 3.50 rad/s 4.20 s 6 6( 0.139 rad/s ) zt c ?= ? = ? =? . At 4.20 st = , 3 3 rad (2.00 rad/s)(4.20 s) ( 0.139 rad/s )(4.20 s) 19.5 rad 4 ?? = + ? ? = . 3 22.00 rad/s 3( 0.139 rad/s )(4.20 s) 9.36 rad/sz? = ? ? = . EVALUATE: ? , z? and z? all increase as t increases. 9.8. IDENTIFY: zz ddt ?? = . 0 av-zt? ? ?? = . When z? is linear in t, av-z? for the time interval 1t to 2t is 1 2 av- 2 1 z z z t t ? ?? += ? . SET UP: From the information given, 2( ) 6.00 rad/s (2.00 rad/s )z t t? = ? + EXECUTE: (a) The angular acceleration is positive, since the angular velocity increases steadily from a negative value to a positive value. (b) It takes 3.00 seconds for the wheel to stop ( 0)z? = . During this time its speed is decreasing. For the next 4.00 s its speed is increasing from 0 rad s to 8.00 rad s+ . Rotation of Rigid Bodies 9-3 (c) The average angular velocity is 6.00 rad s 8.00 rad s 1.00 rad s 2 ? + = . 0 av-zt? ? ?? = then leads to displacement of 7.00 rad after 7.00 s. EVALUATE: When z? and z? have the same sign, the angular speed is increasing; this is the case for 3.00 st = to 7.00 st = . When z? and z? have opposite signs, the angular speed is decreasing; this is the case between 0t = and 3.00 st = . 9.9. IDENTIFY: Apply the constant angular acceleration equations. SET UP: Let the direction the wheel is rotating be positive. EXECUTE: (a) 20 1.50 rad s (0.300 rad s )(2.50 s) 2.25 rad s.z z zt? ? ?= + = + = (b) 2 2 21 10 0 2 2(1.50 rad/s)(2.50 s) (0.300 rad/s )(2.50 s) 4.69 radz zt t? ? ? ?? = + = + = . EVALUATE: 00 1.50 rad/s 2.25 rad/s (2.50 s) 4.69 rad2 2 z z t ? ?? ? + +? ? ? ?? = = =? ? ? ?? ? ? ? , the same as calculated with another equation in part (b). 9.10. IDENTIFY: Apply the constant angular acceleration equations to the motion of the fan. (a) SET UP: 0 (500 rev/min)(1 min/60 s) 8.333 rev/s,z? = = (200 rev/min)(1 min/60 s) 3.333 rev/s,z? = = 4.00 s,t = ?z? = 0z z zt? ? ?= + EXECUTE: 20 3.333 rev/s 8.333 rev/s 1.25 rev/s 4.00 s z z z t ? ?? ? ?= = = ? 0 ?? ?? = 2 2 21 1 0 0 2 2(8.333 rev/s)(4.00 s) ( 1.25 rev/s )(4.00 s) 23.3 revz zt t? ? ? ?? = + = + ? = (b) SET UP: 0z? = (comes to rest); 0 3.333 rev/s;z? = 21.25 rev/s ;z? = ? ?t = 0z z zt? ? ?= + EXECUTE: 0 20 3.333 rev/s 2.67 s1.25 rev/s z z z t ? ? ? ? ?= = =? EVALUATE: The angular acceleration is negative because the angular velocity is decreasing. The average angular velocity during the 4.00 s time interval is 350 rev/min and 0 av-zt? ? ?? = gives 0 23.3 rev,? ?? = which checks. 9.11. IDENTIFY: Apply the constant angular acceleration equations to the motion. The target variables are t and 0.? ?? SET UP: (a) 21.50 rad/s ;z? = 0 0z? = (starts from rest); 36.0 rad/s;z? = ?t = 0z z zt? ? ?= + EXECUTE: 0 236.0 rad/s 0 24.0 s1.50 rad/s z z z t ? ? ? ? ?= = = (b) 0 ?? ?? = 2 2 21 1 0 0 2 20 (1.50 rad/s )(2.40 s) 432 radz zt t? ? ? ?? = + = + = 0 432 rad(1 rev/2 rad) 68.8 rev? ? ?? = = EVALUATE: We could use 10 02 ( )z z t? ? ? ?? = + to calculate 10 2 (0 36.0 rad/s)(24.0 s) 432 rad,? ?? = + = which checks. 9.12. IDENTIFY: In part (b) apply the equation derived in part (a). SET UP: Let the direction the propeller is rotating be positive. EXECUTE: (a) Solving Eq. (9.7) for t gives 0z z z t ? ? ? ?= . Rewriting Eq. (9.11) as 10 0 2( )z zt t? ? ? ?? = + and substituting for t gives 2 20 0 0 0 0 0 0 1 1 1 ( ) ( ) ( ), 2 2 2 z z z z z z z z z z z z z z ? ? ? ?? ? ? ? ? ? ? ? ?? ? ? ? ?? +? ? ? ?? = + ? = ? = ?? ?? ? ? ?? ? ? ?? ? which when rearranged gives Eq. (9.12). (b) ( ) ( ) ( )( )2 22 2 21 102 2 0 1 1 16.0 rad s 12.0 rad s 8.00 rad/s 7.00 radz z z ? ? ?? ? ? ? ? ?= ? = ? =? ? ? ?? ? ?? ? 9-4 Chapter 9 EVALUATE: We could also use 00 2 z z t ? ?? ? +? ?? = ? ?? ? to calculate 0.500 st = . Then 0z z zt? ? ?= + gives 28.00 rad/sz? = , which agrees with our results in part (b). 9.13. IDENTIFY: Use a constant angular acceleration equation and solve for 0 .z? SET UP: Let the direction of rotation of the flywheel be positive. EXECUTE: 210 0 2z zt? ? ? ?? = + gives 20 1 10 2 260.0 rad (2.25 rad/s )(4.00 s) 10.5 rad/s4.00 sz zt ? ?? ??= ? = ? = . EVALUATE: At the end of the 4.00 s interval, 0 19.5 rad/sz z zt? ? ?= + = . 0 0 10.5 rad/s 19.5 rad/s (4.00 s) 60.0 rad 2 2 z z t ? ?? ? + +? ? ? ?? = = =? ? ? ?? ? ? ? , which checks. 9.14. IDENTIFY: Apply the constant angular acceleration equations. SET UP: Let the direction of the rotation of the blade be positive. 0 0z? = . EXECUTE: 0z z z? ? ?= + gives 20 140 rad/s 0 23.3 rad/s6.00 s z z z t ? ?? ? ?= = = . 0 0 0 140 rad/s ( ) (6.00 s) 420 rad 2 2 z z t ? ?? ? + +? ? ? ?? = = =? ? ? ?? ? ? ? EVALUATE: We could also use 210 0 2z zt t? ? ? ?? = + . This equation gives 2 21 0 2 (23.3 rad/s )(6.00 s) 419 rad? ?? = = , in agreement with the result obtained above. 9.15. IDENTIFY: Apply constant angular acceleration equations. SET UP: Let the direction the flywheel is rotating be positive. 0 0200 rev, 500 rev min 8.333 rev s, 30.0 sz t? ? ?? = = = = . EXECUTE: (a) 00 gives 5.00 rev s 300 rpm2 z z zt ? ?? ? ?+? ?? = = =? ?? ? (b) Use the information in part (a) to find :z? 0z z zt? ? ?= + gives 20.1111 rev sz? = ? . Then 0,z? = 2 00.1111 rev s , 8.333 rev sz z? ?= ? = in 0z z zt? ? ?= + gives 75.0 st = and 00 2 z z t ? ?? ? +? ?? = ? ?? ? gives 0 312 rev? ?? = . EVALUATE: The mass and diameter of the flywheel are not used in the calculation. 9.16. IDENTIFY: Use the constant angular acceleration equations, applied to the first revolution and to the first two revolutions. SET UP: Let the direction the disk is rotating be positive. 1 rev 2 rad?= . Let t be the time for the first revolution. The time for the first two revolutions is 0.750 st + . EXECUTE: (a) 210 0 2z zt t? ? ? ?? = + applied to the first and to the first two revolutions gives 2122 rad zt? ?= and 21 24 rad ( 0.750 s)z t? ?= + . Eliminating z? between these equations gives 222 rad4 rad ( 0.750 s)tt ?? = + . 2 22 ( 0.750 s)t t= + . 2 ( 0.750 s)t t= ± + . The positive root is 0.750 s 1.81 s 2 1 t = =? . (b) 2122 rad zt? ?= and 1.81 st = gives 23.84 rad/sz? = EVALUATE: At the start of the second revolution, 20 (3.84 rad/s )(1.81 s) 6.95 rad/sz? = = . The distance the disk rotates in the next 0.750 s is 2 2 21 10 0 2 2(6.95 rad/s)(0.750 s) (3.84 rad/s )(0.750 s) 6.29 radz zt t? ? ? ?? = + = + = , which is two revolutions. 9.17. IDENTIFY: Apply Eq.(9.12) to relate z? to 0? ?? . SET UP: Establish a proportionality. EXECUTE: From Eq.(9.12), with 0 0, z? = the number of revolutions is proportional to the square of the initial angular velocity, so tripling the initial angular velocity increases the number of revolutions by 9, to 9.00 rev. EVALUATE: We don't have enough information to calculate z? ; all we need to know is that it is constant. Rotation of Rigid Bodies 9-5 9.18. IDENTIFY: In each case we apply constant acceleration equations to determine ( )t? and ( )z t? . SET UP: Let 0 0? = . The following table gives the revolutions and the angle ? (in degrees) through which the wheel has rotated for each instant in time (in seconds) and each of the three situations: t (a) rev ? (b) rev ? (c) rev ? 0.05 0.50 180 0.03 11.3 0.44 158 0.10 1.00 360 0.13 45 0.75 270 0.15 1.50 540 0.28 101 0.94 338 0.20 2.00 720 0.50 180 1.00 360 EXECUTE: The ? and z? graphs for each case are given in Figures 9.18 a?c. EVALUATE: The slope of the ( )t? graph is ( )z t? and the slope of the ( )z t? graph is ( )z t? . Figure 9.18 9-6 Chapter 9 9.19. IDENTIFY: Apply the constant angular acceleration equations separately to the time intervals 0 to 2.00 s and 2.00 s until the wheel stops. (a) SET UP: Consider the motion from 0t = to 2.00 s:t = 0 ?;? ?? = 0 24.0 rad/s;z? = 230.0 rad/s ;z? = 2.00 st = EXECUTE: 2 2 21 10 0 2 2(24.0 rad/s)(2.00 s) (30.0 rad/s )(2.00 s)z zt t? ? ? ?? = + = + 0 48.0 rad 60.0 rad 108 rad? ?? = + = Total angular displacement from 0t = until stops: 108 rad 432 rad 540 rad+ = Note: At 2.00 s,t = 20 24.0 rad/s (30.0 rad/s )(2.00 s) 84.0 rad/s;z z zt? ? ?= + = + = angular speed when breaker trips. (b) SET UP: Consider the motion from when the circuit breaker trips until the wheel stops. For this calculation let 0t = when the breaker trips. ?;t = 0 432 rad;? ?? = 0;z? = 0 84.0 rad/sz? = (from part (a)) 0 0 2 z z t ? ?? ? +? ?? = ? ?? ? EXECUTE: 0 0 2( ) 2(432 rad) 10.3 s 84.0 rad/s 0z z t ? ? ? ? ?= = =+ + The wheel stops 10.3 s after the breaker trips so 2.00 s 10.3 s 12.3 s+ = from the beginning. (c) SET UP: ?;z? = consider the same motion as in part (b): 0z z zt? ? ?= + EXECUTE: 20 0 84.0 rad/s 8.16 rad/s 10.3 s z z z t ? ?? ? ?= = = ? EVALUATE: The angular acceleration is positive while the wheel is speeding up and negative while it is slowing down. We could also use 2 20 02 ( )z z z? ? ? ? ?= + ? to calculate 2 2 2 20 0 0 (84.0 rad/s) 8.16 rad/s 2( ) 2(432 rad) z z z ? ?? ? ? ? ?= = = ?? for the acceleration after the breaker trips. 9.20. IDENTIFY: The linear distance the elevator travels, its speed and the magnitude of its acceleration are equal to the tangential displacement, speed and acceleration of a point on the rim of the disk. s r?= , v r?= and a r?= . In these equations the angular quantities must be in radians. SET UP: 1 rev 2 rad?= . 1 rpm 0.1047 rad/s= . rad 180? = ° . For the disk, 1.25 mr = . EXECUTE: (a) 0.250 m/sv = so 0.250 m/s 0.200 rad/s 1.91 rpm 1.25 m v r ? = = = = . (b) 218 1.225 m/sa g= = . 2 21.225 m/s 0.980 rad/s 1.25 m a r ? = = = . (c) 3.25 ms = . 3.25 m 2.60 rad 149 1.25 m s r ? = = = = ° . EVALUATE: When we use s r?= , v r?= and tana r?= to solve for ? , ? and ? , the results are in rad, rad/s and 2rad/s . 9.21. IDENTIFY: When the angular speed is constant, / t? ?= . tanv r?= , tana r?= and 2rada r?= . In these equations radians must be used for the angular quantities. SET UP: The radius of the earth is 6E 6.38 10 mR = × and the earth rotates once in 1 day 86,400 s= . The orbit radius of the earth is 111.50 10 m× and the earth completes one orbit in 71 y 3.156 10 s= × . When ? is constant, / t? ?= . EXECUTE: (a) 1 rev 2 rad? ?= = in 73.156 10 st = × . 772 rad 1.99 10 rad/s3.156 10 s ?? ?= = ×× . (b) 1 rev 2 rad? ?= = in 86,400 st = . 52 rad 7.27 10 rad/s 86,400 s ?? ?= = × (c) 11 7 4(1.50 10 m)(1.99 10 rad/s) 2.98 10 m/sv r? ?= = × × = × . (d) 6 5(6.38 10 m)(7.27 10 rad/s) 464 m/sv r? ?= = × × = . (e) 2 6 5 2 2rad (6.38 10 m)(7.27 10 rad/s) 0.0337 m/sa r? ?= = × × = . tan 0a r?= = . 0? = since the angular velocity is constant. EVALUATE: The tangential speeds associated with these motions are large even though the angular speeds are very small, because the radius for the circular path in each case is quite large. Rotation of Rigid Bodies 9-7 9.22. IDENTIFY: Linear and angular velocities are related by v r?= . Use 0z z zt? ? ?= + to calculate z? . SET UP: /v r? = gives? in rad/s. EXECUTE: (a) 31.25 m/s 50.0 rad/s,25.0 10 m? =× 3 1.25 m/s 21.6 rad/s. 58.0 10 m? =× (b) (1.25 m/s) (74.0 min) (60 s/min) = 5.55 km. (c) 3 221.55 rad/s 50.0 rad/s 6.41 10 rad/s . (74.0 min) (60 s/min)z ? ??= = ? × EVALUATE: The width of the tracks is very small, so the total track length on the disc is huge. 9.23. IDENTIFY: Use constant acceleration equations to calculate the angular velocity at the end of two revolutions. v r?= . SET UP: 2 rev 4 rad.?= 0.200 m.r = EXECUTE: (a) 2 20 02 ( ).z z z? ? ? ? ?= + ? 202 ( ) 2(3.00 rad/s )(4 rad) 8.68 rad/s.z z? ? ? ? ?= ? = = 2 2 2 rad (0.200 m)(8.68 rad/s) 15.1 m/s .a r?= = = (b) (0.200 m)(8.68 rad/s) 1.74 m/s.v r?= = = 2 2 2 rad (1.74 m/s) 15.1 m/s . 0.200 m v a r = = = EVALUATE: 2r? and 2 /v r are completely equivalent expressions for rada . 9.24. IDENTIFY: 2rada r?= , with ? in rad/s. Solve for ? . SET UP: 1 rpm (2 / 60) rad/s?= EXECUTE: 2 4 5rad (400,000)(9.80 m/s ) 1.25 10 rad/s 1.20 10 rpm 0.0250 m a r ? = = = × = × EVALUATE: In 2rada r?= , ? must be in rad/s. 9.25. IDENTIFY and SET UP: Use constant acceleration equations to find ? and ? after each displacement. The use Eqs.(9.14) and (9.15) to find the components of the linear acceleration. EXECUTE: (a) at the start 0t = flywheel starts from rest so 0 0z z? ?= = 2 2 tan (0.300 m)(0.600 rad/s ) 0.180 m/sa r?= = = 2 rad 0a r?= = 2 2 2 rad tan 0.180 m/sa a a= + = (b) 0 60? ?? = ° 2 tan 0.180 m/sa r?= = Calculate :? 0 60 ( rad/180 ) 1.047 rad;? ? ?? = ° ° = 0 0;z? = 20.600 rad/s ;z? = ?z? = 2 2 0 02 ( )z z z? ? ? ? ?= + ? 2 02 ( ) 2(0.600 rad/s )(1.047 rad) 1.121 rad/sz z? ? ? ?= ? = = and .z? ?= Then 2 2 2rad (0.300 m)(1.121 rad/s) 0.377 m/s .a r?= = = 2 2 2 2 2 2 2 rad tan (0.377 m/s ) (0.180 m/s ) 0.418 m/sa a a= + = + = (c) 0 120? ?? = ° 2 tan 0.180 m/sa r?= = Calculate :? 0 120 ( rad/180 ) 2.094 rad;? ? ?? = ° ° = 0 0;z? = 20.600 rad/s ;z? = ?z? = 2 2 0 02 ( )z z z? ? ? ? ?= + ? 2 02 ( ) 2(0.600 rad/s )(2.094 rad) 1.585 rad/sz z? ? ? ?= ? = = and .z? ?= Then 2 2 2rad (0.300 m)(1.585 rad/s) 0.754 m/s .a r?= = = 2 2 2 2 2 2 2 rad tan (0.754 m/s ) (0.180 m/s ) 0.775 m/sa a a= + = + = EVALUATE: ? is constant so tan? is constant. ? increases so rada increases. 9-8 Chapter 9 9.26. IDENTIFY: Apply constant angular acceleration equations. v r?= . A point on the rim has both tangential and radial components of acceleration. SET UP: tana r?= and 2rada r?= . EXECUTE: (a) 20 = 0.250 rev/s (0.900 rev/s )(0.200 s) 0.430 rev/sz z zt? ? ?= + + = (Note that since 0 z? and z? are given in terms of revolutions, it?s not necessary to convert to radians). (b) av- (0.340 rev s)(0.2 s) 0.068 revz t? ? = = . (c) Here, the conversion to radians must be made to use Eq. (9.13), and ( )( )0.750 m 0.430 rev/s 2 rad rev 1.01 m s. 2 v r? ?? ?= = =? ?? ? (d) Combining equations (9.14) and (9.15), 2 2 2 2 2 rad tan ( ) ( )a a a r r? ?= + = + . 2 22 2((0.430 rev/s)(2 rad/rev)) (0.375 m) (0.900 rev/s )(2 rad/rev)(0.375 m)a ? ?? ? ? ?= +? ? ? ? . 23.46 m sa = . EVALUATE: If the angular acceleration is constant, tana is constant but rada increases as ? increases. 9.27. IDENTIFY: Use Eq.(9.15) and solve for r. SET UP: 2rada r?= so 2rad / ,r a ?= where ? must be in rad/s EXECUTE: 2 2rad 3000 3000(9.80 m/s ) 29,400 m/sa g= = = 1 min 2 rad (5000 rev/min) 523.6 rad/s 60 s 1 rev ?? ? ?? ?= =? ?? ?? ?? ? Then 2 rad 2 2 29,400 m/s 0.107 m. (523.6 rad/s) a r ?= = = EVALUATE: The diameter is then 0.214 m, which is larger than 0.127 m, so the claim is not realistic. 9.28. IDENTIFY: In part (b) apply the result derived in part (a). SET UP: 2rada r?= and v r?= ; combine to eliminate r. EXECUTE: (a) 2 2 rad = .va r v? ? ?? ? ?= =? ?? ? (b) From the result of part (a), 2 rad 0.500 m s 0.250 rad s. 2.00 m s a v ? = = = EVALUATE: 2rada r?= and v r?= both require that ? be in rad/s, so in rada v?= , ? is in rad/s. 9.29. IDENTIFY: v r?= and 2 2rad /a r v r?= = . SET UP: 2 rad 1 rev? = , so rad/s 30 rev/min? = . EXECUTE: (a) ( ) 312.7 10 m rad/s(1250 rev min ) 0.831 m s.30 rev/min 2r ?? ?? ?×= =? ?? ? (b) 2 2 2 3 (0.831 m s) 109 m s . (12.7 10 m) 2 v r ? = =× EVALUATE: In v r?= , ? must be in rad/s. 9.30. IDENTIFY: tana r?= , v r?= and 2rad /a v r= . 0 av-zt? ? ?? = . SET UP: When z? is constant, 0av- 2 z z z ? ?? += . Let the direction the wheel is rotating be positive. EXECUTE: (a) 2 2tan 10.0 m s 50.0 rad s 0.200 m a r ? ?= = = ? (b) At 3.00 st = , 50.0 m sv = and 50.0 m s 250 rad s 0.200 m v r ? = = = and at 0,t = 250.0 m s ( 10.0 m s )(0 3.00 s) 80.0 m sv = + ? ? = , 400 rad s.? = (c) av- (325 rad s)(3.00 s) 975 rad 155 revzt? = = = . Rotation of Rigid Bodies 9-9 (d) 2rad (9.80 m/s )(0.200 m) 1.40 m/s.v a r= = = This speed will be reached at time 50.0 m/s 1.40 m/s 4.86 s10.0 m/s ? = after 3.00 st = , or at 7.86 st = . (There are many equivalent ways to do this calculation.) EVALUATE: At 0t = , 2 4 2rad 3.20 10 m/sa r?= = × . At 3.00 st = , 4 2rad 1.25 10 m/sa = × . For rada g= the wheel must be rotating more slowly than at 3.00 s so it occurs some time after 3.00 s. 9.31. IDENTIFY and SET UP: Use Eq.(9.15) to relate ? to rada and m=?F a! ! to relate rada to rad.F Use Eq.(9.13) to relate ? and v, where v is the tangential speed. EXECUTE: (a) 2rada r?= and 2rad radF ma mr?= = 2 2 rad,2 2 rad,1 1 640 rev/min 2.29 423 rev/min F F ? ? ? ? ? ?= = =? ? ? ?? ?? ? (b) v r?= 2 2 1 1 640 rev/min 1.51 423 rev/min v v ? ?= = = (c) v r?= 1 min 2 rad (640 rev/min) 67.0 rad/s 60 s 1 rev ?? ? ?? ?= =? ?? ?? ?? ? Then (0.235 m)(67.0 rad/s) 15.7 m/s.v r?= = = 2 2 2 rad (0.235 m)(67.0 rad/s) 1060 m/sa r?= = = 2 rad 2 1060 m/s 108; 9.80 m/s a g = = 108a g= EVALUATE: In parts (a) and (b), since a ratio is used the units cancel and there is no need to convert ? to rad/s. In part (c), v and rada are calculated from ,? and ? must be in rad/s. 9.32. IDENTIFY: v r?= and tana r?= . SET UP: The linear acceleration of the bucket equals tana for a point on the rim of the axle. EXECUTE: (a) v R?= . 7.5 rev 1 min 2 rad2.00 cm s min 60 s 1 rev R ?? ?? ?? ?= ? ?? ?? ?? ?? ?? ? gives 2.55 cmR = . 2 5.09 cmD R= = . (b) tana R?= . 2 2tan 0.400 m s 15.7 rad s 0.0255 m a R ? = = = . EVALUATE: In v R?= and tana R?= , ? and ? must be in radians. 9.33. IDENTIFY: Apply v r?= . SET UP: Points on the chain all move at the same speed, so r r f fr r? ?= . EXECUTE: The angular velocity of the rear wheel is rr 5.00 m s 15.15 rad s.0.330 m v r ? = = = The angular velocity of the front wheel is f 0.600 rev s 3.77 rad s? = = . ( )r f f r 2.99 cmr r ? ?= = . EVALUATE: The rear sprocket and wheel have the same angular velocity and the front sprocket and wheel have the same angular velocity. r? is the same for both, so the rear sprocket has a smaller radius since it has a larger angular velocity. The speed of a point on the chain is 2r r (2.99 10 m)(15.15 rad/s) 0.453 m/sv r? ?= = × = . The linear speed of the bicycle is 5.50 m/s. 9.34. IDENTIFY and SET UP: Use Eq.(9.16). Treat the spheres as point masses and ignore I of the light rods. EXECUTE: The object is shown in Figure 9.34a. (a) 2 2(0.200 m) (0.200 m) 0.2828 mr = + = 2 24(0.200 kg)(0.2828 m)i iI m r= =? 20.0640 kg mI = ? Figure 9.34a 9-10 Chapter 9 (b) The object is shown in Figure 9.34b. 0.200 mr = 2 24(0.200 kg)(0.200 m)i iI m r= =? 20.0320 kg mI = ? Figure 9.34b (c) The object is shown in Figure 9.34c. 0.2828 mr = 2 22(0.200 kg)(0.2828 m)i iI m r= =? 20.0320 kg mI = ? Figure 9.34c EVALUATE: In general I depends on the axis and our answer for part (a) is larger than for parts (b) and (c). It just happens that I is the same in parts (b) and (c). 9.35. IDENTIFY: Use Table 9.2. The correct expression to use in each case depends on the shape of the object and the location of the axis. SET UP: In each case express the mass in kg and the length in m, so the moment of inertia will be in 2kg m? . EXECUTE: (a) (i) 2 2 21 13 3 (2.50 kg)(0.750 m) 0.469 kg mI ML= = = ? . (ii) 2 2 21 112 4 (0.469 kg m ) 0.117 kg mI ML= = ? = ? . (iii) For a very thin rod, all of the mass is at the axis and 0I = . (b) (i) 2 2 22 25 5 (3.00 kg)(0.190 m) 0.0433 kg mI MR= = = ? . (ii) 2 2 2523 3 (0.0433 kg m ) 0.0722 kg mI MR= = ? = ? . (c) (i) 2 2 2(8.00 kg)(0.0600 m) 0.0288 kg mI MR= = = ? . (ii) 2 2 21 12 2 (8.00 kg)(0.0600 m) 0.0144 kg mI MR= = = ? . EVALUATE: I depends on how the mass of the object is distributed relative to the axis. 9.36. IDENTIFY: Treat each block as a point mass, so for each block 2I mr= , where r is the distance of the block from the axis. The total I for the object is the sum of the I for each of its pieces. SET UP: In part (a) two blocks are a distance / 2L from the axis and the third block is on the axis. In part (b) two blocks are a distance / 4L from the axis and one is a distance 3 / 4L from the axis. EXECUTE: (a) 2 2122 ( / 2)I m L mL= = . (b) 2 2 2 2 1 11 2 ( / 4) (3 / 4) (2 9) 16 16 I m L m L mL mL= + = + = . EVALUATE: For the same object I is in general different for different axes. 9.37. IDENTIFY: I for the object is the sum of the values of I for each part. SET UP: For the bar, for an axis perpendicular to the bar, use the appropriate expression from Table 9.2. For a point mass, 2I mr= , where r is the distance of the mass from the axis. EXECUTE: (a) 2 2 bar balls bar balls 1 2 12 2 L I I I M L m ? ?= + = + ? ?? ? . ( )( ) ( )( )2 2 21 4.00 kg 2.00 m 2 0.500 kg 1.00 m 2.33 kg m 12 I = + = ? (b) ( )( ) ( ) ( )2 22 2 2bar ball1 1 4.00 kg 2.00 m 0.500 kg 2.00 m 7.33 kg m3 3I m L m L= + = + = ? (c) 0I = because all masses are on the axis. (d) All the mass is a distance 0.500 md = from the axis and 2 2 2 2 2 bar ball Total2 (5.00 kg)(0.500 m) 1.25 kg mI m d m d M d= + = = = ? . EVALUATE: I for an object depends on the location and direction of the axis. Rotation of Rigid Bodies 9-11 9.38. IDENTIFY and SET UP: According to Eq.(9.16), I for the entire object equals the sum of I for each piece, the rod plus the end caps. The object is shown in Figure 9.38. EXECUTE: rod cap2I I I= + ( )2 2 21 1 112 12 22( )( / 2)I ML m L M m L= + = + Figure 9.38 EVALUATE: Table 9.2 was used for rodI and 2I mr= for the end caps, since they are treated as point particles. 9.39. IDENTIFY and SET UP: 2i iI m r=? implies rim spokesI I I= + EXECUTE: 2 2 2rim (1.40 kg)(0.300 m) 0.126 kg mI MR= = = ? Each spoke can be treated as a slender rod with the axis through one end, so ( )2 2 281spokes 3 38 (0.280 kg)(0.300 m) 0.0672 kg mI ML= = = ? 2 2 2 rim spokes 0.126 kg m 0.0672 kg m 0.193 kg mI I I= + = ? + ? = ? EVALUATE: Our result is smaller than 2 2 2tot (3.64 kg)(0.300 m) 0.328 kg m ,m R = = ? since the mass of each spoke is distributed between 0r = and .r R= 9.40. IDENTIFY: Compare this object to a uniform disk of radius R and mass 2M . SET UP: With an axis perpendicular to the round face of the object at its center, I for a uniform disk is the same as for a solid cylinder. EXECUTE: (a) The total I for a disk of mass 2M and radius R, 2 212 (2 )I M R MR= = . Each half of the disk has the same I , so for the half-disk, 212I MR= . (b) The same mass M is distributed the same way as a function of distance from the axis. (c) The same method as in part (a) says that I for a quarter-disk of radius R and mass M is half that of a half-disk of radius R and mass 2M , so 2 21 1 12 2 2( [2 ] )I M R MR= = . EVALUATE: I depends on how the mass of the object is distributed relative to the axis, and this is the same for any segment of a disk. 9.41. IDENTIFY: I for the compound disk is the sum of I of the solid disk and of the ring. SET UP: For the solid disk, 21 d d2I m r= . For the ring, 2 21r r 1 22 ( )I m r r= + , where 1 250.0 cm, 70.0 cmr r= = . The mass of the disk and ring is their area times their area density. EXECUTE: d rI I I= + . Disk: 2 2d d(3.00 g cm ) 23.56 kgm r?= = . 2 2d d d1 2.945 kg m2I m r= = ? . Ring: 2 2 2r 2 1(2.00 g cm ) ( ) 15.08 kgm r r?= ? = . 2 2 2r r 1 21 ( ) 5.580 kg m2I m r r= + = ? . 2 d r 8.52 kg mI I I= + = ? . EVALUATE: Even though r dm m< , r dI I> since the mass of the ring is farther from the axis. 9.42. IDENTIFY: 212K I ?= . Use Table 9.2b to calculate I . SET UP: 2112I ML= . 1 rpm 0.1047 rad/s= EXECUTE: (a) 2 2112 (117 kg)(2.08 m) 42.2 kg mI = = ? . 0.1047 rad/s(2400 rev/min) 251 rad/s1 rev/min? ? ?= =? ?? ? . 2 2 2 61 1 2 2 (42.2 kg m )(251 rad/s) 1.33 10 JK I ?= = ? = × . (b) 2 211 1 1 112K M L ?= , 2 212 2 2 212K M L ?= . 1 2L L= and 1 2K K= , so 2 21 1 2 2M M? ?= . 1 1 2 1 2 1 (2400 rpm) 2770 rpm 0.750 M M M M ? ?= = = EVALUATE: The rotational kinetic energy is proportional to the square of the angular speed and directly proportional to the mass of the object. 9.43. IDENTIFY: 212K I ?= . Use Table 9.2 to calculate I . SET UP: 225I MR= . For the moon, 227.35 10 kgM = × and 61.74 10 mR = × . The moon moves through 1 rev 2 rad?= in 27.3 d. 41 d 8.64 10 s= × . 9-12 Chapter 9 EXECUTE: (a) 22 6 2 34 225 (7.35 10 kg)(1.74 10 m) 8.90 10 kg mI = × × = × ? . 6 4 2 rad 2.66 10 rad/s (27.3 d)(8.64 10 s/d) ?? ?= = ×× . 2 34 2 6 2 231 1 2 2 (8.90 10 kg m )(2.66 10 rad/s) 3.15 10 JK I ? ?= = × ? × = × . (b) 23 20 3.15 10 J 158 years 5(4.0 10 J) × =× . Considering the expense involved in tapping the moon?s rotational energy, this does not seem like a worthwhile scheme for only 158 years worth of energy. EVALUATE: The moon has a very large amount of kinetic energy due to its motion. The earth has even more, but changing the rotation rate of the earth would change the length of a day. 9.44. IDENTIFY: 212K I ?= . Use Table 9.2 to relate I to the mass M of the disk. SET UP: 45.0 rpm 4.71 rad/s= . For a uniform solid disk, 212I MR= . EXECUTE: (a) 22 22 2(0.250 J) 0.0225 kg m(4.71 rad/s) K I ?= = = ? . (b) 212I MR= and 2 2 2 2 2(0.0225 kg m ) 0.500 kg (0.300 m) I M R ?= = = . EVALUATE: No matter what the shape is, the rotational kinetic energy is proportional to the mass of the object. 9.45. IDENTIFY: 212K I ?= , with ? in rad/s. Solve for I . SET UP: 1 rev/min (2 / 60) rad/s?= . 500 JK? = ? EXECUTE: i 650 rev/min 68.1 rad/s? = = . f 520 rev/min 54.5 rad/s? = = . 2 21f i f i2 ( )K K K I ? ?? = ? = ? and 2 2 2 2 2 f i 2( ) 2( 500 J) 0.600 kg m (54.5 rad/s) (68.1 rad/s) K I ? ? ? ?= = = ?? ? . EVALUATE: In 212K I ?= , ? must be in rad/s. 9.46. IDENTIFY: The work done on the cylinder equals its gain in kinetic energy. SET UP: The work done on the cylinder is PL , where L is the length of the rope. 1 0K = . 212 2K I ?= . 2 21 1 2 2 w I mr r g ? ?= = ? ?? ? . EXECUTE: 2 2 2 2 1 1 (40.0 N)(6.00 m s) , or 14.7 N. 2 2 2(9.80 m s )(5.00 m) w w v PL v P g g L = = = = EVALUATE: The linear speed v of the end of the rope equals the tangential speed of a point on the rim of the cylinder. When K is expressed in terms of v, the radius r of the cylinder doesn't appear. 9.47. IDENTIFY and SET UP: Combine Eqs.(9.17) and (9.15) to solve for K . Use Table 9.2 to get I . EXECUTE: 212K I ?= 2 rad ,a R?= so 2rad / (3500 m/s ) /1.20 m 54.0 rad/sa R? = = = For a disk, 2 2 21 12 2 (70.0 kg)(1.20 m) 50.4 kg mI MR= = = ? Thus 2 2 2 41 12 2 (50.4 kg m )(54.0 rad/s) 7.35 10 JK I ?= = ? = × EVALUATE: The limit on rada limits ? which in turn limits K . 9.48. IDENTIFY: Repeat the calculation in Example 9.9, but with a different expression for I . SET UP: For the solid cylinder in Example 9.9, 212I MR= . For the thin-walled, hollow cylinder, 2I MR= . EXECUTE: (a) With 2,I MR= the expression for v is 2 . 1 g h v M m = + (b) This expression is smaller than that for the solid cylinder; more of the cylinder?s mass is concentrated at its edge, so for a given speed, the kinetic energy of the cylinder is larger. A larger fraction of the potential energy is converted to the kinetic energy of the cylinder, and so less is available for the falling mass. EVALUATE: When M is much larger than m, v is very small. When M is much less than m, v becomes 2v gh= , the same as for a mass that falls freely from a height h. Rotation of Rigid Bodies 9-13 9.49. IDENTIFY: Apply conservation of energy to the system of stone plus pulley. v r?= relates the motion of the stone to the rotation of the pulley. SET UP: For a uniform solid disk, 212I MR= . Let point 1 be when the stone is at its initial position and point 2 be when it has descended the desired distance. Let y+ be upward and take 0y = at the initial position of the stone, so 1 0y = and 2y h= ? , where h is the distance the stone descends. EXECUTE: (a) 21p p2K I ?= . 2 2 21 1p p2 2 (2.50 kg)(0.200 m) 0.0500 kg mI M R= = = ? . p 2 p 2 2(4.50 J) 13.4 rad/s 0.0500 kg m K I ? = = =? . The stone has speed (0.200 m)(13.4 rad/s) 2.68 m/sv R?= = = . The stone has kinetic energy 2 21 1s 2 2 (1.50 kg)(2.68 m/s) 5.39 JK mv= = = . 1 1 2 2K U K U+ = + gives 2 20 K U= + . 0 4.50 J 5.39 J ( )mg h= + + ? . 29.89 J 0.673 m(1.50 kg)(9.80 m/s )h = = . (b) tot p s 9.89 JK K K= + = . p tot 4.50 J 45.5% 9.89 J K K = = . EVALUATE: The gravitational potential energy of the pulley doesn?t change as it rotates. The tension in the wire does positive work on the pulley and negative work of the same magnitude on the stone, so no net work on the system. 9.50. IDENTIFY: 21p 2K I ?= for the pulley and 21b 2K mv= for the bucket. The speed of the bucket and the rotational speed of the pulley are related by v R?= . SET UP: 1p b2K K= EXECUTE: 2 2 2 21 1 1 12 2 2 4( )I mv mR? ?= = . 212I mR= . EVALUATE: The result is independent of the rotational speed of the pulley and the linear speed of the mass. 9.51. IDENTIFY: The general expression for I is Eq.(9.16). 212K I ?= . SET UP: R will be multiplied by f. EXECUTE: (a) In the expression of Eq. (9.16), each term will have the mass multiplied by 3f and the distance multiplied by ,f and so the moment of inertia is multiplied by 3 2 5( ) . f f f= (b) 5 8 (2.5 J)(48) 6.37 10 J.= × EVALUATE: Mass and volume are proportional to each other so both scale by the same factor. 9.52. IDENTIFY: The work the person does is the negative of the work done by gravity. grav grav,1 grav,2W U U= ? . grav cmU Mgy= . SET UP: The center of mass of the ladder is at its center, 1.00 m from each end. cm,1 (1.00 m)sin53.0 0.799 my = =° . cm,2 1.00 my = . EXECUTE: 2grav (9.00 kg)(9.80 m/s )(0.799 m 1.00 m) 17.7 JW = ? = ? . The work done by the person is 17.7 J. EVALUATE: The gravity force is downward and the center of mass of the ladder moves upward, so gravity does negative work. The person pushes upward and does positive work. 9.53. IDENTIFY: cmU Mgy= . 2 1U U U? = ? . SET UP: Half the rope has mass 1.50 kg and length 12.0 m. Let 0y = at the top of the cliff and take y+ to be upward. The center of mass of the hanging section of rope is at its center and cm,2 6.00 my = ? . EXECUTE: 22 1 cm,2 cm,1( ) (1.50 kg)(9.80 m/s )( 6.00 m 0) 88.2 JU U U mg y y? = ? = ? = ? ? = ? . EVALUATE: The potential energy of the rope decreases when part of the rope moves downward. 9.54. IDENTIFY: Apply Eq.(9.19), the parallel-axis theorem. SET UP: The center of mass of the hoop is at its geometrical center. EXECUTE: In Eq. (9.19), 2 2 2cm and , so 2 .PI MR d R I MR= = = EVALUATE: I is larger for an axis at the edge than for an axis at the center. Some mass is closer than distance R from the axis but some is also farther away. Since I for each piece of the hoop is proportional to the square of the distance from the axis, the increase in distance has a larger effect. 9.55. IDENTIFY: Use Eq.(9.19) to relate I for the wood sphere about the desired axis to I for an axis along a diameter. SET UP: For a thin-walled hollow sphere, axis along a diameter, 223 .I MR= For a solid sphere with mass M and radius R, 22cm 5 ,I MR= for an axis along a diameter. 9-14 Chapter 9 EXECUTE: Find d such that 2cmPI I Md= + with 223 :PI MR= 2 2 22 2 3 5MR MR Md= + The factors of M divide out and the equation becomes ( ) 2 22 23 5 R d? = (10 6) /15 2 / 15 0.516 .d R R R= ? = = The axis is parallel to a diameter and is 0.516R from the center. EVALUATE: cm cm(lead) (wood)I I> even though M and R are the same since for a hollow sphere all the mass is a distance R from the axis. Eq.(9.19) says cm ,PI I> so there must be a d where cm(wood) (lead).PI I= 9.56. IDENTIFY: Using the parallel-axis theorem to find the moment of inertia of a thin rod about an axis through its end and perpendicular to the rod. SET UP: The center of mass of the rod is at its center, and 21cm 12I ML= . EXECUTE: 2 2 2 2 cm .12 2 3p M L M I I Md L M L? ?= + = + =? ?? ? EVALUATE: I is larger when the axis is not at the center of mass. 9.57. IDENTIFY and SET UP: Use Eq.(9.19). The cm of the sheet is at its geometrical center. The object is sketched in Figure 9.57. EXECUTE: 2cm .PI I Md= + From part (c) of Table 9.2, 2 21 cm 12 ( ).I M a b= + The distance d of P from the cm is 2 2( / 2) ( / 2) .d a b= + Figure 9.57 Thus ( ) ( )2 2 2 2 2 2 21 1 1 1 1cm 12 4 4 12 4( ) ( )PI I Md M a b M a b M a b= + = + + + = + + = 2 21 3 ( )M a b+ EVALUATE: cm4 .PI I= For an axis through P mass is farther from the axis. 9.58. IDENTIFY: Consider the plate as made of slender rods placed side-by-side. SET UP: The expression in Table 9.2(a) gives I for a rod and an axis through the center of the rod. EXECUTE: (a) I is the same as for a rod with length a: 2112I Ma= . (b) I is the same as for a rod with length b: 2112I Mb= . EVALUATE: I is smaller when the axis is through the center of the plate than when it is along one edge. 9.59. IDENTIFY: Use the equations in Table 9.2. I for the rod is the sum of I for each segment. The parallel-axis theorem says 2p cmI I Md= + . SET UP: The bent rod and axes a and b are shown in Figure 9.59. Each segment has length / 2L and mass / 2M . EXECUTE: (a) For each segment the moment of inertia is for a rod with mass / 2M , length / 2L and the axis through one end. For one segment, 2 2 s 1 1 3 2 2 24 M L I ML? ?? ?= =? ?? ?? ?? ? . For the rod, 2 a s 1 2 12 I I ML= = . (b) The center of mass of each segment is at the center of the segment, a distance of / 4L from each end. For each segment, 2 2 cm 1 1 12 2 2 96 M L I ML? ?? ?= =? ?? ?? ?? ? . Axis b is a distance / 4L from the cm of each segment, so for each segment the parallel axis theorem gives I for axis b to be 2 2 2 s 1 1 96 2 4 24 M L I ML ML? ?= + =? ?? ? and 2 b s 1 2 12 I I ML= = . Rotation of Rigid Bodies 9-15 EVALUATE: I for these two axes are the same. Figure 9.59 9.60. IDENTIFY: Apply the parallel-axis theorem. SET UP: ( ) ( )2cmIn Eq 9 19 and 212 M . . , I L d L h= = ? . EXECUTE: 2 2 2 2 2 2 21 1 1 1 12 2 12 4 3P L I M L h M L L Lh h M L Lh h ? ?? ? ? ? ? ?= + ? = + ? + = ? +? ?? ? ? ? ? ?? ? ? ? ? ?? ?? ? , which is the same as found in Example 9.11. EVALUATE: Example 9.11 shows that this result gives the expected result for 0h = , h L= and / 2h L= . 9.61. IDENTIFY: Apply Eq.(9.20). SET UP: (2 )dm dV rL dr? ? ?= = , where L is the thickness of the disk. 2M L R? ?= . EXECUTE: The analysis is identical to that of Example 9.12, with the lower limit in the integral being zero and the upper limit being R. The result is 212I MR= . EVALUATE: Our result agrees with Table 9.2(f). 9.62. IDENTIFY: Eq.(9.20), 2 I r dm= ? SET UP: Figure 9.62 Take the x-axis to lie along the rod, with the origin at the left end. Consider a thin slice at coordinate x and width dx, as shown in Figure 9.62. The mass per unit length for this rod is / ,M L so the mass of this slice is ( / ) .dm M L dx= EXECUTE: 2 2 3 2130 0( / ) ( / ) ( / )( /3) L L I x M L dx M L x dx M L L ML= = = =? ? EVALUATE: This result agrees with Table 9.2. 9.63. IDENTIFY: Apply Eq.(9.20). SET UP: For this case, .dm dx?= EXECUTE: (a) 2 2 0 0 2 2 LL x L M dm x dx ?? ?= = = =? ? (b) 4 4 2 2 0 0 ( ) . 24 4 LL x L MI x x dx L ?? ?= = = =? This is larger than the moment of inertia of a uniform rod of the same mass and length, since the mass density is greater further away from the axis than nearer the axis. (c) 2 3 4 4 2 2 2 3 2 2 0 0 0 ( ) ( 2 ) 2 2 3 4 12 6 LL L x x x L M I L x xdx L x Lx x dx L L L? ? ? ?? ?= ? = ? + = ? + = =? ?? ?? ? . This is a third of the result of part (b), reflecting the fact that more of the mass is concentrated at the right end. EVALUATE: For a uniform rod with an axis at one end, 213I ML= . The result in (b) is larger than this and the result in (a) is smaller than this. 9.64. IDENTIFY: We know that v r?= and v! is tangential. We know that 2rada r?= and rada! is in toward the center of the wheel. See if the vector product expressions give these results. SET UP: sinAB ?× =A B , where ? is the angle between A! and B! . EXECUTE: (a) For a counterclockwise rotation,?! will be out of the page. 9-16 Chapter 9 (b) The upward direction crossed into the radial direction is, by the right-hand rule, counterclockwise. !? and r! are perpendicular, so the magnitude of × r! !? is r v? = . (c) !? is perpendicular to v! and so × v! !? has magnitude rad,v a? = and from the right-hand rule, the upward direction crossed into the counterclockwise direction is inward, the direction of rad.a ! EVALUATE: If the wheel rotates clockwise, the directions of !? and v! are reversed, but rada! is still inward. 9.65. IDENTIFY: Apply t? ?= . SET UP: For alignment, the earth must move through 60° more than Mars, in the same time t. e 360 / yr? = ° . M 360 /(1.9 yr)? = ° . EXECUTE: e M 60? ?= + ° . e M 60t t? ?= + ° . 2 e M 60 60 (1/[0.9yr /1.9 yr ]) 0.352 yr 128 days 360 360 360 1 yr 1.9 yr t ? ?= = = = =? ? ° 60° ° ° ° ° . EVALUATE: Earth has a larger angular velocity than Mars, and completes one orbit in less time. 9.66. IDENTIFY and SET UP: Use Eqs.(9.3) and (9.5). As long as 0,z? > z? increases. At the t when 0,z? = z? is at its maximum positive value and then starts to decrease when z? becomes negative. 2 3( ) ;t t t? ? ?= ? 23.20 rad/s ,? = 30.500 rad/s? = EXECUTE: (a) 2 3 2( )( ) 2 3z d d t t t t t dt dt ? ? ?? ? ??= = = ? (b) 2(2 3 ) ( ) 2 6zz d d t t t t dt dt ? ? ?? ? ??= = = ? (c) The maximum angular velocity occurs when 0.z? = 2 6 0t? ?? = implies 2 3 2 3.20 rad/s 2.133 s 6 3 3(0.500 rad/s ) t ? ? ? ?= = = = At this t, 2 2 3 22 3 2(3.20 rad/s )(2.133 s) 3(0.500 rad/s )(2.133 s)z t t? ? ?= ? = ? = 6.83 rad/s The maximum positive angular velocity is 6.83 rad/s and it occurs at 2.13 s. EVALUATE: For large t both z? and z? are negative and z? increases in magnitude. In fact, z? ? ?? at .t ? ? So the answer in (c) is not the largest angular speed, just the largest positive angular velocity. 9.67. IDENTIFY: The angular acceleration ? of the disk is related to the linear acceleration a of the ball by a R?= . Since the acceleration is not constant, use 0 0 t z z zdt? ? ?? = ? and 0 0t zdt? ? ?? = ? to relate ? , z? , z? and t for the disk. 0 0z? = . SET UP: 11 1 n nt dt t n += +? . In a R?= , ? is in 2rad/s . EXECUTE: (a) 2 31.80 m/s 0.600 m/s 3.00 s a A t = = = (b) 3 3(0.600 m/s ) (2.40 rad/s ) 0.250 m a t t R ? = = = (c) 3 3 2 0 (2.40 rad/s ) (1.20 rad/s ) t z tdt t? = =? . 15.0 rad/sz? = for 315.0 rad/s 3.54 s1.20 rad/st = = . (d) 3 2 3 30 0 0 (1.20 rad/s ) (0.400 rad/s ) t t zdt t dt t? ? ?? = = =? ? . For 3.54 st = , 0 17.7 rad? ?? = . EVALUATE: If the disk had turned at a constant angular velocity of 15.0 rad/s for 3.54 s it would have turned through an angle of 53.1 rad in 3.54 s. It actually turns through less than half this because the angular velocity is increasing in time and is less than 15.0 rad/s at all but the end of the interval. 9.68. IDENTIFY and SET UP: The translational kinetic energy is 212K mv= and the kinetic energy of the rotating flywheel is 212 .K I ?= Use the scale speed to calculate the actual speed v. From that calculate K for the car and then solve for ? that gives this K for the flywheel. Rotation of Rigid Bodies 9-17 EXECUTE: (a) toy toy scale real v L v L = toy toy scale real 0.150 m (700 km/h) 35.0 km/h 3.0 m L v v L ? ? ? ?= = =? ? ? ?? ?? ? toy (35.0 km/h)(1000 m/1 km)(1 h/3600 s) 9.72 m/sv = = (b) 2 21 12 2 (0.180 kg)(9.72 m/s) 8.50 JK mv= = = (c) 212K I ?= gives that 5 22 2(8.50 J) 652 rad/s4.00 10 kg m K I ? ?= = =× ? EVALUATE: 212K I ?= gives ? in rad/s. 652 rad/s 6200 rev/min= so the rotation rate of the flywheel is very large. 9.69. IDENTIFY: tana r?= , 2rada r?= . Apply the constant acceleration equations and m=?F a! ! . SET UP: tana and rada are perpendicular components of a! , so 2 2rad tana a a= + . EXECUTE: (a) 2 2tan 3.00 m s 0.050 rad s 60.0 m a r ? = = = (b) 2(0.05 rad s )(6.00 s) 0.300 rad s.t? = = (c) 2 2 2rad (0.300 rad s) (60.0 m) 5.40 m s .a r?= = = (d) The sketch is given in Figure 9.69. (e) 22 2 2 2 2 2rad tan (5.40 m s ) (3.00 m s ) 6.18 m s ,a a a= + = + = and the magnitude of the force is 2(1240 kg)(6.18 m s ) 7.66 kN.F ma= = = (f) rad tan 5.40 arctan arctan 60.9 . 3.00 a a ? ? ? ?= = °? ? ? ?? ?? ? EVALUATE: tana is constant and rada increases as ? increases. At 0t = , a! is parallel to v! . As t increases, a ! moves toward the radial direction and the angle between a ! increases toward 90° . Figure 9.69 9.70. IDENTIFY: Apply conservation of energy to the system of drum plus falling mass, and compare the results for earth and for Mars. SET UP: 21drum 2K I ?= . 21mass 2K mv= . v R?= so if drumK is the same, ? is the same and v is the same on both planets. Therefore, massK is the same. Let 0y = at the initial height of the mass and take y+ upward. Configuration 1 is when the mass is at its initial position and 2 is when the mass has descended 5.00 m, so 1 0y = and 2y h= ? , where h is the height the mass descends. EXECUTE: (a) 1 1 2 2K U K U+ = + gives drum mass0 K K mgh= + ? . drum massK K+ are the same on both planets, so E E M Mmg h mg h= . 2 E M E 2 M 9.80 m/s (5.00 m) 13.2 m 3.71 m/s g h h g ? ? ? ?= = =? ? ? ?? ?? ? . 9-18 Chapter 9 (b) M M drum massmg h K K= + . 21 M M drum2 mv mg h K= ? and 2drum M M 2 2(250.0 J) 2 2(3.71 m/s )(13.2 m) 8.04 m/s 15.0 kg K v g h m = ? = ? = EVALUATE: We did the calculations without knowing the moment of inertia I of the drum, or the mass and radius of the drum. 9.71. IDENTIFY and SET UP: All points on the belt move with the same speed. Since the belt doesn?t slip, the speed of the belt is the same as the speed of a point on the rim of the shaft and on the rim of the wheel, and these speeds are related to the angular speed of each circular object by .v r?= EXECUTE: Figure 9.71 (a) 1 1 1v r?= 1 (60.0 rev/s)(2 rad/1 rev) 377 rad/s? ?= = 2 1 1 1 (0.45 10 m)(377 rad/s) 1.70 m/sv r? ?= = × = (b) 1 2v v= 1 1 2 2r r? ?= 2 1 2 1( / ) (0.45 cm/2.00 cm)(377 rad/s) 84.8 rad/sr r? ?= = = EVALUATE: The wheel has a larger radius than the shaft so turns slower to have the same tangential speed for points on the rim. 9.72. IDENTIFY: The speed of all points on the belt is the same, so 1 1 2 2r r? ?= applies to the two pulleys. SET UP: The second pulley, with half the diameter of the first, must have twice the angular velocity, and this is the angular velocity of the saw blade. rad/s 30 rev/min? = . EXECUTE: (a) 2 rad s 0.208 m(2(3450 rev min)) 75.1 m s.30 rev min 2v ?? ?? ?= =? ?? ?? ?? ? (b) 2 2 4 2 rad rad s 0.208 m 2(3450 rev min) 5.43 10 m s , 30 rev min 2 a r ?? ? ?? ? ? ?= = = ×? ?? ? ? ?? ? ? ?? ?? ? so the force holding sawdust on the blade would have to be about 5500 times as strong as gravity. EVALUATE: In v r?= and 2rada r?= , ? must be in rad/s. 9.73. IDENTIFY and SET UP: Use Eq.(9.15) to relate rada to ? and then use a constant acceleration equation to replace .? EXECUTE: (a) 2rad ,a r?= 2rad,1 1 ,a r?= 2rad,2 2a r?= 2 2 rad rad,2 rad,1 2 1( )a a a r ? ?? = ? = ? One of the constant acceleration equations can be written 2 2 2 1 2 12 ( ),z z? ? ? ? ?= + ? or 2 22 1 2 12 ( )z z z? ? ? ? ?? = ? Thus rad 2 1 2 12 ( ) 2 ( ),z za r r? ? ? ? ? ?? = ? = ? as was to be shown. (b) 2 2 2rad 2 1 85.0 m/s 25.0 m/s 8.00 rad/s 2 ( ) 2(0.250 m)(15.0 rad)z a r ? ? ? ? ?= = =? Then 2 2tan (0.250 m)(8.00 rad/s ) 2.00 m/sa r?= = = EVALUATE: 2? is proportional to z? and 0( )? ?? so rada is also proportional to these quantities. rada increases while r stays fixed, z? increases, and z? is positive. IDENTIFY and SET UP: Use Eq.(9.17) to relate K and ? and then use a constant acceleration equation to replace .? EXECUTE: (c) 212 ;K I ?= 212 22 ,K I ?= 211 12K I ?= 2 21 1 2 1 2 1 2 1 2 12 2( ) (2 ( )) ( ),z zK K K I I I? ? ? ? ? ? ? ?? = ? = ? = ? = ? as was to be shown. (d) 22 2 1 45.0 J 20.0 J 0.208 kg m ( ) (8.00 rad/s )(15.0 rad)z K I ? ? ? ? ?= = = ?? EVALUATE: z? is positive, ? increases, and K increases. Rotation of Rigid Bodies 9-19 9.74. IDENTIFY: wood lead.I I I= + m V?= , where ? is the volume density and m A?= , where ? is the area density. SET UP: For a solid sphere, 225I mR= . For the hollow sphere (foil), 223I mR= . For a sphere, 343V R?= and 24A R?= . 3w w w w 43m V R? ? ?= = . 2 L L L L 4m A R? ? ?= = . EXECUTE: 2 2 3 2 2 2 4 ww L w L L2 2 2 4 2 8( 4 )5 3 5 3 3 3 5 R I m R m R R R R R R ?? ? ? ? ? ?? ? ? ?= + = + = +? ? ? ?? ? ? ? . 3 4 2 28 (800 kg m )(0.20 m)(0.20 m) 20 kg m 0.70 kg m 3 5 I ? ? ?= + = ?? ?? ? . EVALUATE: W 26.8 kgm = and 2W 0.429 kg mI = ? . L 10.1 kgm = and 2L 0.268 kg mI = ? . Even though the foil is only 27% of the total mass its contribution to I is about 38% of the total. 9.75. IDENTIFY: Estimate the shape and dimensions of your body and apply the approximate expression from Table 9.2. SET UP: I approximate my body as a vertical cylinder with mass 80 kg, length 1.7 m, and diameter 0.30 m (radius 0.15 m) EXECUTE: 2 2 21 1 (80 kg) (0.15 m) 0.9 kg m 2 2 I mR= = = ? EVALUATE: I depends on your mass and width but not on your height. 9.76. IDENTIFY: Treat the V like two thin 0.160 kg bars, each 25 cm long. SET UP: For a slender bar with the axis at one end, 213I mL= . EXECUTE: 2 2 3 21 12 2 (0.160 kg)(0.250 m) 6.67 10 kg m 3 3 I mL ?? ? ? ?= = = × ?? ? ? ?? ? ? ? EVALUATE: The value of I is independent of the angle between the two sides of the V; the angle 70.0° didn't enter into the calculation. 9.77. IDENTIFY: 212K I ?= . 2rada r?= . m V?= . SET UP: For a disk with the axis at the center, 212I mR= . 2V t R?= , where 0.100 mt = is the thickness of the flywheel. 37800 kg m? = is the density of the iron. EXECUTE: (a) 90.0 rpm 9.425 rad s? = = . 6 5 2 2 2 2 2(10 0 10 J) 2 252 10 kg m (9 425 rad s) K . I . .? ×= = = × ? . 2m V R t? ??= = . 2 41 1 2 2 I mR tR??= = . This gives 1 4(2 ) 3.68 mR I t??= = and the diameter is 7.36 m. (b) 2 2rad 327m sa R?= = EVALUATE: In 212K I ?= , ? must be in rad/s. rada is about 33g ; the flywheel material must have large cohesive strength to prevent the flywheel from flying apart. 9.78. IDENTIFY: 212K I ?= . To have the same K for any ? the two parts must have the same I . Use Table 9.2 for I . SET UP: For a solid sphere, 22solid solid5I M R= . For a hollow sphere, 22hollow hollow3I M R= . EXECUTE: solid hollowI I= gives 2 22 2solid hollow5 3M R M R= and 3 3hollow solid5 5M M M= = . EVALUATE: The hollow sphere has less mass since all its mass is distributed farther from the rotation axis. 9.79. IDENTIFY: 212K I ?= . 2 radT ?? = , where T is the period of the motion. For the earth's orbital motion it can be treated as a point mass and 2I MR= . SET UP: The earth's rotational period is 24 h 86,164 s= . Its orbital period is 71 yr 3.156 10 s= × . 245.97 10 kgM = × . 66.38 10 mR = × . EXECUTE: (a) 2 2 24 6 2 29 2 2 2 2 (0.3308)(5.97 10 kg)(6.38 10 m) 2.14 10 J. (86,164 s) I K T ? ? × ×= = = × (b) 2 2 24 11 2 33 7 2 1 2 2 (5.97 10 kg)(1.50 10 m) 2.66 10 J. 2 (3.156 10 s) R M T ? ? × ×? ? = = ×? ? ×? ? (c) Since the Earth?s moment of inertia is less than that of a uniform sphere, more of the Earth?s mass must be concentrated near its center. EVALUATE: These kinetic energies are very large, because the mass of the earth is very large. 9-20 Chapter 9 9.80. IDENTIFY: Using energy considerations, the system gains as kinetic energy the lost potential energy, mgR . SET UP: The kinetic energy is 2 21 1 2 2 K I mv?= + , with 212I mR= for the disk. v R?= . EXECUTE: 2 2 21 1 1( ) ( ) 2 2 2 K I m R I mR? ?= + = + . 212Using and solving for ,? mR ?= 2 43 g R ? = and 4 . 3 g R ? = EVALUATE: The small object has speed 2 2 3 v gR= . If it was not attached to the disk and was dropped from a height h, it would attain a speed 2 g R . Being attached to the disk reduces its final speed by a factor of 2 3 . 9.81. IDENTIFY: Use Eq.(9.20) to calculate I . Then use 212K I ?= to calculate K . (a) SET UP: The object is sketched in Figure 9.81. Consider a small strip of width dy and a distance y below the top of the triangle. The length of the strip is ( / ) .x y h b= Figure 9.81 EXECUTE: The strip has area x dy and the area of the sign is 12 ,bh so the mass of the strip is 21 2 2 2x dy yb dy M dm M M y dy bh h bh h ? ? ? ?? ? ? ?= = =? ? ? ?? ? ? ?? ?? ? ? ?? ? 2 2 31 3 4 2 ( ) 3 Mb dI dm x y dy h ? ?= = ? ?? ? 2 2 3 4 2 04 40 0 2 2 1 1 3 3 6 h h hMb MbI dI y dy y Mb h h h ? ?= = = =? ?? ?? ? (b) 2 216 2.304 kg mI Mb= = ? 2.00 rev/s 4.00 rad/s? ?= = 21 2 182 JK I ?= = EVALUATE: From Table (9.2), if the sign were rectangular, with length b, then 213 .I Mb= Our result is one-half this, since mass is closer to the axis for the triangular than for the rectangular shape. 9.82. IDENTIFY: Apply conservation of energy to the system. SET UP: For the falling mass 212K mv= . For the wheel 212K I ?= . EXECUTE: (a) The kinetic energy of the falling mass after 2.00 m is ( )( )221 12 2 8.00 kg 5.00 m/s 100 J.K mv= = = The change in its potential energy while falling is ( )( )( )28.00 kg 9.8 m/s 2.00 m 156.8 Jmgh = = . The wheel must have the ?missing? 56.8 J in the form of rotational kinetic energy. Since its outer rim is moving at the same speed as the falling mass, 5.00 m/s , v r?= gives 5.00 m/s 13.51 rad/s 0.370 m v r ? = = = . 21 ; therefore 2 K I ?= ( ) ( ) 2 22 2 56.8 J2 0.622 kg m 13.51 rad s K I ?= = = ? . (b) The wheel?s mass is 2(280 N) (9.8 m s ) 28.6 kg= . The wheel with the largest possible moment of inertia would have all this mass concentrated in its rim. Its moment of inertia would be ( )( )22 228.6 kg 0.370 m 3.92 kg mI MR= = = ? . The boss?s wheel is physically impossible. Rotation of Rigid Bodies 9-21 EVALUATE: If the mass falls from rest in free-fall its speed after it has descended 2.00 m is 2 (2.00 m) 6.26 m/sv g= = . Its actual speed is less because some of the energy of the system is in the form of rotational kinetic energy of the wheel. 9.83. IDENTIFY: Use conservation of energy. The stick rotates about a fixed axis so 212 .K I ?= Once we have ? use v r?= to calculate v for the end of the stick. SET UP: The object is sketched in Figure 9.83. Take the origin of coordinates at the lowest point reached by the stick and take the positive y-direction to be upward. Figure 9.83 EXECUTE: (a) Use Eq.(9.18): cmU Mgy= 2 1 cm2 cm1( )U U U Mg y y? = ? = ? The center of mass of the meter stick is at its geometrical center, so cm1 1.00 my = and cm2 0.50 my = Then 2(0.160 kg)(9.80 m/s )(0.50 m 1.00 m) 0.784 JU? = ? = ? (b) Use conservation of energy: 1 1 other 2 2K U W K U+ + = + Gravity is the only force that does work on the meter stick, so other 0.W = 1 0.K = Thus 2 1 2 ,K U U U= ? = ?? where U? was calculated in part (a). 21 2 22K I ?= so 21 22 I U? = ?? and 2 2( ) /U I? = ?? For stick pivoted about one end, 213I ML= where 1.00 m,L = so 2 2 2 6( ) 6(0.784 J) 5.42 rad/s (0.160 kg)(1.00 m) U ML ? ??= = = (c) (1.00 m)(5.42 rad/s) 5.42 m/sv r?= = = (d) For a particle in free-fall, with y+ upward, 0 0;yv = 0 1.00 m;y y? = ? 29.80 m/s ;ya = ? ?yv = 2 2 0 02 ( )y y yv v a y y= + ? 2 02 ( ) 2( 9.80 m/s )( 1.00 m) 4.43 m/sy yv a y y= ? ? = ? ? ? = ? EVALUATE: The magnitude of the answer in part (c) is larger. 1,gravU is the same for the stick as for a particle falling from a height of 1.00 m. For the stick ( )2 2 2 21 1 1 122 2 3 6( / ) .K I ML v L Mv?= = = For the stick and for the particle, 2K is the same but the same K gives a larger v for the end of the stick than for the particle. The reason is that all the other points along the stick are moving slower than the end opposite the axis. 9.84. IDENTIFY: Apply conservation of energy to the system of cylinder and rope. SET UP: Taking the zero of gravitational potential energy to be at the axle, the initial potential energy is zero (the rope is wrapped in a circle with center on the axle).When the rope has unwound, its center of mass is a distance R? below the axle, since the length of the rope is 2 R? and half this distance is the position of the center of the mass. Initially, every part of the rope is moving with speed 0 ,R? and when the rope has unwound, and the cylinder has angular speed ,? the speed of the rope is R? (the upper end of the rope has the same tangential speed at the edge of the cylinder). 2(1 2)I MR= for a uniform cylinder, 9-22 Chapter 9 EXECUTE: 1 2 2K K U= + . 2 2 2 20 .4 2 4 2 M m M m R R mg R? ? ?? ? ? ?+ = + ?? ? ? ?? ? ? ? Solving for? gives ( ) ( )20 4 2 mg R M m ?? ?= + + , and the speed of any part of the rope is .v R?= EVALUATE: When 0m ? , 0? ?? . When m M>> , 20 2 gR ?? ?= + and 20 2v v gR?= + . This is the final speed when an object with initial speed 0v descends a distance R? . 9.85. IDENTIFY: Apply conservation of energy to the system consisting of blocks A and B and the pulley. SET UP: The system at points 1 and 2 of its motion is sketched in Figure 9.85. Figure 9.85 Use the work-energy relation 1 1 other 2 2.K U W K U+ + = + Use coordinates where y+ is upward and where the origin is at the position of block B after it has descended. The tension in the rope does positive work on block A and negative work of the same magnitude on block B, so the net work done by the tension in the rope is zero. Both blocks have the same speed. EXECUTE: Gravity does work on block B and kinetic friction does work on block A. Therefore other k .f AW W m gd?= = ? 1 0K = (system is released from rest) 1 1 ;B B BU m gy m gd= = 2 2 0B BU m gy= = 2 2 21 1 1 2 2 2 22 2 2 .A BK m v m v I ?= + + But (blocks) (pulley),v R?= so 2 2 /v R? = and 2 2 2 21 1 1 2 2 2 22 2 2( ) ( / ) ( / )A B A BK m m v I v R m m I R v= + + = + + Putting all this into the work-energy relation gives 2 21 k 22 ( / )B A A Bm gd m gd m m I R v?? = + + 2 2 2 k( / ) 2 ( )A B B Am m I R v gd m m?+ + = ? k 2 2 2 ( ) / B A A B g d m m v m m I R ??= + + EVALUATE: If B Am m>> and 2/ ,I R then 2 2 ;v gd= block B falls freely. If I is very large, 2v is very small. Must have kB Am m?> for motion, so the weight of B will be larger than the friction force on A. 2/I R has units of mass and is in a sense the ?effective mass? of the pulley. 9.86. IDENTIFY: Apply conservation of energy to the system of two blocks and the pulley. SET UP: Let the potential energy of each block be zero at its initial position. The kinetic energy of the system is the sum of the kinetic energies of each object. v R?= , where v is the common speed of the blocks and ? is the angular velocity of the pulley. EXECUTE: The amount of gravitational potential energy which has become kinetic energy is ( )( )( )24.00 kg 2.00 kg 9.80 m s 5.00 m 98.0 J.K = ? = In terms of the common speed v of the blocks, the kinetic energy of the system is 2 2 1 2 1 1 ( ) 2 2 v K m m v I R ? ?= + + ? ?? ? . 2 2 2 2 1 (0.480 kg m ) 4.00 kg 2 00 kg (12.4 kg). 2 (0.160 m) K v . v ? ??= + + =? ?? ? Solving for v gives 98.0 J 2.81 m s. 12.4 kg v = = Rotation of Rigid Bodies 9-23 EVALUATE: If the pulley is massless, 21298.0 J (4.00 kg 2.00 kg)v= + and 5.72 m/sv = . The moment of inertia of the pulley reduces the final speed of the blocks. 9.87. IDENTIFY and SET UP: Apply conservation of energy to the motion of the hoop. Use Eq.(9.18) to calculate grav.U Use 212K I ?= for the kinetic energy of the hoop. Solve for .? The center of mass of the hoop is at its geometrical center. Take the origin to be at the original location of the center of the hoop, before it is rotated to one side, as shown in Figure 9.87. Figure 9.87 cm1 cos (1 cos )y R R R? ?= ? = ? cm2 0y = (at equilibrium position hoop is at original position) EXECUTE: 1 1 other 2 2K U W K U+ + = + other 0W = (only gravity does work) 1 0K = (released from rest), 212 22K I ?= For a hoop, 2cm ,I MR= so 2 2I Md MR= + with d R= and 22 ,I MR= for an axis at the edge. Thus 2 2 2 21 2 2 22 (2 ) .K MR MR? ?= = 1 cm1 (1 cos ),U Mgy MgR ?= = ? 2 cm2 0U mgy= = Thus 1 1 other 2 2K U W K U+ + = + gives 2 2 2(1 cos )MgR MR? ?? = and (1 cos ) /g R? ?= ? EVALUATE: If 0,? = then 2 0.? = As ? increases, 2? increases. 9.88. IDENTIFY: 212K I ?= , with ? in rad/s. energyP t= SET UP: For a solid cylinder, 212I MR= . 1 rev/min (2 / 60) rad/s?= EXECUTE: (a) 3000 rev/min 314 rad/s? = = . 2 212 (1000 kg)(0.900 m) 405 kg mI = = ? 2 2 71 2 (405 kg m )(314 rad/s) 2.00 10 JK = ? = × . (b) 7 3 4 2.00 10 J 1.08 10 s 17.9 min 1.86 10 W K t P ×= = = × =× . EVALUATE: In 212K I ?= , we must use ? in rad/s. 9.89. IDENTIFY: 1 2I I I= + . Apply conservation of energy to the system. The calculation is similar to Example 9.9. SET UP: 1 v R ? = for part (b) and 2 v R ? = for part (c). EXECUTE: (a) 2 2 2 2 2 21 1 2 21 1 1 ((0.80 kg)(2.50 10 m) (1.60 kg)(5 00 10 m) )2 2 2I M R M R . ? ?= + = × + × 3 22.25 10 kg m .I ?= × ? (b) The method of Example 9.9 yields 2 1 2 1 ( ) gh v I mR = + . 2 3 2 2 2(9.80 m s )(2.00 m) 3.40 m s. (1 ((2.25 10 kg m ) (1.50 kg)(0.025 m) )) v ?= =+ × ? The same calculation, with 2R instead of 1R gives 4.95 m s.v = EVALUATE: The final speed of the block is greater when the string is wrapped around the larger disk. v R?= , so when 2R R= the factor that relates v to ? is larger. For 2R R= a larger fraction of the total kinetic energy resides 9-24 Chapter 9 with the block. The total kinetic energy is the same in both cases (equal to mgh ), so when 2R R= the kinetic energy and speed of the block are greater. 9.90. IDENTIFY: Apply conservation of energy to the motion of the mass after it hits the ground. SET UP: From Example 9.9, the speed of the mass just before it hits the ground is 2 1 / 2 gh v M m = + . EXECUTE: (a) In the case that no energy is lost, the rebound height h? is related to the speed v by 2 2 v h g ? = , and with the form for v given in Example 9.9, . 1 2 hh M m ? = + (b) Considering the system as a whole, some of the initial potential energy of the mass went into the kinetic energy of the cylinder. Considering the mass alone, the tension in the string did work on the mass, so its total energy is not conserved. EVALUATE: If m M>> , h h? = and the mass does rebound to its initial height. 9.91. IDENTIFY: Apply conservation of energy to relate the height of the mass to the kinetic energy of the cylinder. SET UP: First use (cylinder) 250 JK = to find ? for the cylinder and v for the mass. EXECUTE: 2 21 12 2 (10.0 kg)(0.150 m) 0.1125 kg mI MR 2= = = ? 21 2K I ?= so 2 / 66.67 rad/sK I? = = 10.0 m/sv R?= = SET UP: Use conservation of energy 1 1 2 2K U K U+ = + to solve for the distance the mass descends. Take 0y = at lowest point of the mass, so 2 0y = and 1 ,y h= the distance the mass descends. EXECUTE: 1 2 0K U= = so 1 2.U K= 2 21 1 2 2 ,mgh mv I ?= + where 12.0 kgm = For the cylinder, 212I MR= and / ,v R? = so 2 21 12 4 .I Mv? = 2 21 1 2 4mgh mv Mv= + 2 1 7.23 m 2 2 v M h g m ? ?= + =? ?? ? EVALUATE: For the cylinder ( )2 2 2 21 1 1 1cyl 2 2 2 4( / ) .K I MR v R Mv?= = = 21 mass 2 ,K mv= so mass cyl(2 / ) [2(12.0 kg)/10.0 kg](250 J) 600 J.K m M K= = = The mass has 600 J of kinetic energy when the cylinder has 250 J of kinetic energy and at this point the system has total energy 850 J since 2 0.U = Initially the total energy of the system is 1 1 850 J,U mgy mgh= = = so the total energy is shown to be conserved. 9.92. IDENTIFY: Energy conservation: Loss of U of box equals gain in K of system. Both the cylinder and pulley have kinetic energy of the form 212K I ?= . 2 2 2box box box pulley pulley cylinder cylinder1 1 12 2 2m gh m v I I? ?= + + . SET UP: Box Boxpulley cylinder p cylinder and v v r r ? ?= = . EXECUTE: 2 2 2 2 2 B B B B P p C C p C 1 1 1 1 1 2 2 2 2 2 Bv vm gh m v m r m r r r ? ? ? ?? ? ? ?= + +? ? ? ?? ? ? ?? ?? ? ? ? ? ?? ? . 2 2 2B B B P B C B 1 1 1 2 4 4 m gh m v m v m v= + + and 2 B B 1 1 1 1 B p C2 4 4 4 (3.00 kg)(9.80 m s )(1.50 m) 3.68 m s 1.50 kg (7.00 kg) m gh v m m m = = =+ + + . EVALUATE: If the box was disconnected from the rope and dropped from rest, after falling 1.50 m its speed would be 2 (1.50 m) 5.42 m/sv g= = . Since in the problem some of the energy of the system goes into kinetic energy of the cylinder and of the pulley, the final speed of the box is less than this. 9.93. IDENTIFY: disk holeI I I= ? , where holeI is I for the piece punched from the disk. Apply the parallel-axis theorem to calculate the required moments of inertia. SET UP: For a uniform disk, 212I MR= . Rotation of Rigid Bodies 9-25 EXECUTE: (a) The initial moment of inertia is 210 2 .I MR= The piece punched has a mass of 16 M and a moment of inertia with respect to the axis of the original disk of 2 2 21 9 . 16 2 4 2 512 M R R MR ? ?? ? ? ?+ =? ?? ? ? ?? ? ? ?? ?? ? The moment of inertia of the remaining piece is then 2 2 2 1 9 247 . 2 512 512 I MR MR MR= ? = (b) 2 2 2 23831 12 2 512 ( / 2) ( /16)( / 4) .I MR M R M R MR= + ? = EVALUATE: For a solid disk and an axis at a distance / 2R from the disk's center, the parallel-axis theorem gives 2 2 23 3841 2 4 512I MR MR MR= = = . For both choices of axes the presence of the hole reduces I , but the effect of the hole is greater in part (a), when it is farther from the axis. 9.94. IDENTIFY: In part (a) use the parallel-axis theorem to relate the moment of inertia cmI for an axis through the center of the sphere to PI , the moment of inertia for an axis at the pivot. SET UP: I for a uniform solid sphere and the axis through its center is 225 .MR I for a slender rod and an axis at one end is 213 mL , where m is the mass of the rod and L is its length. EXECUTE: (a) From the parallel-axis theorem, the moment of inertia is 2 2(2 5) ,PI MR ML= + and 2 2 2 1 . 5 PI R ML L ? ?? ?? ?= +? ?? ?? ?? ?? ?? ?? ? If (0.05) ,R L= the difference is 2(2 5)(0.05) 0.001 0.1%.= = (b) 2rod rod ( ) ( 3 ),I ML m M= which is 0.33% when rod (0.01) .m M= EVALUATE: In both these cases the correction to 2I ML= is very small. 9.95. IDENTIFY: Follow the instructions in the problem to derive the perpendicular-axis theorem. Then apply that result in part (b). SET UP: 2i i i I m r=? . The moment of inertia for the washer and an axis perpendicular to the plane of the washer at its center is 2 21 1 22 ( )M R R+ . In part (b), I for an axis perpendicular to the plane of the square at its center is 2 2 21 1 12 6( )M L L ML+ = . EXECUTE: (a) With respect to O , 2 2 2,i i ir x y= + and so 2 2 2 2 2( ) .O i i i i i i i i i x y i i i i I m r m x y m x m y I I= = + = + = +? ? ? ? (b) Two perpendicular axes, both perpendicular to the washer?s axis, will have the same moment of inertia about those axes, and the perpendicular-axis theorem predicts that they will sum to the moment of inertia about the washer axis, which is 2 21 1 22 ( ),I M R R= + and so xI yI= 2 21 1 24 ( ).M R R= + (c) 210 6 .I mL= Since 10 12, and , both and must be x y x y x yI I I I I I I= + = 2.mL EVALUATE: The result in part (c) says that I is the same for an axis that bisects opposite sides of the square as for an axis along the diagonal of the square, even though the distribution of mass relative to the two axes is quite different in these two cases. 9.96. IDENTIFY: Apply the parallel-axis theorem to each side of the square. SET UP: Each side has length a and mass / 4,M and the moment of inertia of each side about an axis perpendicular to the side and through its center is 2 21 1 112 4 48Ma Ma= . EXECUTE: The moment of inertia of each side about the axis through the center of the square is, from the perpendicular axis theorem, ( )22 248 4 2 12Ma a MaM+ = . The total moment of inertia is the sum of the contributions from the four sides, or 2 2 4 . 12 3 Ma Ma× = EVALUATE: If all the mass of a side were at its center, a distance / 2a from the axis, we would have 2 214 4 2 4 M a I Ma? ?? ?= =? ?? ?? ?? ? . If all the mass was divided equally among the four corners of the square, a distance / 2a from the axis, we would have 2 214 4 22 M a I Ma? ?? ?= =? ?? ?? ?? ? . The actual I is between these two values. 9-26 Chapter 9 9.97. IDENTIFY: Use Eq.(9.20) to calculate I . (a) SET UP: Let L be the length of the cylinder. Divide the cylinder into thin cylindrical shells of inner radius r and outer radius .r dr+ An end view is shown in Figure 9.97. r? ?= The mass of the thin cylindrical shell is 2(2 ) 2 dm dV r dr L Lr dr? ? ? ??= = = Figure 9.97 EXECUTE: ( )2 4 5 51 25 50 2 2RI r dm L r dr L R LR?? ?? ??= = = =? ? Relate M to :? ( )2 3 31 23 302 2 ,RM dm L r dr L R LR?? ?? ??= = = =? ? so 3 3 / 2.LR M?? = Using this in the above result for I gives 2 2325 5(3 / 2) .I M R MR= = (b) EVALUATE: For a cylinder of uniform density 212 .I MR= The answer in (a) is larger than this. Since the density increases with distance from the axis the cylinder in (a) has more mass farther from the axis than for a cylinder of uniform density. 9.98. IDENTIFY: Write K in terms of the period T and take derivatives of both sides of this equation to relate /dK dt to /dT dt . SET UP: 2 T ?? = and 212K I ?= . The speed of light is 83.00 10 m/sc = × . EXECUTE: (a) 2 2 2 I K T ?= . 2 3 4dK I dT dt T dt ?= ? . The rate of energy loss is 2 3 4 I dT T dt ? . Solving for the moment of inertia in terms of the power ,I P 3 31 3 38 2 2 13 1 (5 10 W)(0.0331 s) 1 s 1.09 10 kg m 4 4 4.22 10 s PT I dT dt? ? ? ×= = = × ?× (b) 38 2 3 30 5 5(1.08 10 kg m ) 9.9 10 m, about 10 km. 2 2(1.4)(1.99 10 kg) I R M × ?= = = ×× (c) 3 6 32 2 (9.9 10 m) 1.9 10 m s 6.3 10 . (0.0331 s) R v c T ? ? ?×= = = × = × (d) 17 33 6.9 10 kg m ,(4 3) M M V R ? ?= = = × which is much higher than the density of ordinary rock by 14 orders of magnitude, and is comparable to nuclear mass densities. EVALUATE: I is huge because M is huge. A small rate of change in the period corresponds to a large release of energy. 9.99. IDENTIFY: In part (a), do the calculations as specified in the hint. In part (b) calculate the mass of each shell of inner radius 1R and outer radius 2R and sum to get the total mass. In part (c) use the expression in part (a) to calculate I for each shell and sum to get the total I . SET UP: m V?= . For a solid sphere, 343V R?= . EXECUTE: (a) Following the hint, the moment of inertia of a uniform sphere in terms of the mass density is 2 582 5 15 ,I MR R??= = and so the difference in the moments of inertia of two spheres with the same density ? but different radii 5 52 1 2 1 and is (8 15)( ).R R I R R? ?= ? (b) A rather tedious calculation, summing the product of the densities times the difference in the cubes of the radii that bound the regions and multiplying by 24 4 3, gives 5.97 10 kg.M? = × (c) A similar calculation, summing the product of the densities times the difference in the fifth powers of the radii that bound the regions and multiplying by 22 2 28 15, gives 8.02 10 kg m 0.334 .I MR? = × ? = EVALUATE: The calculated value of 20.334I MR= agrees closely with the measured value of 20.3308MR . This simple model is fairly accurate. Rotation of Rigid Bodies 9-27 9.100. IDENTIFY: Apply Eq.(9.20) SET UP: Let z be the coordinate along the vertical axis. ( ) zRr z h = . 2 2 2 R z dm h ??= and 4 4 42 R dI z dz h ??= . EXECUTE: 4 4 4 5 4 4 00 1 2 10 10 h h R RI dI z dz z R h . h h ?? ?? ??? ?= = = =? ?? ? The volume of a right circular cone is 2 21 1 3 3, the mass is and soV R h R h? ??= 2 2 23 3 . 10 3 10 R h I R MR ??? ?= =? ?? ? EVALUATE: For a uniform cylinder of radius R and for an axis through its center, 212I MR= . I for the cone is less, as expected, since the cone is constructed from a series of parallel discs whose radii decrease from R to zero along the vertical axis of the cone. 9.101. IDENTIFY: Follow the steps outlined in the problem. SET UP: /z d dt? ?= . 2 2/z zd dt? ?= . EXECUTE: (a) 0 ds r d r d d? ? ?? ?= = + so 20( ) 2s r ?? ? ?= + . ? must be in radians. (b) Setting 20 2 s vt r ?? ?= = + gives a quadratic in ? . The positive solution is 2 0 0 1 ) 2(t r vt r? ?? ? ?= + ?? ? . (The negative solution would be going backwards, to values of r smaller than 0r .) (c) Differentiating, 2 0 ( ) , 2 z d v t dt r vt ?? ?= = + ( ) 2 3 22 0 . 2 z z d v dt r vt ? ?? ?= = ? + The angular acceleration z? is not constant. (d) 0 25.0 mm.r = ? must be measured in radians, so ( )( )1.55 m rev 1 rev 2 rad 0.247 m rad. ? ? ? ?= = Using ( )t? from part (b), the total angle turned in 74.0 min 4440 s= is ( )( )( ) ( )27 3 371 2 2.47 10 m/rad 1.25 m/s 4440 s 25.0 10 m 25.0 10 m 2.47 10 m/rad? ? ? ?? ? ?= × + × ? ×? ?× ? ? 5= 1.337 10 rad? × , which is 42.13 10 rev× . (e) The graphs are sketched in Figure 9.101. EVALUATE: z? must decrease as r increases, to keep v r?= constant. For z? to decrease in time, z? must be negative. Figure 9.101 10-1 DYNAMICS OF ROTATIONAL M OTION 10.1. IDENTIFY: Use Eq.(10.2) to calculate the magnitude of the torque and use the right-hand rule illustrated in Fig.(10.4) to calculate the torque direction. (a) SET UP: Consider Figure 10.1a. EXECUTE: Fl? = sin (4.00 m)sin90l r ?= = ° 4.00 ml = (10.0 N)(4.00 m) 40.0 N m? = = ? Figure 10.1a This force tends to produce a counterclockwise rotation about the axis; by the right-hand rule the vector ?! is directed out of the plane of the figure. (b) SET UP: Consider Figure 10.1b. EXECUTE: Fl? = sin (4.00 m)sin120l r ?= = ° 3.464 ml = (10.0 N)(3.464 m) 34.6 N m? = = ? Figure 10.1b This force tends to produce a counterclockwise rotation about the axis; by the right-hand rule the vector ?! is directed out of the plane of the figure. (c) SET UP: Consider Figure 10.1c. EXECUTE: Fl? = sin (4.00 m)sin30l r ?= = ° 2.00 ml = (10.0 N)(2.00 m) 20.0 N m? = = ? Figure 10.1c This force tends to produce a counterclockwise rotation about the axis; by the right-hand rule the vector ?! is directed out of the plane of the figure. (d) SET UP: Consider Figure 10.1d. EXECUTE: Fl? = sin (2.00 m)sin 60 1.732 ml r ?= = ° = (10.0 N)(1.732 m) 17.3 N m? = = ? Figure 10.1d This force tends to produce a clockwise rotation about the axis; by the right-hand rule the vector ?! is directed into the plane of the figure. (e) SET UP: Consider Figure 10.1e. EXECUTE: Fl? = 0r = so 0l = and 0? = Figure 10.1e 10 10-2 Chapter 10 (f) SET UP: Consider Figure 10.1f. EXECUTE: Fl? = sin ,l r ?= 180 ,? = ° so 0l = and 0? = Figure 10.1f EVALUATE: The torque is zero in parts (e) and (f) because the moment arm is zero; the line of action of the force passes through the axis. 10.2. IDENTIFY: Fl? = with sinl r ?= . Add the two torques to calculate the net torque. SET UP: Let counterclockwise torques be positive. EXECUTE: 1 1 1 (8.00 N)(5.00 m) 40.0 N mFl? = ? = ? = ? ? . 2 2 2 (12.0 N)(2.00 m)sin30.0 12.0 N mF l? = + = = + ?° . 1 2 28.0 N m? ? ?= + = ? ?? . The net torque is 28.0 N m? , clockwise. EVALUATE: Even though 1 2F F< , the magnitude of 1? is greater than the magnitude of 2 ,? because 1F has a larger moment arm. 10.3. IDENTIFY and SET UP: Use Eq.(10.2) to calculate the magnitude of each torque and use the right-hand rule (Fig.10.4) to determine the direction. Consider Figure 10.3 Figure 10.3 Let counterclockwise be the positive sense of rotation. EXECUTE: 2 21 2 3 (0.090 m) (0.090 m) 0.1273 mr r r= = = + = 1 1 1Fl? = ? 1 1 1sin (0.1273 m)sin135 0.0900 ml r ?= = ° = 1 (18.0 N)(0.0900 m) 1.62 N m? = ? = ? ? 1?! is directed into paper 2 2 2F l? = + 2 2 2sin (0.1273 m)sin135 0.0900 ml r ?= = ° = 2 (26.0 N)(0.0900 m) 2.34 N m? = + = + ? 2?! is directed out of paper 3 3 3F l? = + 3 3 3sin (0.1273 m)sin90 0.1273 ml r ?= = ° = 3 (14.0 N)(0.1273 m) 1.78 N m? = + = + ? 3?! is directed out of paper 1 2 3 1.62 N m 2.34 N m 1.78 N m 2.50 N m? ? ? ?= + + = ? ? + ? + ? = ?? EVALUATE: The net torque is positive, which means it tends to produce a counterclockwise rotation; the vector torque is directed out of the plane of the paper. In summing the torques it is important to include + or ? signs to show direction. 10.4. IDENTIFY: Use sinFl rF? ?= = to calculate the magnitude of each torque and use the right-hand rule to determine the direction of each torque. Add the torques to find the net torque. Dynamics of Rotational Motion 10-3 SET UP: Let counterclockwise torques be positive. For the 11.9 N force ( 1F ), 0r = . For the 14.6 N force ( 2F ), 0.350 mr = and 40.0? = ° . For the 8.50 N force ( 3F ), 0.350 mr = and 90.0? = ° EXECUTE: 1 0? = . 2 (14.6 N)(0.350 m)sin 40.0 3.285 N m? = ? = ? ?° . 3 (8.50 N)(0.350 m)sin90.0 2.975 N m? = + = + ?° . 3.285 N m 2.975 N m 0.31 N m? = ? ? + ? = ? ?? .The net torque is 0.31 N m? and is clockwise. EVALUATE: If we treat the torques as vectors, 2?! is into the page and 3?! is out of the page. 10.5. IDENTIFY and SET UP: Calculate the torque using Eq.(10.3) and also determine the direction of the torque using the right-hand rule. (a) ? ?( 0.450 m) (0.150 m) ;= ? +r i j! ? ?( 5.00 N) (4.00 N) .= ? +F i j! The sketch is given in Figure 10.5. Figure 10.5 EXECUTE: (b) When the fingers of your right hand curl from the direction of r! into the direction of F! (through the smaller of the two angles, angle )? your thumb points into the page (the direction of ,?! the -direction).z? (c) ? ? ? ?( 0.450 m) +(0.150 m) ( 5.00 N) (4.00 N)? ? ? ? ?= × = ? × ? +? ? ? ?r F i j i j !!! ? ? ? ? ? ? ? ?(2.25 N m) (1.80 N m) (0.750 N m) (0.600 N m)? = + ? × ? ? × ? ? × + ? ×i i i j j i j j! ? ? ? ?× = × = 0i i j j ? ? ?,×i j = k ? ? ?× = ?j i k Thus ? ? ?(1.80 N m) (0.750 N m)( ) ( 1.05 N m) .? = ? ? ? ? ? = ? ?k k k! EVALUATE: The calculation gives that ?! is in the -direction.z? This agrees with what we got from the right- hand rule. 10.6. IDENTIFY: Use sinFl rF? ?= = for the magnitude of the torque and the right-hand rule for the direction. SET UP: In part (a), 0.250 mr = and 37? = ° EXECUTE: (a) (17.0 N)(0.250 m)sin37 2.56 N m? = = ?° . The torque is counterclockwise. (b) The torque is maximum when 90? = ° and the force is perpendicular to the wrench. This maximum torque is (17.0 N)(0.250 m) 4.25 N m= ? . EVALUATE: If the force is directed along the handle then the torque is zero. The torque increases as the angle between the force and the handle increases. 10.7. IDENTIFY: Apply z zI? ?=? . SET UP: 0 0z? = . 2 rad/rev(400 rev/min) 41.9 rad/s60 s/minz ?? ? ?= =? ?? ? EXECUTE: ( )20 41.9 rad/s2.50 kg m 13.1 N m. 8.00 s z z z zI ? I t ? ?? ?= = = ? = ? EVALUATE: In z zI? ?= , z? must be in 2rad/s . 10.8. IDENTIFY: Use a constant acceleration equation to calculate z? and then apply z zI? ?=? . SET UP: 2 223 2 , where 8.40 kg, 2.00 kgI MR mR M m= + = = , so 20.600 kg mI = ? . 0 75.0 rpm 7.854 rad s; 50.0 rpm 5.236 rad s; 30.0 sz z? ? t= = = = = . EXECUTE: z 2 0 gives 0.08726 rad sz z z? ? ? t ?= + = ? . 0.0524 N mz z? I ?= = ? ? EVALUATE: The torque is negative because its direction is opposite to the direction of rotation, which must be the case for the speed to decrease. 10.9. IDENTIFY: Use z zI? ?=? to calculate ? . Use a constant angular acceleration kinematic equation to relate z? , z? and t. SET UP: For a solid uniform sphere and an axis through its center, 225I MR= . Let the direction the sphere is spinning be the positive sense of rotation. The moment arm for the friction force is 0.0150 ml = and the torque due to this force is negative. 10-4 Chapter 10 EXECUTE: (a) 222 5 (0.0200 N)(0.0150 m) 14.8 rad/s (0.225 kg)(0.0150 m) z z I ?? ?= = = ? (b) 0 22.5 rad/sz z? ?? = ? . 0z z zt? ? ?= + gives 0 222.5 rad/s 1.52 s14.8 rad/s z z z t ? ? ? ? ?= = =? . EVALUATE: The fact that z? is negative means its direction is opposite to the direction of spin. The negative z? causes z? to decrease. 10.10. IDENTIFY: Apply z zI? ?=? to the wheel. The acceleration a of a point on the cord and the angular acceleration ? of the wheel are related by a R?= . SET UP: Let the direction of rotation of the wheel be positive. The wheel has the shape of a disk and 212I MR= . The free-body diagram for the wheel is sketched in Figure 10.10a for a horizontal pull and in Figure 10.10b for a vertical pull. P is the pull on the cord and F is the force exerted on the wheel by the axle. EXECUTE: (a) 221 2 (40.0 N)(0.250 m) 34.8 rad/s (9.20 kg)(0.250 m) z z I ?? = = = . 2 2(0.250 m)(34.8 rad/s ) 8.70 m/sa R?= = = . (b) xF P= ? , yF Mg= ? . 2 2 2 2 2( ) (40.0 N) ([9.20 kg][9.80 m/s ]) 98.6 NF P Mg= + = + = . 2(9.20 kg)(9.80 m/s ) tan 40.0 N y x F Mg F P ? = = = and 66.1? = ° . The force exerted by the axle has magnitude 98.6 N and is directed at 66.1° above the horizontal, away from the direction of the pull on the cord. (c) The pull exerts the same torque as in part (a), so the answers to part (a) don?t change. In part (b), F P Mg+ = and 2(9.20 kg)(9.80 m/s ) 40.0 N 50.2 NF Mg P= ? = ? = . The force exerted by the axle has magnitude 50.2 N and is upward. EVALUATE: The weight of the wheel and the force exerted by the axle produce no torque because they act at the axle. Figure 10.10 10.11. IDENTIFY: Use a constant angular acceleration equation to calculate z? and then apply z zI? ?=? to the motion of the cylinder. k kf n?= . SET UP: ( )( )22 21 12 2 8.25 kg 0.0750 m 0.02320 kg mI mR= = = ? . Let the direction the cylinder is rotating be positive. 0 0220 rpm 23.04 rad/s; 0; 5.25 rev 33.0 radz z? ? ? ?= = = ? = = . EXECUTE: ( )2 20 02z z z? ? ? ? ?= + ? gives 28.046 rad/sz? = ? . k kz f? ? f R ? nR? = = ? = ? . Then z zI? ?=? gives k z? nR I ?? = and k 7.47 Nz I ? n ? R = = . EVALUATE: The friction torque is directed opposite to the direction of rotation and therefore produces an angular acceleration that slows the rotation. 10.12. IDENTIFY: Apply m=?F a! ! to the stone and z zI? ?=? to the pulley. Use a constant acceleration equation to find a for the stone. SET UP: For the motion of the stone take y+ to be downward. The pulley has 212I MR= . a R?= . Dynamics of Rotational Motion 10-5 EXECUTE: (a) 210 0 2y yy y v t a t? = + gives ( )21212.6 m 3.00 sya= and 22.80 m sya = . Then y yF ma=? applied to the stone gives mg T ma? = . z zI? ?=? applied to the pulley gives ( )2 21 12 2 /TR MR MR a R?= = . 12T Ma= . Combining these two equations to eliminate T gives 2 2 2 10.0 kg 2.80 m/s 2.00 kg 2 2 9.80 m/s 2.80 m/s M a M g a ? ?? ? ? ?= = =? ?? ? ? ?? ?? ?? ? ? ? . (b) ( )( )21 1 10.0 kg 2.80 m/s 14.0 N 2 2 T Ma= = = EVALUATE: The tension in the wire is less than the weight 19.6 Nmg = of the stone, because the stone has a downward acceleration. 10.13. IDENTIFY: Use the kinematic information to solve for the angular acceleration of the grindstone. Assume that the grindstone is rotating counterclockwise and let that be the positive sense of rotation. Then apply Eq.(10.7) to calculate the friction force and use k kf n?= to calculate k.? SET UP: 0 850 rev/min(2 rad/1 rev)(1 min/60 s) 89.0 rad/sz? ?= = 7.50 s;t = 0z? = (comes to rest); ?z? = EXECUTE: 0z z zt? ? ?= + 20 89.0 rad/s 11.9 rad/s 7.50 sz ? ?= = ? SET UP: Apply z zI? ?=? to the grindstone. The free-body diagram is given in Figure 10.13. Figure 10.13 The normal force has zero moment arm for rotation about an axis at the center of the grindstone, and therefore zero torque. The only torque on the grindstone is that due to the friction force kf exerted by the ax; for this force the moment arm is l R= and the torque is negative. EXECUTE: k kz f R nR? ?= ? = ?? 21 2I MR= (solid disk, axis through center) Thus z zI? ?=? gives ( )21k 2 znR MR? ?? = 2 k (50.0 kg)(0.260 m)( 11.9 rad/s ) 0.483 2 2(160 N) zMR n ?? ?= ? = ? = EVALUATE: The friction torque is clockwise and slows down the counterclockwise rotation of the grindstone. 10.14. IDENTIFY: Apply y yF ma=? to the bucket, with y+ downward. Apply z zI? ?=? to the cylinder, with the direction the cylinder rotates positive. SET UP: The free-body diagram for the bucket is given in Fig.10.14a and the free-body diagram for the cylinder is given in Fig.10.14b. 212I MR= . (bucket) (cylinder)a R?= EXECUTE: (a) For the bucket, mg T ma? = . For the cylinder, z zI? ?=? gives 212TR MR ?= . /a R? = then gives 12T Ma= . Combining these two equations gives 12mg Ma ma? = and 2 215.0 kg (9.80 m/s ) 7.00 m/s / 2 15.0 kg 6.0 kg mg a m M ? ?= = =? ?+ +? ? . 2 2( ) (15.0 kg)(9.80 m/s 7.00 m/s ) 42.0 NT m g a= ? = ? = . (b) 2 20 02 ( )y y yv v a y y= + ? gives 22(7.00 m/s )(10.0 m) 11.8 m/syv = = . (c) 27.00 m/sya = , 0 0yv = , 0 10.0 my y? = . 210 0 2y yy y v t t?? = + gives 0 22( ) 2(10.0 m) 1.69 s7.00 m/sy y y t a ?= = = 10-6 Chapter 10 (d) y yF ma=? applied to the cylinder gives 0n T Mg? ? = and 242.0 N (12.0 kg)(9.80 m/s ) 160 Nn T mg= + = + = . EVALUATE: The tension in the rope is less than the weight of the bucket, because the bucket has a downward acceleration. If the rope were cut, so the bucket would be in free-fall, the bucket would strike the water in 2 2(10.0 m) 1.43 s 9.80 m/s t = = and would have a final speed of 14.0 m/s. The presence of the cylinder slows the fall of the bucket. Figure 10.14 10.15. IDENTIFY: Apply m=?F a! ! to each book and apply z zI? ?=? to the pulley. Use a constant acceleration equation to find the common acceleration of the books. SET UP: 1 2.00 kgm = , 2 3.00 kgm = . Let 1T be the tension in the part of the cord attached to 1m and 2T be the tension in the part of the cord attached to 2m . Let the -directionx+ be in the direction of the acceleration of each book. a R?= . EXECUTE: (a) 210 0 2x xx x v t a t? = + gives 202 22( ) 2(1.20 m) 3.75 m/s(0.800 s)x x x a t ?= = = . 21 3.75 m/sa = so 1 1 1 7.50 NT m a= = and ( )2 2 1 18.2 NT m g a= ? = . (b) The torque on the pulley is ( )2 1 0.803 N m,T T R? = ? and the angular acceleration is 2 2 1 50 rad/s , so 0.016 kg m .a R I? ? ?= = = = ? EVALUATE: The tensions in the two parts of the cord must be different, so there will be a net torque on the pulley. 10.16. IDENTIFY: Apply m?F = a! ! to each box and z zI? ?=? to the pulley. The magnitude a of the acceleration of each box is related to the magnitude of the angular acceleration ? of the pulley by a R?= . SET UP: The free-body diagrams for each object are shown in Figure 10.16a-c. For the pulley, 0.250 mR = and 21 2I MR= . 1T and 2T are the tensions in the wire on either side of the pulley. 1 12.0 kgm = , 2 5.00 kgm = and 2.00 kgM = . F! is the force that the axle exerts on the pulley. For the pulley, let clockwise rotation be positive. EXECUTE: (a) x xF ma=? for the 12.0 kg box gives 1 1T m a= . y yF ma=? for the 5.00 kg weight gives 2 2 2m g T m a? = . z zI? ?=? for the pulley gives 212 1 2( ) ( )T T R MR ?? = . a R?= and 12 1 2T T Ma? = . Adding these three equations gives 12 1 2 2( )m g m m M a= + + and 2 22 1 1 2 2 5.00 kg (9.80 m/s ) 2.72 m/s 12.0 kg 5.00 kg 1.00 kg m a g m m M ? ? ? ?= = =? ? ? ?+ + + +? ?? ? . Then 2 1 1 (12.0 kg)(2.72 m/s ) 32.6 NT m a= = = . 2 2 2m g T m a? = gives 2 2 2 2 ( ) (5.00 kg)(9.80 m/s 2.72 m/s ) 35.4 NT m g a= ? = ? = . The tension to the left of the pulley is 32.6 N and below the pulley it is 35.4 N. (b) 22.72 m/sa = (c) For the pulley, x xF ma=? gives 1 32.6 NxF T= = and y yF ma=? gives 2 2 (2.00 kg)(9.80 m/s ) 35.4 N 55.0 NyF Mg T= + = + = . Dynamics of Rotational Motion 10-7 EVALUATE: The equation 12 1 2 2( )m g m m M a= + + says that the external force 2m g must accelerate all three objects. Figure 10.16 10.17. IDENTIFY: Apply z zI? ?=? to the post and m?F = a! ! to the hanging mass. The acceleration a! of the mass has the same magnitude as the tangential acceleration tana r?= of the point on the post where the string is attached; 1.75 m 0.500 m 1.25 mr = ? = . SET UP: The free-body diagrams for the post and mass are given in Figures 10.17a and b. The post has 21 3I ML= , with 15.0 kgM = and 1.75 mL = . EXECUTE: (a) z zI? ?=? for the post gives ( )213Tr ML ?= . a r?= so ar? = and 2 23 ML T a r ? ?= ? ?? ? . y yF ma=? for the mass gives mg T ma? = . These two equations give 2 2( /[3 ])mg m ML r a= + and 2 2 2 2 2 2 5.00 kg (9.80 m/s ) 3.31 m/s /[3 ] 5.00 kg [15.0 kg][1.75 m] /3[1.25 m] m a g m ML r ? ?? ?= = =? ?? ?+ +? ? ? ? . 2 23.31 m/s 2.65 rad/s 1.25 m a r ? = = = . (b) No. As the post rotates and the point where the string is attached moves in an arc of a circle, the string is no longer perpendicular to the post. The torque due to this tension changes and the acceleration due to this torque is not constant. (c) From part (a), 23.31 m/sa = . The acceleration of the mass is not constant. It changes as ? for the post changes. EVALUATE: At the instant the cable breaks the tension in the string is less than the weight of the mass because the mass accelerates downward and there is a net downward force on it. Figure 10.17 10.18. IDENTIFY: Apply z zI? ?=? to the rod. SET UP: For the rod and axis at one end, 213I Ml= . EXECUTE: 21 3 3 . Fl F I Ml Ml ?? = = = EVALUATE: Note that ? decreases with the length of the rod, even though the torque increases. 10.19. IDENTIFY: Since there is rolling without slipping, cmv R?= . The kinetic energy is given by Eq.(10.8). The velocities of points on the rim of the hoop are as described in Figure 10.13 in chapter 10. SET UP: 3.00 rad/s? = and 0.600 mR = . For a hoop rotating about an axis at its center, 2I MR= . 10-8 Chapter 10 EXECUTE: (a) cm (0.600 m)(3.00 rad/s) 1.80 m/sv R?= = = . (b) 2 2 2 2 2 2 21 1 1 1cm cm cm cm2 2 2 2 ( )( / ) (2.20 kg)(1.80 m/s) 7.13 JK Mv I Mv MR v R Mv?= + = + = = = (c) (i) cm2 3.60 m/sv v= = . v! is to the right. (ii) 0v = (iii) 2 2 2 2cm tan cm cm( ) 2 2.55 m/sv v v v R v?= + = + = = . v! at this point is at 45° below the horizontal. (d) To someone moving to the right at cmv v= , the hoop appears to rotate about a stationary axis at its center. (i) 1.80 m/sv R?= = , to the right. (ii) 1.80 m/sv = , to the left. (iii) 1.80 m/sv = , downward. EVALUATE: For the special case of a hoop, the total kinetic energy is equally divided between the motion of the center of mass and the rotation about the axis through the center of mass. In the rest frame of the ground, different points on the hoop have different speed. 10.20. IDENTIFY: Only gravity does work, so other 0W = and conservation of energy gives i i f fK U K U+ = + . 2 21 1 f cm cm2 2K Mv I ?= + . SET UP: Let f 0y = , so f 0U = and i 0.750 my = . The hoop is released from rest so i 0K = . cmv R?= . For a hoop with an axis at its center, 2cmI MR= . EXECUTE: (a) Conservation of energy gives i fU K= . 2 2 2 2 2 21 1f 2 2 ( )K MR MR MR? ? ?= + = , so 2 2 iMR Mgy? = . 2 i (9.80 m/s )(0.750 m) 33.9 rad/s 0.0800 m gy R ? = = = . (b) (0.0800 m)(33.9 rad/s) 2.71 m/sv R?= = = EVALUATE: An object released from rest and falling in free-fall for 0.750 m attains a speed of 2 (0.750 m) 3.83 m/sg = . The final speed of the hoop is less than this because some of its energy is in kinetic energy of rotation. Or, equivalently, the upward tension causes the magnitude of the net force of the hoop to be less than its weight. 10.21. IDENTIFY: Apply Eq.(10.8). SET UP: For an object that is rolling without slipping, cmv R?= . EXECUTE: The fraction of the total kinetic energy that is rotational is ( ) ( ) ( ) 2 cm 2 2 2 2 2 cm cm cm cm cm 1 2 1 1 1 2 1 2 1( / ) / 1 ( / ) I Mv I M I v MR I ? ? ?= =+ + (a) 2cm (1 2) , so the above ratio is 1 3.I MR= (b) 2cm (2 5)I MR= so the above ratio is 2 7 . (c) 2cm (2 3)I MR= so the ratio is 2 5 . (d) 2cm (5 8)I MR= so the ratio is 5 13. EVALUATE: The moment of inertia of each object takes the form 2I MR?= . The ratio of rotational kinetic energy to total kinetic energy can be written as 1 1 1/ 1 ? ? ?=+ + . The ratio increases as ? increases. 10.22. IDENTIFY: Apply m=?F a! ! to the translational motion of the center of mass and z zI? ?=? to the rotation about the center of mass. SET UP: Let x+ be down the incline and let the shell be turning in the positive direction. The free-body diagram for the shell is given in Fig.10.22. From Table 9.2, 22cm 3I mR= . EXECUTE: x xF ma=? gives cmsinmg f ma? ? = . z zI? ?=? gives 223( )fR mR ?= . With cm /a R? = this becomes 2 cm3f ma= . Combining the equations gives 2 cm cm3sinmg ma ma? ? = and 2 2 cm 3 sin 3(9.80 m/s )(sin38.0 ) 3.62 m/s 5 5 g a ?= = =° . 22 2cm3 3 (2.00 kg)(3.62 m/s ) 4.83 Nf ma= = = . The friction is static since there is no slipping at the point of contact. cos 15.45 Nn mg ?= = . s 4.83 N 0.31315.45 N f n ? = = = . (b) The acceleration is independent of m and doesn?t change. The friction force is proportional to m so will double; 9.66 Nf = . The normal force will also double, so the minimum s? required for no slipping wouldn?t change. Dynamics of Rotational Motion 10-9 EVALUATE: If there is no friction and the object slides without rolling, the acceleration is sing ? . Friction and rolling without slipping reduce a to 0.60 times this value. Figure 10.22 10.23. IDENTIFY: Apply ext cmm=?F a! ! and cmz zI? ?=? to the motion of the ball. (a) SET UP: The free-body diagram is given in Figure 10.23a. EXECUTE: y yF ma=? cosn mg ?= and s s cosf mg? ?= x xF ma=? ssin cosmg mg ma? ? ?? = s(sin cos )g a? ? ?? = (eq. 1) Figure 10.23a SET UP: Consider Figure 10.23b. n and mg act at the center of the ball and provide no torque Figure 10.23b EXECUTE: s cos ;f mg R? ? ? ?= =? 225I mR= cmz zI? ?=? gives 22s 5cosmg R mR? ? ?= No slipping means / ,a R? = so 2s 5cosg a? ? = (eq.2) We have two equations in the two unknowns a and s.? Solving gives 57 sina g ?= and 2 2 s 7 7tan tan 65.0 0.613? ?= = ° = (b) Repeat the calculation of part (a), but now 223 .I mR= 35 sina g ?= and 2 2s 5 5tan tan65.0 0.858? ?= = ° = The value of s? calculated in part (a) is not large enough to prevent slipping for the hollow ball. (c) EVALUATE: There is no slipping at the point of contact. More friction is required for a hollow ball since for a given m and R it has a larger I and more torque is needed to provide the same .? Note that the required s? is independent of the mass or radius of the ball and only depends on how that mass is distributed. 10.24. IDENTIFY: Apply conservation of energy to the motion of the marble. SET UP: 2 21 12 2K mv I ?= + , with 225I MR= . cm for no slippingv R?= . Let 0y = at the bottom of the bowl. The marble at its initial and final locations is sketched in Figure 10.24. 10-10 Chapter 10 EXECUTE: (a) Motion from the release point to the bottom of the bowl: 2 21 1 2 2 mgh mv I ?= + . 2 2 21 1 2 2 2 5 v mgh mv mR R ? ?? ?= + ? ?? ?? ?? ? and 10 7 v gh= . Motion along the smooth side: The rotational kinetic energy does not change, since there is no friction torque on the marble, 2 rot rot 1 2 mv K mgh K?+ = + . 2 10 7 5 2 2 7 ghv h h g g ? = = = (b) mgh mgh ?= so h h? = . EVALUATE: (c) With friction on both halves, all the initial potential energy gets converted back to potential energy. Without friction on the right half some of the energy is still in rotational kinetic energy when the marble is at its maximum height. Figure 10.24 10.25. IDENTIFY: Apply conservation of energy to the motion of the wheel. SET UP: The wheel at points 1 and 2 of its motion is shown in Figure 10.25. Take y = 0 at the center of the wheel when it is at the bottom of the hill. Figure 10.25 The wheel has both translational and rotational motion so its kinetic energy is 2 21 1cm cm2 2 .K I Mv?= + EXECUTE: 1 1 other 2 2K U W K U+ + = + other fric 3500 JW W= = ? (the friction work is negative) 2 21 1 1 1 12 2 ;K I Mv?= + v R?= and 20.800I MR= so 2 2 2 2 2 21 1 1 1 1 12 2(0.800) 0.900K MR MR MR? ? ?= + = 2 0,K = 1 0,U = 2U Mgh= Thus 2 21 fric0.900MR W Mgh? + = 2/ 392 N/(9.80 m/s ) 40.0 kgM w g= = = 2 2 1 fric0.900MR Wh Mg ? += 2 2 2 (0.900)(40.0 kg)(0.600 m) (25.0 rad/s) 3500 J 11.7 m (40.0 kg)(9.80 m/s ) h ?= = EVALUATE: Friction does negative work and reduces h. 10.26. IDENTIFY: Apply z zI? ?=? and m=?F a! ! to the motion of the bowling ball. SET UP: cma R?= . s sf n?= . Let x+ be directed down the incline. EXECUTE: (a) The free-body diagram is sketched in Figure 10.26. The angular speed of the ball must decrease, and so the torque is provided by a friction force that acts up the hill. (b) The friction force results in an angular acceleration, given by .I fR? = m=?F a! ! applied to the motion of the center of mass gives cm,sinmg f ma? ? = and the acceleration and angular acceleration are related by cma R?= . Combining, ( )2sin 1 7 5Img ma mamR? ? ?= + =? ?? ? . ( )cm 5 7 sina g ?= . Dynamics of Rotational Motion 10-11 (c) From either of the above relations between f and cm ,a cm s s 2 2 sin cos 5 7 f ma mg n mg? ? ? ?= = ? = . ( )s 2 7 tan .? ?? EVALUATE: If s 0? = , cm sina mg ?= . cma is less when friction is present. The ball rolls farther uphill when friction is present, because the friction removes the rotational kinetic energy and converts it to gravitational potential energy. In the absence of friction the ball retains the rotational kinetic energy that is has initially. Figure 10.26 10.27. (a) IDENTIFY: Use Eq.(10.7) to find z? and then use a constant angular acceleration equation to find .z? SET UP: The free-body diagram is given in Figure 10.27. EXECUTE: Apply z zI? ?=? to find the angular acceleration: zFR I ?= 2 2 (18.0 N)(2.40 m) 0.02057 rad/s 2100 kg mz FR I ? = = =? Figure 10.27 SET UP: Use the constant z? kinematic equations to find .z? ?;z? = 0 z? (initially at rest); 20.02057 rad/s ;z? = 15.0 st = EXECUTE: 20 0 (0.02057 rad/s )(15.0 s) 0.309 rad/sz z zt? ? ?= + = + = (b) IDENTIFY and SET UP: Calculate the work from Eq.(10.21), using a constant angular acceleration equation to calculate 0,? ?? or use the work-energy theorem. We will do it both ways. EXECUTE: (1) zW ? ?= ? (Eq.(10.21)) 2 2 21 1 0 0 2 20 (0.02057 rad/s )(15.0 s) 2.314 radz zt t? ? ? ? ?? = ? = + = + = (18.0 N)(2.40 m) 43.2 N mzt FR= = = ? Then (43.2 N m)(2.314 rad) 100 J.zW ? ?= ? = ? = or (2) tot 2 1W K K= ? (the work-energy relation from chapter 6) tot ,W W= the work done by the child 1 0;K = 2 2 21 12 2 2 (2100 kg m )(0.309 rad/s) 100 JK I ?= = ? = Thus 100 J,W = the same as before. EVALUATE: Either method yields the same result for W. (c) IDENTIFY and SET UP: Use Eq.(6.15) to calculate avP EXECUTE: av 100 J 6.67 W15.0 s W P t ?= = =? EVALUATE: Work is in joules, power is in watts. 10.28. IDENTIFY: Apply P ??= and W ? ?= ? . SET UP: P must be in watts, ?? must be in radians, and ? must be in rad/s. 1 rev 2 rad?= . 1 hp 746 W= . rad/s 30 rev/min? = . EXECUTE: (a) ( )( ) ( ) 175 hp 746 W / hp 519 N m. rad/s 2400 rev/min 30 rev/min P? ??= = = ?? ?? ?? ? (b) ( )( )519 N m 2 rad 3260 JW ? ? ?= ? = ? = 10-12 Chapter 10 EVALUATE: 40 rev/s? = , so the time for one revolution is 0.025 s. 51.306 10 WP = × , so in one revolution, 3260 JW Pt= = , which agrees with our previous result. 10.29. IDENTIFY: Apply z zI? ?=? and constant angular acceleration equations to the motion of the wheel. SET UP: 1 rev 2 rad?= . rad/s 30 rev/min? = . EXECUTE: (a) 0 z zz zI I t ? ?? ? ?= = . ( )( )( )( )( )2 rad s1 2 1.50 kg 0.100 m 1200 rev min 30 rev min 0.377 N m 2.5 sz ? ? ? ?? ?? ?= = ? (b) ( )( ) av 600 rev/min 2.5 s 25.0 rev 157 rad. 60 s/min t? ? = = = (c) (0.377 N m)(157 rad) 59.2 JW ? ?= ? = ? = . (d) ( ) 22 21 1 rad/s (1/ 2)(1.5 kg)(0.100 m) (1200 rev/min) 59.2 J 2 2 30 rev/min K I ?? ? ?? ?= = =? ?? ?? ?? ? . the same as in part (c). EVALUATE: The agreement between the results of parts (c) and (d) illustrates the work-energy theorem 10.30. IDENTIFY: The power output of the motor is related to the torque it produces and to its angular velocity by z zP ? ?= , where z? must be in rad/s. SET UP: The work output of the motor in 60.0 s is 2 (9.00 kJ) 6.00 kJ 3 = , so 6.00 kJ 100 W 60.0 s P = = . 2500 rev/min 262 rad/sz? = = . EXECUTE: 100 W 0.382 N m 262 rad/sz z P? ?= = = ? EVALUATE: For a constant power output, the torque developed decreases and the rotation speed of the motor increases. 10.31. IDENTIFY: Apply FR? = and P ??= . SET UP: 1 hp 746 W= . rad/s 30 rev/min? = EXECUTE: (a) With no load, the only torque to be overcome is friction in the bearings (neglecting air friction), and the bearing radius is small compared to the blade radius, so any frictional torque can be neglected. (b) / (1.9 hp)(746 W/hp) 65.6 N. rad/s (2400 rev/min) (0.086 m) 30 rev/min P F R R ? ? ?= = = =? ?? ?? ? EVALUATE: In P I ?= , ? must be in watts and ? must be in rad/s. 10.32. IDENTIFY: Apply z zI? ?=? to the motion of the propeller and then use constant acceleration equations to analyze the motion. W ? ?= ? . SET UP: 2 2 21 12 2 (117 kg)(2.08 m) 42.2 kg mI mL= = = ? . EXECUTE: (a) 221950 N m 46.2 rad/s .42.2 kg mI ?? ?= = =? (b) 2 20 02 ( )z z z? ? ? ? ?= + ? gives 22 2(46.2 rad/s )(5.0 rev)(2 rad/rev) 53.9 rad/s.? ?? ?= = = (c) 4(1950 N m)(5.00 rev)(2 rad/rev) 6.13 10 J.W ?? ?= = ? = × (d) 0 2 53.9 rad/s 1.17 s 46.2 rad/s z z z t ? ? ? ?= = = . 4 av 6.13 10 J 52.5 kW 1.17 s W P t ×= = =? . EVALUATE: P ??= . ? is constant and? is linear in t, so avP is half the instantaneous power at the end of the 5.00 revolutions. We could also calculate W from 2 2 2 41 12 2 (42.2 kg m )(53.9 rad/s) 6.13 10 JW K I ?= ? = = ? = × . 10.33. (a) IDENTIFY and SET UP: Use Eq.(10.23) and solve for .z? ,z zP ? ?= where z? must be in rad/s EXECUTE: (4000 rev/min)(2 rad/1 rev)(1 min/60 s) 418.9 rad/sz? ?= = 51.50 10 W 358 N m 418.9 rad/sz z P? ? ×= = = ? Dynamics of Rotational Motion 10-13 (b) IDENTIFY and SET UP: Apply m=?F a! ! to the drum. Find the tension T in the rope using z? from part (a). The system is sketched in Figure 10.33. EXECUTE: v constant implies 0a = and T w= z TR? = implies / 358 N m/0.200 m 1790 NzT R?= = ? = Thus a weight 1790 Nw = can be lifted. Figure 10.33 (c) IDENTIFY and SET UP: Use .v R?= EXECUTE: The drum has 418.9 rad/s,? = so (0.200 m)(418.9 rad/s) 83.8 m/sv = = EVALUATE: The rate at which T is doing work on the drum is (1790 N)(83.8 m/s) 150 kW.P Tv= = = This agrees with the work output of the motor. 10.34. IDENTIFY: L I ?= and disk womanI I I= + . SET UP: 0.50 rev/s 3.14 rad/s? = = . 21disk disk2I m R= and 2woman womanI m R= . EXECUTE: 2 2(55 kg 50.0 kg)(4.0 m) 1680 kg mI = + = ? . 2 3 2(1680 kg m )(3.14 rad/s) 5.28 10 kg m /sL = ? = × ? EVALUATE: The disk and the woman have similar values of I , even though the disk has twice the mass. 10.35. (a) IDENTIFY: Use sinL mvr ?= (Eq.(10.25)): SET UP: Consider Figure 10.35. EXECUTE: sinL mvr ?= = (2.00 kg)(12.0 m/s)(8.00 m)sin143.1° 2115 kg m /sL = ? Figure 10.35 To find the direction of L ! apply the right-hand rule by turning r ! into the direction of v ! by pushing on it with the fingers of your right hand. Your thumb points into the page, in the direction of .L ! (b) IDENTIFY and SET UP: By Eq.(10.26) the rate of change of the angular momentum of the rock equals the torque of the net force acting on it. EXECUTE: 2 2(8.00 m)cos36.9 125 kg m /smg? = ° = ? To find the direction of ?! and hence of / ,d dtL! apply the right-hand rule by turning r! into the direction of the gravity force by pushing on it with the fingers of your right hand. Your thumb points out of the page, in the direction of / .d dtL ! EVALUATE: L! and /d dtL! are in opposite directions, so L is decreasing. The gravity force is accelerating the rock downward, toward the axis. Its horizontal velocity is constant but the distance l is decreasing and hence L is decreasing. 10.36. IDENTIFY: z zL I ?= SET UP: For a particle of mass m moving in a circular path at a distance r from the axis, 2I mr= and v r?= . For a uniform sphere of mass M and radius R and an axis through its center, 225I MR= . The earth has mass 24 E 5.97 10 kgm = × , radius 6E 6.38 10 mR = × and orbit radius 111.50 10 mr = × . The earth completes one rotation on its axis in 24 h 86,400 s= and one orbit in 71 y 3.156 10 s= × . EXECUTE: (a) 2 24 11 2 40 272 rad(5.97 10 kg)(1.50 10 m) 2.67 10 kg m /s3.156 10 sz z zL I mr ?? ? ? ?= = = × × = × ?? ?×? ? . The radius of the earth is much less than its orbit radius, so it is very reasonable to model it as a particle for this calculation. (b) ( )2 24 6 2 33 22 25 5 2 rad(5.97 10 kg)(6.38 10 m) 7.07 10 kg m /s86,400 sz zL I MR ?? ? ? ?= = = × × = × ?? ?? ? EVALUATE: The angular momentum associated with each of these motions is very large. 10-14 Chapter 10 10.37. IDENTIFY and SET UP: Use L I ?= EXECUTE: The second hand makes 1 revolution in 1 minute, so (1.00 rev/min)(2 rad/1 rev)(1 min/60 s) 0.1047 rad/s? ?= = For a slender rod, with the axis about one end, 2 3 2 5 21 1 3 3 (6.00 10 kg)(0.150 m) 4.50 10 kg mI ML ? ?= = × = × ? Then 5 2 6 2(4.50 10 kg m )(0.1047 rad/s) 4.71 10 kg m /s.L I ? ? ?= = × ? = × ? EVALUATE: L! is clockwise. 10.38. IDENTIFY: /z d dt? ?= . z zL I ?= and z zdL dt? = . SET UP: For a hollow, thin-walled sphere rolling about an axis through its center, 223I MR= . 0.240 mR = . EXECUTE: (a) 21.50 rad/sA = and 41.10 rad/sB = , so that ( )t? will have units of radians. (b) (i) 32 4z d At Bt dt ?? = = + . At 3.00 st = , 2 4 32(1.50 rad/s )(3.00 s) 4(1.10 rad/s )(3.00 s) 128 rad/sz? = + = . 2 2 22 2 3 3( ) (12.0 kg)(0.240 m) (128 rad/s) 59.0 kg m /sz zL MR ?= = = ? . (ii) 2(2 12 )z zz dL d I I A Bt dt dt ?? = = = + and 2 2 4 22 3 (12.0 kg)(0.240 m) (2[1.50 rad/s ] 12[1.10 rad/s ][3.00 s] ) 56.1 N mz? = + = ? . EVALUATE: The angular speed of rotation is increasing. This increase is due to an acceleration z? that is produced by the torque on the sphere. When I is constant, as it is here, /z z z zdL dt Id dt I? ? ?= = = and Equations (10.29) and (10.7) are identical. 10.39. IDENTIFY: Apply conservation of angular momentum. SET UP: For a uniform sphere and an axis through its center, 225I MR= . EXECUTE: The moment of inertia is proportional to the square of the radius, and so the angular velocity will be proportional to the inverse of the square of the radius, and the final angular velocity is 2 25 31 2 1 2 2 rad 7.0 10 km 4.6 10 rad s. (30 d)(86,400 s d) 16 km R R ?? ? ? ? ? ?? ? ×= = = ×? ? ? ?? ?? ?? ?? ? EVALUATE: 21 12 2K I L? ?= = . L is constant and ? increases by a large factor, so there is a large increase in the rotational kinetic energy of the star. This energy comes from potential energy associated with the gravity force within the star. 10.40. IDENTIFY and SET UP: L! is conserved if there is no net external torque. Use conservation of angular momentum to find ? at the new radius and use 212K I ?= to find the change in kinetic energy, which is equal to the work done on the block. EXECUTE: (a) Yes, angular momentum is conserved. The moment arm for the tension in the cord is zero so this force exerts no torque and there is no net torque on the block. (b) 1 2L L= so 1 1 2 2.I I? ?= Block treated as a point mass, so 2,I mr= where r is the distance of the block from the hole. 2 2 1 1 2 2mr mr? ?= 2 2 1 2 1 2 0.300 m (1.75 rad/s) 7.00 rad/s 0.150 m r r ? ?? ? ? ?= = =? ? ? ?? ?? ? (c) 2 2 2 21 1 11 1 1 1 1 12 2 2K I mr mv? ?= = = 1 1 1 (0.300 m)(1.75 rad/s) 0.525 m/sv r?= = = 2 21 1 1 12 2 (0.0250 kg)(0.525 m/s) 0.00345 JK mv= = = 21 2 22K mv= 2 2 2 (0.150 m)(7.00 rad/s) 1.05 m/sv r ?= = = 2 21 1 2 22 2 (0.0250 kg)(1.05 m/s) 0.01378 JK mv= = = 2 1 0.01378 J 0.00345 J 0.0103 JK K K? = ? = ? = Dynamics of Rotational Motion 10-15 (d) totW K= ? But tot ,W W= the work done by the tension in the cord, so 0.0103 JW = EVALUATE: Smaller r means smaller I . L I ?= is constant so ? increases and K increases. The work done by the tension is positive since it is directed inward and the block moves inward, toward the hole. 10.41. IDENTIFY: Apply conservation of angular momentum to the motion of the skater. SET UP: For a thin-walled hollow cylinder 2I mR= . For a slender rod rotating about an axis through its center, 21 12I ml= . EXECUTE: i fL L= so i i f fI I? ?= . 2 2 21 i 120.40 kg m (8.0 kg)(1.8 m) 2.56 kg mI = ? + = ? . 2 2 2f 0.40 kg m (8.0 kg)(0.25 m) 0.90 kg mI = ? + = ? . 2 i f i 2 f 2.56 kg m (0.40 rev/s)=1.14 rev/s 0.90 kg m I I ? ?? ? ? ??= =? ? ? ??? ?? ? . EVALUATE: 21 12 2K I L? ?= = . ? increases and L is constant, so K increases. The increase in kinetic energy comes from the work done by the skater when he pulls in his hands. 10.42. IDENTIFY: Apply conservation of angular momentum to the diver. SET UP: The number of revolutions she makes in a certain time is proportional to her angular velocity. The ratio of her untucked to tucked angular velocity is 2 2(3.6 kg m ) /(18 kg m )? ? . EXECUTE: If she had tucked, she would have made 2 2(2 rev)(3.6 kg m ) (18 kg m ) 0.40 rev? ? = in the last 1.0 s, so she would have made (0.40 rev)(1.5 1.0) 0.60 rev= in the total 1.5 s. EVALUATE: Untucked she rotates slower and completes fewer revolutions. 10.43. IDENTIFY and SET UP: There is no net external torque about the rotation axis so the angular momentum L I ?= is conserved. EXECUTE: (a) 1 2L L= gives 1 1 2 2 ,I I? ?= so 2 1 2 1( / )I I? ?= 2 2 21 1 1 tt 2 2 (120 kg)(2.00 m) 240 kg mI I MR= = = = ? 2 2 2 2 2 2 tt p 240 kg m 240 kg m (70 kg)(2.00 m) 520 kg mI I I mR= + = ? + = ? + = ? 2 2 2 1 2 1( / ) (240 kg m /520 kg m )(3.00 rad/s) 1.38 rad/sI I? ?= = ? ? = (b) 2 2 21 11 1 12 2 (240 kg m )(3.00 rad/s) 1080 JK I ?= = ? = 2 2 21 1 2 2 22 2 (520 kg m )(1.38 rad/s) 495 JK I ?= = ? = EVALUATE: The kinetic energy decreases because of the negative work done on the turntable and the parachutist by the friction force between these two objects. The angular speed decreases because I increases when the parachutist is added to the system. 10.44. IDENTIFY: Apply conservation of angular momentum to the collision. SET UP: Let the width of the door be l. The initial angular momentum of the mud is ( / 2)mv l , since it strikes the door at its center. For the axis at the hinge, 21door 3I Ml= and 2mud ( / 2)I m l= . EXECUTE: ( )( ) ( )22 2 1 3 2 mv lL I Ml m l ? = = + . ( )( )( ) ( )( )( ) ( )( )2 2 0.500 kg 12.0 m s 0.500 m 0.223 rad s. 1 3 40.0 kg 1.00 m 0.500 kg 0.500 m ? = =+ Ignoring the mass of the mud in the denominator of the above expression gives 0.225 rad s,? = so the mass of the mud in the moment of inertia does affect the third significant figure. EVALUATE: Angular momentum is conserved but there is a large decrease in the kinetic energy of the system. 10.45. (a) IDENTIFY and SET UP: Apply conservation of angular momentum ,L! with the axis at the nail. Let object A be the bug and object B be the bar. Initially, all objects are at rest and 1 0.L = Just after the bug jumps, it has angular momentum in one direction of rotation and the bar is rotating with angular velocity B? in the opposite direction. EXECUTE: 2 A A B BL m v r I ?= ? where 1.00 mr = and 213B BI m r= 1 2L L= gives 213A A B Bm v r m r ?= 3 0.120 rad/sA AB B m v m r ? = = 10-16 Chapter 10 (b) 1 0;K = 2 21 12 2 2A A B BK m v I ?= + = ( )2 2 2 41 1 12 2 3(0.0100 kg)(0.200 m/s) [0.0500 kg][1.00 m] (0.120 rad/s) 3.2 10 J.?+ = × The increase in kinetic energy comes from work done by the bug when it pushes against the bar in order to jump. EVALUATE: There is no external torque applied to the system and the total angular momentum of the system is constant. There are internal forces, forces the bug and bar exert on each other. The forces exert torques and change the angular momentum of the bug and the bar, but these changes are equal in magnitude and opposite in direction. These internal forces do positive work on the two objects and the kinetic energy of each object and of the system increases. 10.46. IDENTIFY: Apply conservation of angular momentum to the system of earth plus asteroid. SET UP: Take the axis to be the earth?s rotation axis. The asteroid may be treated as a point mass and it has zero angular momentum before the collision, since it is headed toward the center of the earth. For the earth, z zL I ?= and 225I MR= ,where M is the mass of the earth and R is its radius. The length of a day is 2 radT ??= , where ? is the earth?s angular rotation rate. EXECUTE: Conservation of angular momentum applied to the collision between the earth and asteroid gives 2 2 22 2 1 25 5( )MR mR MR? ?= + and 1 225 2 m M ? ? ? ? ??= ? ?? ? . 2 11.250T T= gives 2 1 1 1.250 ? ?= and 1 21.250? ?= . 1 2 2 0.250 ? ? ? ? = . 25 (0.250) 0.100m M M= = . EVALUATE: If the asteroid hit the surface of the earth tangentially it could have some angular momentum with respect to the earth?s rotation axis, and could either speed up or slow down the earth?s rotation rate. 10.47. IDENTIFY: Apply conservation of angular momentum to the collision. SET UP: The system before and after the collision is sketched in Figure 10.47. Let counterclockwise rotation be positive. The bar has 21 23I m L= . EXECUTE: (a) Conservation of angular momentum: 211 0 1 23m v d m vd m L?= ? + . 2 2 1 90.0 N (3.00 kg)(10.0 m s)(1.50 m) (3.00 kg)(6.00 m s)(1.50 m) (2.00 m) 3 9.80 m s ?? ?= ? + ? ?? ? 5 88 rad s.? = . (b) There are no unbalanced torques about the pivot, so angular momentum is conserved. But the pivot exerts an unbalanced horizontal external force on the system, so the linear momentum is not conserved. EVALUATE: Kinetic energy is not conserved in the collision. Figure 10.47 10.48. IDENTIFY: d dt?=L! ! , so dL! is in the direction of ?! . SET UP: The direction of ?! is given by the right-hand rule, as described in Figure 10.26 in the textbook. EXECUTE: The sketches are given in Figures 10.48a?d. Dynamics of Rotational Motion 10-17 EVALUATE: In figures (a) and (c) the precession is counterclockwise and in figures (b) and (d) it is clockwise. When the direction of either ?! or ?! reverses, the direction of precession reverses. Figure 10.48 10.49. IDENTIFY: The precession angular velocity is wr I ?? = , where ? is in rad/s. Also apply m=?F a ! ! to the gyroscope. SET UP: The total mass of the gyroscope is r f 0.140 kg 0.0250 kg 0.165 kgm m+ = + = . 2 rad 2 rad 2.856 rad/s 2.20 sT ? ?? = = = . EXECUTE: (a) 2p tot (0.165 kg)(9.80 m/s ) 1.62 NF w= = = (b) 2 3 4 2 (0.165 kg)(9.80 m/s )(0.0400 m) 189 rad/s 1.80 10 rev/min (1.20 10 kg m )(2.856 rad/s) wr I ? ?= = = = ×? × ? (c) If the figure in the problem is viewed from above, ? ! is in the direction of the precession and L ! is along the axis of the rotor, away from the pivot. EVALUATE: There is no vertical component of acceleration associated with the motion, so the force from the pivot equals the weight of the gyroscope. The larger ? is, the slower the rate of precession. 10.50. IDENTIFY: The precession angular speed is related to the acceleration due to gravity by Eq.(10.33), with w mg= . SET UP: E 0.50 rad/s? = , Eg g= and M 0.165g g= . For the gyroscope, m, r, I , and ? are the same on the moon as on the earth. EXECUTE: mgr I ?? = . constant mr g I ? ? = = , so E M E Mg g ? ?= . M M E E E 0.165 (0.165)(0.50 rad/s) 0.0825 rad/s g g ? ?? = ? = ? = =? ?? ? . EVALUATE: In the limit that 0g ? the precession rate 0? . 10.51. IDENTIFY and SET UP: Apply Eq.(10.33). EXECUTE: (a) halved (b) doubled (assuming that the added weight is distributed in such a way that r and I are not changed) (c) halved (assuming that w and r are not changed) (d) doubled (e) unchanged. EVALUATE: ? is directly proportional to w and r and is inversely proportional to I and ? . 10.52. IDENTIFY: Apply Eq.(10.33), where wr? = . SET UP: 1 day 86,400 s= . 71 yr 3.156 10 s= × . The earth has mass 245.97 10 kgM = × and radius 66.38 10 mR = × . For a uniform sphere and an axis through its center, 225I MR= . EXECUTE: (a) 2(2 /5) .I MR? ? ?= ? = ? Using 2 rad 86,400 s ?? = and 72 rad(26,000 y)(3.156 10 s/y) ?? = × , and the mass and radius of the earth from Appendix F, 5.4 N m? = ? . EVALUATE: If the torque is applied by the sun, 111.5 10 mr = × and 113.6 10 NF? = × . 10-18 Chapter 10 10.53. IDENTIFY: Apply z zI? ?=? and constant acceleration equations to the motion of the grindstone. SET UP: Let the direction of rotation of the grindstone be positive. The friction force is kf n?= and produces torque fR . 2 rad 1 min 4 rad 1 rev 60 s ?? ?? ?? ?= =? ?? ?? ?? ? . 2 21 2 1.69 kg mI MR= = ? . EXECUTE: (a) The net torque must be 20 4 rad/s(1.69 kg m ) 2.36 N m. 9.00 s z zI I t ? ? ?? ? ?= = = ? = ? This torque must be the sum of the applied force FR and the opposing frictional torques f? at the axle and kfR nR?= due to the knife. f k1 ( )F nRR ? ? ?= + + . ( )1 (2.36 N m) (6.50 N m) (0.60)(160 N)(0.260 m) 67.6 N. 0.500 m F = ? + ? + = (b) To maintain a constant angular velocity, the net torque ? is zero, and the force is F ? 1 (6.50 N m 24.96 N m) 62.9 N. 0.500 m F ? = ? + ? = (c) The time t needed to come to a stop is found by taking the magnitudes in Eq.(10.27), with f? ?= constant; ( )2 f f (4 rad/s) 1.69 kg m 3.27 s. 6.50 N m L I t ?? ? ? ?= = = =? EVALUATE: The time for a given change in? is proportional to? , which is in turn proportional to the net torque, so the time in part (c) can also be found as ( ) 2.36 N m9.00 s .6.50 N mt ?= ? 10.54. IDENTIFY: Apply z zI? ?=? and use the constant acceleration equations to relate? to the motion. SET UP: Let the direction the wheel is rotating be positive. 100 rev/min 10.47 rad/s= EXECUTE: (a) 0z z zt? ? ?= + gives 20 10.47 rad/s 0 5.23 rad/s2.00 s z z z t ? ?? ? ?= = = . 2 2 5.00 N m 0.956 kg m 5.23 rad/s z z I ? ? ?= = = ?? (b) 0 10.47 rad/sz? = , 0z? = , 125 st = . 0z z zt? ? ?= + gives 20 0 10.47 rad/s 0.0838 rad/s125 s z z z t ? ?? ? ?= = = ? 2 2(0.956 kg m )( 0.0838 rad/s ) 0.0801 N mz zI? ?= = ? ? = ? ?? (c) 0 10.47 rad/s 0 (125 s) 654 rad 104 rev 2 2 z z t ? ?? + +? ? ? ?= = = =? ? ? ?? ? ? ? EVALUATE: The applied net torque ( 5.00 N m? ) is much larger than the magnitude of the friction torque ( 0.0801 N m? ), so the time of 2.00 s that it takes the wheel to reach an angular speed of 100 rev/min is much less than the 125 s it takes the wheel to be brought to rest by friction. 10.55. IDENTIFY and SET UP: Apply .v r?= v is the tangential speed of a point on the rim of the wheel and equals the linear speed of the car. EXECUTE: (a) 60 mph 26.82 m/sv = = 12 in. 0.3048 mr = = 88.0 rad/s 14.0 rev/s 840 rpm v r ? = = = = (b) Same ? as in part (a) since speedometer reads same. 15 in. 0.381 mr = = (0.381 m)(88.0 rad/s) 33.5 m/s 75 mphv r?= = = = (c) 50 mph 22.35 m/sv = = 10 in. 0.254 mr = = 88.0 rad/s. v r ? = = This is the same as for 60 mph with correct tires, so speedometer read 60 mph. EVALUATE: For a given ,? v increases when r increases. Dynamics of Rotational Motion 10-19 10.56. IDENTIFY: The kinetic energy of the disk is 2 21 1cm2 2K Mv I ?= + . As it falls its gravitational potential energy decreases and its kinetic energy increases. The only work done on the disk is the work done by gravity, so 1 1 2 2K U K U+ = + . SET UP: 2 21cm 2 12 ( )I M R R= + , where 1 0.300 mR = and 2 0.500 mR = . cm 2v R ?= . Take 1 0y = , so 2 1.20 my = ? . EXECUTE: 1 1 2 2K U K U+ = + . 1 0K = , 1 0U = . 2 2K U= ? . 2 21 1cm cm 22 2Mv I Mgy?+ = ? . ( )2 2 2 21 1cm 1 2 cm cm2 4 1 [ / ] 0.340I M R R v Mv? = + = . Then 2cm 20.840Mv Mgy= ? and 2 2 cm (9.80 m/s )( 1.20 m) 3.74 m/s 0.840 0.840 gy v ? ? ?= = = EVALUATE: A point mass in free-fall acquires a speed of 4.85 m/s after falling 1.20 m. The disk has a value of cmv that is less than this, because some of the original gravitational potential energy has been converted to rotational kinetic energy. 10.57. IDENTIFY: Use z zI? ?=? to find the angular acceleration just after the ball falls off and use conservation of energy to find the angular velocity of the bar as it swings through the vertical position. SET UP: The axis of rotation is at the axle. For this axis the bar has 21 bar12I m L= , where bar 3.80 kgm = and 0.800 mL = . Energy conservation gives 1 1 2 2K U K U+ = + . The gravitational potential energy of the bar doesn?t change. Let 1 0y = , so 2 / 2y L= ? . EXECUTE: (a) ball ( / 2)z m g L? = and 2 21ball bar bar ball12 ( / 2)I I I m L m L= + = + . z zI? ?=? gives ball ball 2 21 bar ball ball bar12 ( / 2) 2 ( / 2) /3z m g L g m m L m L L m m ? ? ?= = ? ?+ +? ? and 2 22(9.80 m/s ) 2.50 kg 16.3 rad/s 0.800 m 2.50 kg [3.80 kg]/3z ? ? ?= =? ?+? ? . (b) As the bar rotates, the moment arm for the weight of the ball decreases and the angular acceleration of the bar decreases. (c) 1 1 2 2K U K U+ = + . 2 20 K U= + . 21 bar ball ball2 ( ) ( / 2)I I m g L?+ = ? ? . 2 ball ball 2 2 ball bar ball bar 4 9.80 m/s 4[2.50 kg] / 4 /12 /3 0.800 m 2.50 kg [3.80 kg]/3 m gL g m m L m L L m m ? ? ? ? ?= = =? ? ? ?+ + +? ?? ? 5.70 rad/s? = . EVALUATE: As the bar swings through the vertical, the linear speed of the ball that is still attached to the bar is (0.400 m)(5.70 rad/s) 2.28 m/sv = = . A point mass in free-fall acquires a speed of 2.80 m/s after falling 0.400 m; the ball on the bar acquires a speed less than this. 10.58. IDENTIFY: Use z zI? ?=? to find ,z? and then use the constant z? kinematic equations to solve for t. SET UP: The door is sketched in Figure 10.58. EXECUTE: (220 N)(1.25 m) 275 N mz Fl? = = = ?? From Table 9.2(d), 213I Ml= 2 2 21 3 (750 N/9.80 m/s )(1.25 m) 39.9 kg mI = = ? Figure 10.58 z zI? ?=? so 22275 N m 6.89 rad/s39.9 kg mzz I ?? ?= = =? ? SET UP: 26.89 rad/s ;z? = 0 90 ( rad/180 ) /2 rad;? ? ? ?? = ° ° = 0 0z? = (door initially at rest); ?t = 21 0 0 2z zt t? ? ? ?? = + EXECUTE: 0 22( ) 2( / 2 rad) 0.675 s6.89 rad/szt ? ? ? ? ?= ? = EVALUATE: The forces and the motion are connected through the angular acceleration. 10.59. IDENTIFY: sinrF? ?= SET UP: Let x be the distance from the left end of the rod where the string is attached. For the value of x where ( )f x is a maximum, / 0df dx = . 10-20 Chapter 10 EXECUTE: (a) From geometric consideration, the lever arm and the sine of the angle between and F r""! ! are both maximum if the string is attached at the end of the rod. (b) In terms of the distance x where the string is attached, the magnitude of the torque is 2 2.Fxh x h+ This function attains its maximum at the boundary, where ,x h= so the string should be attached at the right end of the rod. (c) As a function of x, l and h, the torque has magnitude 2 2 . ( 2) xh F x l h ? = ? + Differentiating? with respect to x and setting equal to zero gives maxx 2( 2)(1 (2 ) ).l h l= + This will be the point at which to attach the string unless 2 ,h l> in which case the string should be attached at the furthest point to the right, .x l= EVALUATE: In part (a) the maximum torque is independent of h. In part (b) the maximum torque is independent of l. In part (c) the maximum torque depends on both h and l. 10.60. IDENTIFY: Apply z zI? ?=? , where z? is due to the gravity force on the object. SET UP: rod clayI I I= + . 21rod 3I ML= . In part (b), 2clayI ML= . In part (c), clay 0I = . EXECUTE: (a) A distance 4L from the end with the clay. (b) In this case 2(4 3)I ML= and the gravitational torque is (3 4)(2 )sin (3 2)sin ,L Mg Mg L? ?= so (9 8 )sin .g L? ?= (c) In this case 2(1 3)I ML= and the gravitational torque is ( 4)(2 )sin ( 2)sin ,L Mg Mg L? ?= so (3 2 )sin .g L? ?= This is greater than in part (b). (d) The greater the angular acceleration of the upper end of the cue, the faster you would have to react to overcome deviations from the vertical. EVALUATE: In part (b), I is 4 times larger than in part (c) and ? is 3 times larger. / I? ?= , so the net effect is that ? is smaller in (b) than in (c). 10.61. IDENTIFY: Calculate W using the procedure specified in the problem. In part (c) apply the work-energy theorem. In part (d), tana R?= and z zI? ?=? . 2rada R?= . SET UP: Let ? be the angle the disk has turned through. The moment arm for F is cosR ? . EXECUTE: (a) The torque is cos .FR? ?= 2 0 cos W FR d FR ? ? ?= =? . (b) In Eq.(6.14), dl is the horizontal distance the point moves, and so ,W F dl FR= =? the same as part (a). (c) From 2 22 ( 4) , 4 .K W MR F MR? ?= = = (d) The torque, and hence the angular acceleration, is greatest when 0,? = at which point ( ) 2I F MR? ?= = , and so the maximum tangential acceleration is 2 .F M (e) Using the value for ? found in part (c), 2rad 4 .a R F M?= = EVALUATE: 2tana R?= is maximum initially, when the moment arm for F is a maximum, and it is zero after the disk has rotated one-quarter of a revolution. rada is zero initially and is a maximum at the end of the motion, after the disk has rotated one-quarter of a revolution. 10.62. IDENTIFY: Apply m=?F a! ! to the crate and z zI? ?=? to the cylinder. The motions are connected by (crate) (cylinder).a R?= SET UP: The force diagram for the crate is given in Figure 10.62a. EXECUTE: y yF ma=? T mg ma? = 2 2( ) 50 kg(9.80 m/s 0.80 m/s ) 530 NT m g a= + = + = Figure 10.62a Dynamics of Rotational Motion 10-21 SET UP: The force diagram for the cylinder is given in Figure 10.62b. EXECUTE: z zI? ?=? ,zFl TR I ?? = where 0.12 ml = and 0.25 mR = a R?= so /z a R? = /Fl TR Ia R= + Figure 10.62b 2 20.25 m (2.9 kg m )(0.80 m/s ) 530 N 1200 N 0.12 m (0.25 m)(0.12 m) R Ia F T l Rl ?? ? ? ?= + = + =? ? ? ?? ? ? ? EVALUATE: The tension in the rope is greater than the weight of the crate since the crate accelerates upward. If F were applied to the rim of the cylinder (l = 0.25 m), it would have the value 567 N.F = This is greater than T because it must accelerate the cylinder as well as the crate. And F is larger than this because it is applied closer to the axis than R so has a smaller moment arm and must be larger to give the same torque. 10.63. IDENTIFY: Apply ext cmm=?F a! ! and cmz zI? ?=? to the roll. SET UP: At the point of contact, the wall exerts a friction force f directed downward and a normal force n directed to the right. This is a situation where the net force on the roll is zero, but the net torque is not zero. EXECUTE: (a) Balancing vertical forces, rod cos ,F f w F? = + + and balancing horizontal forces rod ksin . With ,F n f n? ?= = these equations become rod kcos ,F n F w? ?= + + rod sin . F n? = Eliminating n and solving for rodF gives 2 rod k (16.0 kg) (9.80 m/s ) (40.0 N) 266 N. cos sin cos 30 (0.25)sin30 w F F ? ? ? + += = =? ° ? ° (b) With respect to the center of the roll, the rod and the normal force exert zero torque. The magnitude of the net torque is k( ) , and F f R f n?? = may be found by insertion of the value found for rodF into either of the above relations; i . e ., k rod sin 33.2 N. f F? ?= = Then, 2 2 2 (40.0 N 31.54 N)(18.0 10 m) 4.71 rad/s . (0.260 kg m )I ?? ?? ×= = =? EVALUATE: If the applied force F is increased, rodF increases and this causes n and f to increase. The angle ? changes as the amount of paper unrolls and this affects ? for a given F. 10.64. IDENTIFY: Apply z zI? ?=? to the flywheel and m=?F a! ! to the block. The target variables are the tension in the string and the acceleration of the block. (a) SET UP: Apply z zI? ?=? to the rotation of the flywheel about the axis. The free-body diagram for the flywheel is given in Figure 10.64a. EXECUTE: The forces n and Mg act at the axis so have zero torque. z TR? =? zTR I ?= Figure 10.64a 10-22 Chapter 10 SET UP: Apply m=?F a! ! to the translational motion of the block. The free-body diagram for the block is given in Figure 10.64b. EXECUTE: y yF ma=? cos36.9 0n mg? ° = cos36.9n mg= ° k k k cos36.9f n mg? ?= = ° Figure 10.64b x xF ma=? ksin36.9 cos36.9mg T mg ma?° ? ? ° = k(sin36.9 cos36.9 )mg T ma?° ? ° ? = But we also know that block wheel ,a R?= so / .a R? = Using this in the z zI? ?=? equation gives /TR Ia R= and 2( / ) .T I R a= Use this to replace T in the x xF ma=? equation: 2 k(sin36.9 cos36.9 ) ( / )mg I R a ma?° ? ° ? = k 2 (sin36.9 cos36.9 ) / mg a m I R ?° ? °= + 2 2 2 2 (5.00 kg)(9.80 m/s )(sin36.9 (0.25)cos36.9 ) 1.12 m/s 5.00 kg 0.500 kg m /(0.200 m) a ° ? °= =+ ? (b) 2 2 2 0.500 kg m (1.12 m/s ) 14.0 N (0.200 m) T ?= = EVALUATE: If the string is cut the block will slide down the incline with 2 ksin36.9 cos36.9 3.92 m/s .a g g?= ° ? ° = The actual acceleration is less than this because sin36.9mg ° must also accelerate the flywheel. ksin36.9 19.6 N.mg f° ? = T is less than this; there must be more force on the block directed down the incline than up then incline since the block accelerates down the incline. 10.65. IDENTIFY: Apply m=?F a! ! to the block and z zI? ?=? to the combined disks. SET UP: For a disk, 21disk 2I MR= , so I for the disk combination is 3 22.25 10 kg m .I ?= × ? EXECUTE: For a tension T in the string, and .amg T ma TR I I R ?? = = = Eliminating T and solving for a gives 2 2 ,/ 1 / m g a g m I R I mR = =+ + where m is the mass of the hanging block and R is the radius of the disk to which the string is attached. (a) With 1.50m = kg and 2 22.50 10 m, 2.88 m/s .R a?= × = (b) With 1.50m = kg and 2 25.00 10 m, 6.13 m/s .R a?= × = The acceleration is larger in case (b); with the string attached to the larger disk, the tension in the string is capable of applying a larger torque. EVALUATE: /v R? = , where v is the speed of the block and ? is the angular speed of the disks. When R is larger, in part (b), a smaller fraction of the kinetic energy resides with the disks. The block gains more speed as it falls a certain distance and therefore has a larger acceleration. 10.66. IDENTIFY: Apply both m=?F a! ! and z zI? ?=? to the motion of the roller. Rolling without slipping means cm .a R?= Target variables are cma and f. Dynamics of Rotational Motion 10-23 SET UP: The free-body diagram for the roller is given in Figure 10.66. EXECUTE: Apply m=?F a! ! to the translational motion of the center of mass: x xF ma=? cmF f Ma? = Figure 10.66 Apply z zI? ?=? to the rotation about the center of mass: z fR? =? thin-walled hollow cylinder: 2I MR= Then z zI? ?=? implies 2 .fR MR ?= But cm ,R? ?= so cm.f Ma= Using this in the x xF ma=? equation gives cm cmF Ma Ma? = cm / 2 ,a F M= and then cm ( / 2 ) / 2.f Ma M F M F= = = EVALUATE: If the surface were frictionless the object would slide without rolling and the acceleration would be cm / .a F M= The acceleration is less when the object rolls. 10.67. IDENTIFY: Apply m?F = a! ! to each object and apply z zI? ?=? to the pulley. SET UP: Call the 75.0 N weight A and the 125 N weight B. Let AT and BT be the tensions in the cord to the left and to the right of the pulley. For the pulley, 212I MR= , where 50.0 NMg = and 0.300 mR = . The 125 N weight accelerates downward with acceleration a, the 75.0 N weight accelerates upward with acceleration a and the pulley rotates clockwise with angular acceleration ? , where a R?= . EXECUTE: m?F = a! ! applied to the 75.0 N weight gives A A AT w m a? = . m?F = a! ! applied to the 125.0 N weight gives B B Bw T m a? = . z zI? ?=? applied to the pulley gives 212( ) ( )B A zT T R MR ?? = and 12B AT T M? = . Combining these three equations gives ( / 2)B A A Bw w m m M a? = + + and pulley 125 N 75.0 N 0.222 / 2 75.0 N 125 N 25.0 N B A A B w w a g g g w w w ? ?? ?? ?= = =? ? ? ?? ?+ + + +? ?? ? . (1 / ) 1.222 91.65 NA A AT w a g w= + = = . (1 / ) 0.778 97.25 NB B BT w a g w= ? = = . m?F = a! ! applied to the pulley gives that the force F applied by the hook to the pulley is pulley 239 NA BF T T w= + + = . The force the ceiling applies to the hook is 239 N. EVALUATE: The force the hook exerts on the pulley is less than the total weight of the system, since the net effect of the motion of the system is a downward acceleration of mass. 10.68. IDENTIFY: This problem can be done either with conservation of energy or with ext .m=?F a! ! We will do it both ways. (a) SET UP: (1) Conservation of energy: 1 1 other 2 2.K U W K U+ + = + Take position 1 to be the location of the disk at the base of the ramp and 2 to be where the disk momentarily stops before rolling back down, as shown in Figure 10.68a. Figure 10.68a Take the origin of coordinates at the center of the disk at position 1 and take y+ to be upward. Then 1 0y = and 2 sin30 ,y d= ° where d is the distance that the disk rolls up the ramp. ?Rolls without slipping? and neglect rolling friction says 0;fW = only gravity does work on the disk, so other 0W = 10-24 Chapter 10 EXECUTE: 1 1 0U Mgy= = 2 21 1 1 1 cm 12 2K Mv I ?= + (Eq.10.11). But 1 1 /v R? = and 21cm 2 ,I MR= so ( )2 2 2 21 1 1 1cm 1 1 12 2 2 4( / ) .I MR v R Mv? = = Thus 2 2 231 1 1 1 1 12 4 4 .K Mv Mv Mv= + = 2 2 sin30U Mgy Mgd= = ° 2 0K = (disk is at rest at point 2). Thus 23 14 sin30Mv Mgd= ° 2 2 1 2 3 3(2.50 m/s) 0.957 m 4 sin30 4(9.80 m/s )sin30 v d g = = =° ° SET UP: (2) force and acceleration The free-body diagram is given in Figure 10.68b. EXECUTE: Apply x xF ma=? to the translational motion of the center of mass: cmsinMg f Ma? ? = Apply z zI? ?=? to the rotation about the center of mass: ( )212 zf R MR ?= 1 2 zf MR?= Figure 10.68b But cma R?= in this equation gives 1 cm2 .f Ma= Use this in the x xF ma=? equation to eliminate f. 1 cm cm2sinMg Ma Ma? ? = M divides out and 3 cm2 sin .a g ?= 2 22 2cm 3 3sin (9.80 m/s )sin30 3.267 m/sa g ?= = ° = SET UP: Apply the constant acceleration equations to the motion of the center of mass. Note that in our coordinates the positive x-direction is down the incline. 0 2.50 m/sxv = ? (directed up the incline); 23.267 m/s ;xa = + 0xv = (momentarily comes to rest); 0 ?x x? = 2 2 0 02 ( )x x xv v a x x= + ? EXECUTE: 2 2 0 0 2 ( 2.50 m/s) 0.957 m 2 2(3.267 m/s ) x x v x x a ?? = ? = ? = ? (b) EVALUATE: The results from the two methods agree; the disk rolls 0.957 m up the ramp before it stops. The mass M enters both in the linear inertia and in the gravity force so divides out. The mass M and radius R enter in both the rotational inertia and the gravitational torque so divide out. 10.69. IDENTIFY: Apply ext cmm=?F a! ! to the motion of the center of mass and apply cmz zI? ?=? to the rotation about the center of mass. SET UP: ( )2 2122I MR MR= = . The moment arm for T is b. EXECUTE: The tension is related to the acceleration of the yo-yo by (2 ) (2 ) ,m g T m a? = and to the angular acceleration by .aTb I I b ?= = Dividing the second equation by b and adding to the first to eliminate T yields 2 2 2 2 2 2 , (2 ) 2 ( ) 2 m a g g g m I b R b b R b ?= = =+ + + . The tension is found by substitution into either of the two equations: 2 2 2 2 2 ( ) 2 (2 )( ) (2 ) 1 2 . 2 ( ) 2 ( ) (2( ) 1) R b mg T m g a mg mg R b R b b R ? ?= ? = ? = =? ?+ + +? ? EVALUATE: 0a ? when 0b ? . As b R? , 2 /3a g? . Dynamics of Rotational Motion 10-25 10.70. IDENTIFY: Apply conservation of energy to the motion of the shell, to find its linear speed v at points A and B. Apply m?F = a! ! to the circular motion of the shell in the circular part of the track to find the normal force exerted by the track at each point. Since r R<< the shell can be treated as a point mass moving in a circle of radius R when applying m?F = a! ! . But as the shell rolls along the track, it has both translational and rotational kinetic energy. SET UP: 1 1 2 2K U K U+ = + . Let 1 be at the starting point and take 0y = to be at the bottom of the track, so 1 0y h= . 2 21 12 2K mv I ?= + . 223I mr= and /v r? = , so 256K mv= . During the circular motion, 2rad /a v R= . EXECUTE: (a) m?F = a! ! at point A gives 2vn mg m R+ = . The minimum speed for the shell not to fall off the track is when 0n ? and 2v gR= . Let point 2 be A, so 2 2y R= and 22v mR= . Then 1 1 2 2K U K U+ = + gives 5 0 62 ( )mgh mgR m gR= + . 5 170 6 6(2 )h R R= + = . (b) Let point 2 be B, so 2y R= . Then 1 1 2 2K U K U+ = + gives 250 26mgh mgR mv= + . With 176h R= this gives 2 11 5v gR= . Then m?F = a! ! at B gives 2 115vn m mgR= = . (c) Now 212K mv= instead of 256 mv . The shell would be moving faster at A than with friction and would still make the complete loop. (d) In part (c): 210 2(2 )mgh mg R mv= + . 170 6h R= gives 2 53v gR= . m?F = a! ! at point A gives 2vmg n m R+ = and 2 2 3 v n m g mg R ? ?= ? =? ?? ? . In part (a), 0n = , since at this point gravity alone supplies the net downward force that is required for the circular motion. EVALUATE: The normal force at A is greater when friction is absent because the speed of the shell at A is greater when friction is absent than when there is rolling without slipping. 10.71. IDENTIFY: Consider the direction of the net force and the sense of the net torque in each case. SET UP: The free-body diagram in each case is shown in Figure 10.71. EXECUTE: In the first case, ?F and the friction force act in opposite directions, and the friction force causes a larger torque to tend to rotate the yo-yo to the right. The net force to the right is the difference ,F f? so the net force is to the right while the net torque causes a clockwise rotation. For the second case, both the torque and the friction force tend to turn the yo-yo clockwise, and the yo-yo moves to the right. In the third case, friction tends to move the yo-yo to the right, and since the applied force is vertical, the yo-yo moves to the right. EVALUATE: In the first case the torque due to friction must be larger than the torque due to F, so the net torque is clockwise. In the third case the torque due to F must be larger than the torque due to f, so the net torque will be clockwise. Figure 10.71 10.72. IDENTIFY: Apply ext cmm=?F a! ! to the motion of the center of mass and cmz zI? ?=? to the rotation about the center of mass. SET UP: For a hoop, 2I MR= . For a solid disk, 212I MR= . EXECUTE: (a) Because there is no vertical motion, the tension is just the weight of the hoop: ( )( )0.180 kg 9.8 N kg 1.76 NT Mg= = = . (b) Use to find .I? ? ?= The torque is 2, so / / , RT RT I RT MR T MR Mg MR? = = = = so 2 2/ (9.8 m/s ) (0.08 m) 122.5 rad/sg R? = = = . (c) 29.8 m sa R?= = (d) T would be unchanged because the mass M is the same, and a? would be twice as great because I is now 212 .MR EVALUATE: tana for a point on the rim of the hoop or disk equals a for the free end of the string. Since I is smaller for the disk, the same value of T produces a greater angular acceleration. 10-26 Chapter 10 10.73. IDENTIFY: Apply z zI? ?=? to the cylinder or hoop. Find a for the free end of the cable and apply constant acceleration equations. SET UP: tana for a point on the rim equals a for the free end of the cable, and tana R?= . EXECUTE: (a) tan and z zI a R? ? ?= =? gives 2 2 tan1 12 2 aFR MR MR R? ? ?= = ? ?? ? . 2tan 2 200 N 50 m/s 4.00 kg F a M = = = . Distance the cable moves: 210 0 2x xx x v t a t? = + gives ( )2 2150 m 50 m/s2 t= and 1.41 st = . ( )( )20 0 50 m/s 1.41 s 70.5 m sx x xv v a t= + = + = . (b) For a hoop, 2 ,I MR= which is twice as large as before, so tanand a? would be half as large. Therefore the time would be longer by a factor of 2 . For the speed, 2 20 2 ,x x xv v a x= + in which x is the same, so xv would be half as large since xa is smaller. EVALUATE: The acceleration a that is produced depends on the mass of the object but is independent of its radius. But a depends on how the mass is distributed and is different for a hoop versus a cylinder. 10.74. IDENTIFY: Use projectile motion to find the speed v the marble needs at the edge of the pit to make it to the level ground on the other side. Apply conservation of energy to the motion down the hill in order to relate the initial height to the speed v at the edge of the pit. other 0W = so conservation of energy gives i i f fK U K U+ = + . SET UP: In the projectile motion the marble must travel 36 m horizontally while falling vertically 20 m. Let y+ be downward. For the motion down the hill, let f 0y = so f 0U = and iy h= . i 0K = . Rolling without slipping means v R?= . 2 2 2 2 2 271 1 1 2 1cm2 2 2 5 2 10( )K I mv mR mv mv? ?= + = + = . EXECUTE: (a) Projectile motion: 0 0yv = . 29.80 m/sya = . 0 20 my y? = . 210 0 2y yy y v t a t? = + gives 02( ) 2.02 s y y y t a ?= = . Then 0 0xx x v t? = gives 00 36 m 17.8 m/s2.02 sx x x v v t ?= = = = . Motion down the hill: i fU K= . 2710mgh mv= . 2 2 2 7 7(17.8 m/s) 22.6 m 10 10(9.80 m/s ) v h g = = = . (b) 2 21 12 5I mv? = , independent of R. I is proportional to 2R but 2? is proportional to 21/R for a given translational speed v. (c) The object still needs 17.8 m/sv = at the bottom of the hill in order to clear the pit. But now 21f 2K mv= and 2 16.6 m 2 v h g = = . EVALUATE: The answer to part (a) also does not depend on the mass of the marble. But, it does depend on how the mass is distributed within the object. The answer would be different if the object were a hollow spherical shell. In part (c) less height is needed to give the object the same translational speed because in (c) none of the energy goes into rotational motion. 10.75. IDENTIFY: Apply conservation of energy to the motion of the boulder. SET UP: 2 21 12 2K mv I ?= + and v R?= when there is rolling without slipping. 225I mR= . EXECUTE: Break into 2 parts, the rough and smooth sections. Rough: 2 21 11 2 2mgh mv I ?= + . 2 2 2 1 1 1 2 2 2 5 v mgh mv mR R ? ?? ?= + ? ?? ?? ?? ? . 2 1 10 7 v gh= . Smooth: Rotational kinetic energy does not change. 2 22 rot Bottom rot 1 1 2 2 mgh mv K mv K+ + = + . 22 1 B1 10 12 7 2g h gh v ? ?+ =? ?? ? . 2 2 B 1 2 10 10 2 (9.80 m/s )(25 m) 2(9.80 m/s )(25 m) 29.0 m/s 7 7 v gh gh= + = + = . EVALUATE: If all the hill was rough enough to cause rolling without slipping, B 10 (50 m) 26.5 m/s7v g= = . A smaller fraction of the initial gravitational potential energy goes into translational kinetic energy of the center of mass than if part of the hill is smooth. If the entire hill is smooth and the boulder slides without slipping, B 2 (50 m) 31.3 m/sv g= = . In this case all the initial gravitational potential energy goes into the kinetic energy of the translational motion. Dynamics of Rotational Motion 10-27 10.76. IDENTIFY: Apply conservation of energy to the motion of the ball as it rolls up the hill. After the ball leaves the edge of the cliff it moves in projectile motion and constant acceleration equations can be used. (a) SET UP: Use conservation of energy to find the speed 2v of the ball just before it leaves the top of the cliff. Let point 1 be at the bottom of the hill and point 2 be at the top of the hill. Take 0y = at the bottom of the hill, so 1 0y = and 2 28.0 m.y = EXECUTE: 1 1 2 2K U K U= = + 2 2 2 21 1 1 1 1 1 2 2 22 2 2 2mv I mgy mv I? ?+ = + + Rolling without slipping means /v r? = and ( )2 2 2 21 1 2 12 2 5 5( / )I mr v r mv? = = 2 27 7 1 2 210 10mv mgy mv= + 2 10 2 1 27 15.26 m/sv v gy= ? = SET UP: Consider the projectile motion of the ball, from just after it leaves the top of the cliff until just before it lands. Take y+ to be downward. Use the vertical motion to find the time in the air: 0 0,yv = 29.80 m/s ,ya = 0 28.0 m,y y? = ?t = EXECUTE: 210 0 2y yy y v t a t? = + gives 2.39 st = During this time the ball travels horizontally 0 0 (15.26 m/s)(2.39 s) 36.5 m.xx x v t? = = = Just before it lands, 0 23.4 m/sy y yv v a t= + = and 0 15.3 m/sx xv v= = 2 2 28.0 m/sx yv v v= + = (b) EVALUATE: At the bottom of the hill, / (25.0 m/s) / .v r r? = = The rotation rate doesn?t change while the ball is in the air, after it leaves the top of the cliff, so just before it lands (15.3 m/s) / .r? = The total kinetic energy is the same at the bottom of the hill and just before it lands, but just before it lands less of this energy is rotational kinetic energy, so the translational kinetic energy is greater. 10.77. IDENTIFY: Apply conservation of energy to the motion of the wheel. 2 21 12 2K mv I ?= + . SET UP: No slipping means that .v R? = Uniform density means r s2 and m R m R? ? ?= = , where rm is the mass of the rim and sm is the mass of each spoke. For the wheel, rim spokesI I I= + . For each spoke, 21 s3I m R= . EXECUTE: (a) 2 21 1 2 2 mgh mv I ?= + . 2 2rim spokes r s1 6 3I I I m R m R ? ?= + = + ? ?? ? Also, ( )r s 2 6 2 3m m m R R R? ? ? ? ?= + = + = + . Substituting into the conservation of energy equation gives ( ) ( )( )( )2 2 2 21 1 12 3 2 3 2 6 2 2 3 R gh R R R R RR? ? ? ? ? ? ? ? ?? ?? ?+ = + + + ? ?? ?? ?? ? . ( ) ( ) ( )( )( ) ( ) ( ) 2 22 3 9.80 m s 58.0 m3 124 rad s 2 0.210 m 2 gh R ??? ? ? ++= = =+ + and 26.0 m sv R?= = (b) Doubling the density would have no effect because it does not appear in the answer. ? is inversely proportional to R so doubling the diameter would double the radius which would reduce by half, but? v R?= would be unchanged. EVALUATE: Changing the masses of the rim and spokes by different amounts would alter the speed v at the bottom of the hill. 10.78. IDENTIFY: Apply v R?= . SET UP: For the antique bike, v is the same for points on the rim of each wheel and equals the linear speed of the bike. 1 rev 2 rad?= . EXECUTE: (a) The front wheel is turning at 1.00 rev s 2 rad s.? ?= = (0.330 m)(2 rad s) 2.07 sv r? ?= = = . (b) (2.07 m s) (0.655 m) 3.16 rad s 0.503 rev sv r? = = = = (c) (2.07 m s) (0.220 m) 9.41 rad s 1.50 rev sv r? = = = = EVALUATE: Since the front wheel has a larger radius for the antique bike, that wheel doesn't have to rotate at as many rev/s to achieve the same linear speed of the bike. 10-28 Chapter 10 10.79. IDENTIFY: Apply conservation of energy to the motion of the ball. Once the ball leaves the track the ball moves in projectile motion. SET UP: The ball has 225I mR= ; the silver dollar has 212I mR= . For the projectile motion take y+ downward, so 0xa = and ya g= + . EXECUTE: (a) The kinetic energy of the ball when it leaves the track (when it is still rolling without slipping) is 2(7 10)mv and this must be the work done by gravity, W mgh= , so 10 7.v gh= The ball is in the air for a time 2 , so 20 7.t y g x vt hy= = = (b) The answer does not depend on g , so the result should be the same on the moon. (c) The presence of rolling friction would decrease the distance. (e) For the dollar coin, modeled as a uniform disc, 2(3 4) , and so 8 3.K mv x hy= = EVALUATE: The sphere travels a little farther horizontally, because its moment of inertia is a smaller fraction of 2MR than for the disk. The result is independent of the mass and radius of the object but it does depend on how that mass is distributed within the object. 10.80. IDENTIFY and SET UP: Apply conservation of energy to the motion of the ball. The ball ends up with both translational and rotational kinetic energy. Use Fig.(10.13) in the textbook to relate the speed of different points on the ball to cm.v EXECUTE: (a) 2 21 1el 2 2 (400 N m)(0.15 m) 4.50 JU kx= = ? = and 1 el0.800 3.60 JK U= = 2 21 1 1 cm cm2 2K mv I ?= + rolling without slipping says cm /v R? = 22 cm 5I mR= Thus ( ) ( )2 2 2 2 271 1 2 1 11 cm cm cm cm2 2 5 2 5 10( / )K mv mR v R mv mv= + = + = and 1cm 10 10(3.60 J) 9.34 m/s. 7 7(0.0590 kg) K v m = = = (b) Consider Figure 10.80a. From Fig.(10.13) in the textbook, at the top of the ball cm2 18.7 m/sv v= = Figure 10.80a (c) From Fig.(10.13) in the textbook, 0v = at the bottom of the ball. Figure 10.80b (d) The problem says that 2 10.900 3.24 J.U K= = Thus 2 3.24 JU mgh= = and 2 3.24 J 3.24 J 5.60 m (0.0590 kg)(9.80 m/s ) h mg = = = EVALUATE: Not all the potential energy stored in the spring goes into kinetic energy at the base of the ramp or into gravitational potential energy at the top of the ramp because of loss of mechanical energy due to negative work done by friction. If the ball slides without rolling, then 211 cm2K mv= and cm 11.0 m/s.v = cmv is less than this when the ball rolls and some of its total kinetic energy is rotational. 10.81. IDENTIFY: /xv dx dt= , /yv dy dt= . /x xa dv dt= , /y ya dv dt= . SET UP: cos( ) / sin( )d t dt t? ? ?= ? . sin( ) / cos( )d t dt t? ? ?= . EXECUTE: (a) The sketch is shown in Figure 10.81. (b) R is the radius of the wheel (y varies from 0 to 2R) and T is the period of the wheel?s rotation. (c) Differentiating, 2 2 1 cosx R t v T T ? ?? ?? ?= ? ? ?? ?? ?? ? , 22 2 sinx t a R T T ? ?? ? ? ?= ? ? ? ?? ? ? ? and 2 2 siny R t v T T ? ?? ?= ? ?? ? , 22 2 cos .y t a R T T ? ?? ? ? ?= ? ? ? ?? ? ? ? Dynamics of Rotational Motion 10-29 (d) 2 0 when 2x y t v v T ? ?? ?= = =? ?? ? or any multiple of 2 ,? so the times are integer multiples of the period T . The acceleration components at these times are 2 2 4 0, .x y R a a T ?= = (e) 2 2 2 2 2 2 2 2 2 2 4 cos sin ,x y t t R a a a R T T T T ? ? ? ?? ? ? ? ? ?= + = + =? ? ? ? ? ?? ? ? ? ? ? independent of time. This is the magnitude of the radial acceleration for a point moving on a circle of radius R with constant angular velocity 2 / T? . For motion that consists of this circular motion superimposed on motion with constant velocity ( )0 ,=a! the acceleration due to the circular motion will be the total acceleration. EVALUATE: a is independent of time, but v does depend on time. Figure 10.81 10.82. IDENTIFY: Apply the work-energy theorem to the motion of the basketball. 2 21 12 2K mv I ?= + and v R?= . SET UP: For a thin-walled, hollow sphere 223I mR= . EXECUTE: For rolling without slipping, the kinetic energy is ( )( ) ( )2 2 21 2 5 6 ;m I R v mv+ = initially, this is 32.0 J and at the return to the bottom it is 8.0 J. Friction has done 24.0 J? of work, 12.0 J? each going up and down. The potential energy at the highest point was 20.0 J, so the height above the ground was ( )( )220.0 J 3.40 m.0.600 kg 9.80 m s = EVALUATE: All of the kinetic energy of the basketball, translational and rotational, has been removed at the point where the basketball is at its maximum height up the ramp. 10.83. IDENTIFY: Use conservation of energy to relate the speed of the block to the distance it has descended. Then use a constant acceleration equation to relate these quantities to the acceleration. SET UP: For the cylinder, 212 (2 )I M R= , and for the pulley, 212I MR= . EXECUTE: Doing this problem using kinematics involves four unknowns (six, counting the two angular accelerations), while using energy considerations simplifies the calculations greatly. If the block and the cylinder both have speed v, the pulley has angular velocity v/R and the cylinder has angular velocity v/2R , the total kinetic energy is 2 2 2 2 2 2 21 (2 ) 3( 2 ) ( ) . 2 2 2 2 M R MR K Mv v R v R Mv Mv ? ?= + + + =? ?? ? This kinetic energy must be the work done by gravity; if the hanging mass descends a distance y, ,K Mgy= or 2 (2 3) .v gy= For constant acceleration, 2 2 ,v ay= and comparison of the two expressions gives 3.a g= EVALUATE: If the pulley were massless and the cylinder slid without rolling, 2Mg Ma= and / 2a g= . The rotation of the objects reduces the acceleration of the block. 10.84. IDENTIFY: Apply z zI? ?=? to the drawbridge and calculate z? . For part (c) use conservation of energy. SET UP: The free-body diagram for the drawbridge is given in Fig.10.84. For an axis at the lower end, 213I ml= . EXECUTE: (a) z zI? ?=? gives 213(4.00 m)(cos60.0 ) zmg ml ?=° and 223 (4.00 m)(cos60.0 ) 0.919 rad/s(8.00 m)z g? = =° . (b) z? depends on the angle the bridge makes with the horizontal. z? is not constant during the motion and 0z z zt? ? ?= + cannot be used. (c) Use conservation of energy. Take 0y = at the lower end of the drawbridge, so i (4.00 m)(sin60.0 )y = ° and f 0y = . f f i i otherK U K U W+ = + + gives i fU K= , 21i 2mgy I ?= . 2 21 1i 2 3( )mgy ml ?= and 2 i6 6(9.80 m/s )(4.00 m)(sin60.0 ) 1.78 rad/s 8.00 m gy l ? = = =° . 10-30 Chapter 10 EVALUATE: If we incorrectly assume that z? is constant and has the value calculated in part (a), then 2 2 0 02 ( )z z z? ? ? ? ?= + ? gives 139 rad/s? = . The angular acceleration increases as the bridge rotates and the actual angular velocity is larger than this. Figure 10.84 10.85. IDENTIFY: Apply conservation of energy to the motion of the first ball before the collision and to the motion of the second ball after the collision. Apply conservation of angular momentum to the collision between the first ball and the bar. SET UP: The speed of the ball just before it hits the bar is 2 15.34 m/s.v gy= = Use conservation of angular momentum to find the angular velocity ? of the bar just after the collision. Take the axis at the center of the bar. EXECUTE: 21 (5.00 kg)(15.34 m/s)(2.00 m) 153.4 kg mL mvr= = = ? Immediately after the collision the bar and both balls are rotating together. 2 totL I ?= 2 2 2 2 21 1 tot 12 122 (8.00 kg)(4.00 m) 2(5.00 kg)(2.00 m) 50.67 kg mI Ml mr= + = + = ? 2 2 1 153.4 kg mL L= = ? 2 tot/ 3.027 rad/sL I? = = Just after the collision the second ball has linear speed (2.00 m)(3.027 rad/s) 6.055 m/sv r?= = = and is moving upward. 212 mv mgy= gives 1.87 my = for the height the second ball goes. EVALUATE: Mechanical energy is lost in the inelastic collision and some of the final energy is in the rotation of the bar with the first ball stuck to it. As a result, the second ball does not reach the height from which the first ball was dropped. 10.86. IDENTIFY: The rings and the rod exert forces on each other, but there is no net force or torque on the system, and so the angular momentum will be constant. SET UP: For the rod, 2112I ML= . For each ring, 2I mr= , where r is their distance from the axis. EXECUTE: (a) As the rings slide toward the ends, the moment of inertia changes, and the final angular velocity is given by 2 2 4 21 11 112 2 1 1 12 2 3 21 2 212 2 5.00 10 kg m , 2 2.00 10 kg m 4 ML mrI I ML mr ?? ? ? ? ? ? ? ?+ × ?= = = =? ?+ × ?? ? so 2 7.5 rev min.? = (b) The forces and torques that the rings and the rod exert on each other will vanish, but the common angular velocity will be the same, 7.5 rev/min. EVALUATE: Note that conversion from rev/min to rad/s was not necessary. The angular velocity of the rod decreases as the rings move away from the rotation axis. 10.87. IDENTIFY: Apply conservation of angular momentum to the collision. Linear momentum is not conserved because of the force applied to the rod at the axis. But since this external force acts at the axis, it produces no torque and angular momentum is conserved. SET UP: The system before and after the collision is sketched in Figure 10.87. EXECUTE: (a) 1 rod4bm m= EXECUTE: 11 b rod4 ( / 2)L m vr m v L= = 1 1 rod8L m vL= 2 rod b( )L I I ?= + 21 rod rod3I m L= 2 21 b b rod4 ( / 2)I m r m L= = 21 b rod16I m L= Figure 10.87 Dynamics of Rotational Motion 10-31 Thus 1 2L L= gives ( )2 21 1 1rod rod rod8 3 16m vL m L m L ?= + 191 8 48v L?= 6 19 /v L? = (b) 2 21 11 rod2 8K mv m v= = ( )2 2 2 2 21 1 1 1 12 rod b rod rod2 2 2 3 16( ) (6 /19 )K I I I m L m L v L? ?= = + = + ( )( )2 2 219 6 312 rod rod2 48 19 152K m v m v= = Then 23 rod2 152 21 1 rod8 3/19. m vK K m v = = EVALUATE: The collision is inelastic and 2 1.K K< 10.88. IDENTIFY: Apply Eq.(10.29). SET UP: The door has 213I ml= . The torque applied by the force is avrF , where / 2r l= . EXECUTE: av av, av and .rF L rF t rJ?? = ? = ? = The angular velocity? is then ( ) avav av 21 3 2 3 , 2 l F tL rF t F t I I ml ml ? ?? ? ?= = = = where l is the width of the door. Substitution of the given numeral values gives 0.514rad s.? = EVALUATE: The final angular velocity of the door is proportional to both the magnitude of the average force and also to the time it acts. 10.89. (a) IDENTIFY: Apply conservation of angular momentum to the collision between the bullet and the board: SET UP: The system before and after the collision is sketched in Figure 10.89a. Figure 10.89a EXECUTE: 1 2L L= 3 2 1 sin (1.90 10 kg)(360 m/s)(0.125 m) 0.0855 kg m /sL mvr mvl? ?= = = × = ? 2 2 2L I ?= 2 21 2 board bullet 3I I I ML mr= + = + 2 3 2 21 2 3 (0.750 kg)(0.250 m) (1.90 10 kg)(0.125 m) 0.01565 kg mI ?= + × = ? Then 1 2L L= gives that 2 1 2 2 2 0.0855 kg m /s 5.46 rad/s 0.1565 kg m L I ? ?= = =? (b) IDENTIFY: Apply conservation of energy to the motion of the board after the collision. SET UP: The position of the board at points 1 and 2 in its motion is shown in Figure 10.89b. Take the origin of coordinates at the center of the board and y+ to be upward, so cm,1 0y = and cm,2 ,y h= the height being asked for. 1 1 other 2 2K U W K U+ + = + EXECUTE: Only gravity does work, so other 0.W = 21 1 2K I ?= 1 cm,1 0U mgy= = 2 0K = 2 cm,2U mgy mgh= = Figure 10.89b 10-32 Chapter 10 Thus 212 .I mgh? = 2 2 2 3 2 (0.01565 kg m )(5.46 rad/s) 0.0317 m 3.17 cm 2 2(0.750 kg 1.90 10 kg)(9.80 m/s ) I h mg ? ? ?= = = =+ × (c) IDENTIFY and SET UP: The position of the board at points 1 and 2 in its motion is shown in Figure 10.89c. Apply conservation of energy as in part (b), except now we want cm,2 0.250 m.y h= = Solve for the ? after the collision that is required for this to happen. Figure 10.89c EXECUTE: 212 I mgh? = 3 2 2 2 2(0.750 kg 1.90 10 kg)(9.80 m/s )(0.250 m) 0.01565 kg m mgh I ? ?+ ×= = ? 15.34 rad/s? = Now go back to the equation that results from applying conservation of angular momentum to the collision and solve for the initial speed of the bullet. 1 2L L= implies bullet 2 2m vl I ?= 2 2 2 3 bullet (0.01565 kg m )(15.34 rad/s) 1010 m/s (1.90 10 kg)(0.125 m) I v m l ? ? ?= = =× EVALUATE: We have divided the motion into two separate events: the collision and the motion after the collision. Angular momentum is conserved in the collision because the collision happens quickly. The board doesn?t move much until after the collision is over, so there is no gravity torque about the axis. The collision is inelastic and mechanical energy is lost in the collision. Angular momentum of the system is not conserved during this motion, due to the external gravity torque. Our answer to parts (b) and (c) say that a bullet speed of 360 m/s causes the board to swing up only a little and a speed of 1010 m/s causes it to swing all the way over. 10.90. IDENTIFY: Angular momentum is conserved, so 0 0 2 2I I? ?= . SET UP: For constant mass the moment of inertia is proportional to the square of the radius. EXECUTE: 2 20 0 2 2R R? ?= , or ( ) ( )22 2 20 0 0 0 0 0 0 0 0= 2 ,R R R R R R R? ? ? ? ? ?= + ? + ? + ? + ? where the terms in R ?? ? and 2( )?? have been omitted. Canceling the 20 0R ? term gives 0 0 1.1 cm. 2 R ? R ? ?? = ? = ? EVALUATE: 0/R R? and 0/? ?? are each very small so the neglect of terms containing R ?? ? or 2( )?? is an accurate simplifying approximation. 10.91. IDENTIFY: Apply conservation of angular momentum to the collision between the bird and the bar and apply conservation of energy to the motion of the bar after the collision. SET UP: For conservation of angular momentum take the axis at the hinge. For this axis the initial angular momentum of the bird is bird (0.500 m)m v , where bird 0.500 kgm = and 2.25 m/sv = . For this axis the moment of inertia is 2 2 21 1bar3 3 (1.50 kg)(0.750 m) 0.281 kg mI m L= = = ? . For conservation of energy, the gravitational potential energy of the bar is bar cmU m gy= , where cmy is the height of the center of the bar. Take cm,1 0y = , so cm,2 0.375 my = ? . EXECUTE: (a) 1 2L L= gives 21bird bar3(0.500 m) ( )m v m L ?= . bird 2 2 bar 3 (0.500 m) 3(0.500 kg)(0.500 m)(2.25 m/s) 2.00 rad/s (1.50 kg)(0.750 m) m v m L ? = = = . Dynamics of Rotational Motion 10-33 (b) 1 1 2 2U K U K+ = + applied to the motion of the bar after the collision gives 2 21 11 bar 22 2( 0.375 m)I m g I? ?= ? + . 2 2 1 bar 2 (0.375 m)m g I ? ?= + . 2 22 22(2.00 rad/s) (1.50 kg)(9.80 m/s )(0.375 m) 6.58 rad/s0.281 kg m? = + =? EVALUATE: Mechanical energy is not conserved in the collision. The kinetic energy of the bar just after the collision is less than the kinetic energy of the bird just before the collision. 10.92. IDENTIFY: Angular momentum is conserved, since the tension in the string is in the radial direction and therefore produces no torque. Apply m=?F a! ! to the block, with 2rad /a a v r= = . SET UP: The block?s angular momentum with respect to the hole is L mvr= . EXECUTE: The tension is related to the block?s mass and speed, and the radius of the circle, by 2 . v T m r = ( )22 2 2 22 3 3 3 1 . mvrm v r L T mv r m r mr mr = = = = The radius at which the string breaks is ( ) ( )( )( )( ) ( )( ) 222 1 13 max max 0.250 kg 4.00 m s 0.800 m , 0.250 kg 30.0 N mv rL r mT mT = = = from which 0.440 m.r = EVALUATE: Just before the string breaks the speed of the rock is 0.800 m(4.00 m/s) 7.27 m/s 0.440 m ? ? =? ?? ? . We can verify that 7.27 m/sv = and 0.440 mr = do give 30.0 NT = . 10.93. IDENTIFY and SET UP: Apply conservation of angular momentum to the system consisting of the disk and train. SET UP: 1 2 ,L L= counterclockwise positive. The motion is sketched in Figure 10.93. 1 0L = (before you switch on the train?s engine; both the train and the platform are at rest) 2 train diskL L L= + Figure 10.93 EXECUTE: The train is 12 (0.95 m) 0.475 m= from the axis of rotation, so for it 2 2 2 t t t (1.20 kg)(0.475 m) 0.2708 kg mI m R= = = ? rel rel t/ (0.600 m/s)/0.475 s 1.263 rad/sv R? = = = This is the angular velocity of the train relative to the disk. Relative to the earth t rel d.? ? ?= + Thus train t t t rel d( ).L I I? ? ?= = + 2 1L L= says disk trainL L= ? disk d d ,L I ?= where 21d d d2I m R= 21 d d d t rel d2 ( )m R I? ? ?= ? + 2 t rel d 2 2 21 1 d d t2 2 (0.2708 kg m )(1.263 rad/s) 0.30 rad/s. (7.00 kg)(0.500 m) 0.2708 kg m I m R I ?? ?= ? = ? = ?+ + ? EVALUATE: The minus sign tells us that the disk is rotating clockwise relative to the earth. The disk and train rotate in opposite directions, since the total angular momentum of the system must remain zero. Note that we applied 1 2L L= in an inertial frame attached to the earth. 10.94. IDENTIFY: I for the wheel is the sum of the values of I for each of its parts, the rim and each spoke. The total length of wire is constant. The motion is related to the friction torque by z zI? ?=? . SET UP: 04 2R R L?+ = , where R is the radius of the wheel and therefore the length of each of the four spokes. The mass of a piece is proportional to the length of that piece. EXECUTE: (a) 0 4 2 L R ?= + . 2 rim rimI m R= . rim 0 0 0 2 2 4 2 R m M M L ? ? ? ? ?= = ? ?+? ? . 2 3 2 rim 0 0 0 03 2 (5.778 10 ) (2 4) I M L M L ? ? ?= = ×+ . 21 spoke spoke3I m R= . 0spoke 0 0 2 4 R M m M L ?= = + and 2 4 2 spoke 0 0 0 03 1 (3.065 10 ) 3(2 4) I M L M L? ?= = ×+ . 3 2 rim spoke 0 04 (7.00 10 )I I I M L ?= + = × . 10-34 Chapter 10 (b) 0z z zt? ? ?= + gives 0z T ?? = ? . Then z zI? ?=? gives 3 2 0f 0 0(7.00 10 )M L T?? ?= × EVALUATE: If the wire were bent into a circle, without spokes, the moment of inertia would be 2 2 3 20 0 0 0 02 (9.46 10 )(4 2 ) M L M R M L? ?= = ×+ . The actual value of I for the wheel is less than this because the mass in the spokes is closer to the axis than the rim. 10.95. IDENTIFY and SET UP: Use the methods stipulated in the problem. EXECUTE: (a) The initial angular momentum with respect to the pivot is ,mvr and the final total moment of inertia is 2I mr+ , so the final angular velocity is ( )2 .mvr mr I? = + (b) The kinetic energy after the collision is converted to gravitational potential energy, so ( ) ( )2 21 2 mr I M m gh? + = + , or ( )( )2 2 M m gh mr I ? += + . (c) Substitution of 2? ?r= into the result of part (a) gives ( ),m v r m M ? ? ?= ? ?+? ? and into the result of part (b), 2 (1 ),g h r? = which are consistent with the forms for v. EVALUATE: 2I Mr= applies approximately when the pendulum consists of a heavy catcher mounted on a light arm. In the actual apparatus some of the mass is distributed closer to the axis and 2I Mr< . 10.96. IDENTIFY: Apply conservation of momentum to the system of the runner and turntable SET UP: Let the positive sense of rotation be the direction the turntable is rotating initially. EXECUTE: The initial angular momentum is 1 1I mRv? ? , with the minus sign indicating that runner?s motion is opposite the motion of the part of the turntable under his feet. The final angular momentum is 22 ( ), soI mR? + 1 1 2 2 I mRv I mR ?? ?= + . 2 2 2 2 (80 kg m )(0.200 rad s) (55.0 kg)(3.00 m)(2.8 m s) 0.776 rad s (80 kg m ) (55.0 kg)(3.00 m) ? ? ?= = ?? + . EVALUATE: The minus sign indicates that the turntable has reversed its direction of motion. This happened because the man had the larger magnitude of angular momentum initially. 10.97. IDENTIFY: Treat the moon as a point mass, so 2L I mr? ?= = , where r is the distance of the moon from the center of the earth. Conservation of angular momentum says / 0dL dt = . SET UP: 2/ 3.0 cm/y 3.0 10 m/ydr dt ?= = × . The period of the moon?s orbital motion is 627.3 d 2.36 10 s= × . 83.84 10 mr = × . EXECUTE: 2 2/ ( ) (2 ) 0d dr ddL dt mr m r mr dt dt dt ?? ?= = + = , so 2d dr dt r dt ? ?= ? . 6 6 2 rad 2 rad 2.66 10 rad/s 2.36 10 sT ? ?? ?= = = ×× . 6 2 16 8 2(2.66 10 rad/s) (3.0 10 m/y) 4.2 10 rad/s per year 3.84 10 m d dt ? ? ? ?×= ? × = ? ×× . d dt ? is negative, so the angular velocity is decreasing. EVALUATE: 2L mr ?= . If L is constant, then ? decreases when r increases. The fractional changes in r and ? are very, very small. 10.98. IDENTIFY: Follow the method outlined in the hint. SET UP: cmJ m v= ? . cm( )L J x x? = ? . EXECUTE: The velocity of the center of mass will change by cm /v J m? = and the angular velocity will change by cm( )J x x I ? ?? = . The change is velocity of the end of the bat will then be end cm cmv v x?? = ? ? ? = cm cm( )J J x x xm I ?? ? Setting end 0v? = allows cancellation of J cm cmand gives ( ) , I x x x m= ? which when solved for x is 2 2 cm cm (5.30 10 kg m ) (0.600 m) 0.710 m. (0.600 m)(0.800 kg) I x x x m ?× ?= + = + = EVALUATE: The center of percussion is farther from the handle than the center of mass. Dynamics of Rotational Motion 10-35 10.99. IDENTIFY and SET UP: Follow the analysis that led to Eq.(10.33). EXECUTE: In Figure 10.33a in the textbook, if the vector r! and hence the vector L"! are not horizontal but make an angle ? with the horizontal, the torque will still be horizontal (the torque must be perpendicular to the vertical weight). The magnitude of the torque will be cosr? ? , and this torque will change the direction of the horizontal component of the angular momentum, which has magnitude cos L ? . Thus, the situation of Figure 10.35 in the textbook is reproduced, but with horizL "! instead of L "! . Then, the expression found in Eq. (10.33) becomes horiz horiz cos cos dd mgr wr dt dt L I ? ? ? ? ?? = = = = = ? L L L "! "! "! EVALUATE: The torque and the horizontal component of L! both depend on ? by the same factor, cos? . 10.100. IDENTIFY: Apply conservation of energy to the motion of the ball. SET UP: In relating 21 cm2 mv and 212 I ? , instead of cmv R?= use the relation derived in part (a). 225I mR= . EXECUTE: (a) Consider the sketch in Figure 10.100. The distance from the center of the ball to the midpoint of the line joining the points where the ball is in contact with the rails is ( )22 2 ,R d? 2 2cmso 4 v R d?= ? . When 0,d = this reduces to cm ,v R?= the same as rolling on a flat surface. When 2 ,d R= the rolling radius approaches zero, and cm 0 for any .v ?? (b) ( ) ( ) ( ) 2 2 2 2 2 2 cm cm cm 2 22 2 1 1 1 2 2 5 5 2 2 2 10 1 44 v mv K mv I mv mR d RR d ? ? ?? ? ? ?? ?? ? ? ?= + = + = +? ?? ? ?? ?? ??? ? ? ?? ?? ? Setting this equal to mgh and solving for cmv gives the desired result. (c) The denominator in the square root in the expression for cmv is larger than for the case cm0, so d v= is smaller. For a given speed, is larger than in the 0d? = case, so a larger fraction of the kinetic energy is rotational, and the translational kinetic energy, and hence cmv , is smaller. (d) Setting the expression in part (b) equal to 0.95 of that of the 0d = case and solving for the ratio d R gives 1.05.d R = Setting the ratio equal to 0.995 gives 0.37.d R = EVALUATE: If we set 0d = in the expression in part (b), cm 107 g h v = , the same as for a sphere rolling down a ramp. When 2d R? , the expression gives cm 0v = , as it should. Figure 10.100 10.101. IDENTIFY: Apply ext cmm=?F a! ! and cmz zI? ?=? to the motion of the cylinder. Use constant acceleration equations to relate xa to the distance the object travels. Use the work-energy theorem to find the work done by friction. SET UP: The cylinder has 21cm 2I MR= . EXECUTE: (a) The free-body diagram is sketched in Figure 10.101. The friction force is k k k, so .f n Mg a g? ? ?= = = The magnitude of the angular acceleration is ( )k k2 2 . 1 2 MgR gfR I RMR ? ?= = (b) Setting ( )0v at R t R? ? ?= = = ? and solving for t gives 0 0 0 k k k , 2 3 R R R t a R g g g ? ? ? ? ? ? ?= = =+ + and ( ) 2 2 2 2 0 0 k k k 1 1 . 2 2 3 18 R R d at g g g ? ?? ? ? ? ?= = =? ?? ? 10-36 Chapter 10 (c) The final kinetic energy is ( ) ( ) ( )223 4 3 4 ,Mv M at= so the change in kinetic energy is 2 2 2 2 20 k 0 0 k 3 1 1 . 4 3 4 6 R K M g MR MR g ?? ? ?? ? ?? = ? = ?? ?? ? EVALUATE: The fraction of the initial kinetic energy that is removed by friction work is 21 06 21 04 2 3 MR MR ? ? = . This fraction is independent of the initial angular speed 0? . Figure 10.101 10.102. IDENTIFY: The vertical forces must sum to zero. Apply Eq.(10.33). SET UP: Denote the upward forces that the hands exert as and L RF F . ( )L RF F r? = ? , where 0.200 mr = . EXECUTE: The conditions that and L RF F must satisfy are L RF F w+ = and L R IF F r ?? = ? , where the second equation is ,L? = ? divided by r. These two equations can be solved for the forces by first adding and then subtracting, yielding 1 2L I F w r ?? ?= +?? ?? ? and 1 . 2R I F w r ?? ?= ??? ?? ? Using the values 2(8.00 kg)(9.80 m s ) 78.4 N andw mg= = = 2(8.00 kg)(0.325 m) (5.00 rev s 2 rad rev) 132.7 kg m s (0.200 m) I r ? ?×= = ? gives 39.2 N (66.4 N s), 39.2 N (66.4 N s).L RF F= +? ? = ?? ? (a) 0, 39.2 NL RF F? = = = . (b) 0.05 rev s 0.314 rad s, 60.0 N, 18.4 N.L RF F? = = = = (c) 0.3 rev s 1.89 rad s, 165 N, 86.2 NL RF F? = = = = ? , with the minus sign indicating a downward force. (e) 39.2 N 0 gives 0.575 rad s, which is 0.0916 rev s. 66.4 N sR F = ? = =? EVALUATE: The larger the precession rate ? , the greater the torque on the wheel and the greater the difference between the forces exerted by the two hands. 10.103. IDENTIFY: The answer to part (a) can be taken from the solution to Problem 10.92. The work-energy theorem says W K= ? . SET UP: Problem 10.92 uses conservation of angular momentum to show that 1 1 2 2rv r v= . EXECUTE: (a) 2 2 31 1 . T mv r r= (b) and dT r ! ! are always antiparallel. d Tdr? = ?T r! ! . 2 1 1 2 2 2 2 21 1 1 13 2 2 2 1 1 1 . 2 r r r r dr mv W T dr mv r r r r r ? ?= ? = = ?? ?? ?? ? (c) 2 1 1 2( ), sov v r r= 2 2 2 21 2 1 1 2 1 ( ) ( / ) 1 2 2 mv K m v v r r? ?? = ? = ?? ? , which is the same as the work found in part (b). EVALUATE: The work done by T is positive, since T! is toward the hole in the surface and the block moves toward the hole. Positive work means the kinetic energy of the object increases. 11-1 EQUILIBRIUM AND ELASTICITY 11.1. IDENTIFY: Use Eq.(11.3) to calculate cmx . The center of gravity of the bar is at its center and it can be treated as a point mass at that point. SET UP: Use coordinates with the origin at the left end of the bar and the x+ axis along the bar. 1 2.40 kg,m = 2 1.10 kg,m = 3 2.20 kg.m = EXECUTE: 1 1 2 2 3 3cm 1 2 3 (2.40 kg)(0.250 m) 0 (2.20 kg)(0.500 m) 0.298 m 2.40 kg 1.10 kg 2.20 kg m x m x m x x m m m + + + += = =+ + + + . The fulcrum should be placed 29.8 cm to the right of the left-hand end. EVALUATE: The mass at the right-hand end is greater than the mass at the left-hand end. So the center of gravity is to the right of the center of the bar. 11.2. IDENTIFY: Use Eq.(11.3) to calculate cmx of the composite object. SET UP: Use coordinates where the origin is at the original center of gravity of the object and x+ is to the right. With the 1.50 g mass added, cm 2.20 cmx = ? , 1 5.00 gm = and 2 1.50 gm = . 1 0x = . EXECUTE: 2 2cm 1 2 m x x m m = + . 1 2 2 cm 2 5.00 g 1.50 g ( 2.20 cm) 9.53 cm 1.50 g m m x x m ? ? ? ?+ += = ? = ?? ? ? ?? ?? ? . The additional mass should be attached 9.53 cm to the left of the original center of gravity. EVALUATE: The new center of gravity is somewhere between the added mass and the original center of gravity. 11.3. IDENTIFY: The center of gravity of the combined object must be at the fulcrum. Use Eq.(11.3) to calculate cmx SET UP: The center of gravity of the sand is at the middle of the box. Use coordinates with the origin at the fulcrum and x+ to the right. Let 1 25.0 kgm = , so 1 0.500 mx = . Let 2 sandm m= , so 2 0.625 mx = ? . cm 0x = . EXECUTE: 1 1 2 2cm 1 2 0 m x m x x m m += =+ and 1 2 1 2 0.500 m (25.0 kg) 20.0 kg 0.625 m x m m x ? ?= ? = ? =? ??? ? . EVALUATE: The mass of sand required is less than the mass of the plank since the center of the box is farther from the fulcrum than the center of gravity of the plank is. 11.4. IDENTIFY: Apply the first and second conditions for equilibrium to the trap door. SET UP: For 0z? =? take the axis at the hinge. Then the torque due to the applied force must balance the torque due to the weight of the door. EXECUTE: (a) The force is applied at the center of gravity, so the applied force must have the same magnitude as the weight of the door, or 300 N. In this case the hinge exerts no force. (b) With respect to the hinges, the moment arm of the applied force is twice the distance to the center of mass, so the force has half the magnitude of the weight, or 150 N . The hinges supply an upward force of 300 N 150 N 150 N.? = EVALUATE: Less force must be applied when it is applied farther from the hinges. 11.5. IDENTIFY: Apply 0z? =? to the ladder. SET UP: Take the axis to be at point A. The free-body diagram for the ladder is given in Figure 11.5. The torque due to F must balance the torque due to the weight of the ladder. EXECUTE: (8.0 m)sin 40 (2800 N)(10.0 m), so 5.45 kNF F° = = . 11 11-2 Chapter 11 EVALUATE: The force required is greater than the weight of the ladder, because the moment arm for F is less than the moment arm for w. Figure 11.5 11.6. IDENTIFY: Apply the first and second conditions of equilibrium to the board. SET UP: The free-body diagram for the board is given in Figure 11.6. Since the board is uniform its center of gravity is 1.50 m from each end. Apply 0yF =? , with y+ upward. Apply 0? =? with the axis at the end where the first person applies a force and with counterclockwise torques positive. EXECUTE: 0yF =? gives 1 2 0F F w+ ? = and 2 1 160 N 60 N 100 NF w F= ? = ? = . 0? =? gives 2 (1.50 m) 0F x w? = and 2 160 N (1.50 m) (1.50 m) 2.40 m 100 N w x F ? ? ? ?= = =? ? ? ?? ?? ? . The other person lifts with a force of 100 N at a point 2.40 m from the end where the other person lifts. EVALUATE: By considering the axis at the center of gravity we can see that a larger force is applied by the person who pushes closer to the center of gravity. Figure 11.6 11.7. IDENTIFY: Apply 0yF =? and 0z? =? to the board. SET UP: Let y+ be upward. Let x be the distance of the center of gravity of the motor from the end of the board where the 400 N force is applied. EXECUTE: (a) If the board is taken to be massless, the weight of the motor is the sum of the applied forces, 1000 N. The motor is a distance (2.00 m)(600 N) 1.20 m (1000 N) = from the end where the 400 N force is applied, and so is 0.800 m from the end where the 600 N force is applied. (b) The weight of the motor is 400 N 600 N 200 N 800 N.+ ? = Applying 0z? =? with the axis at the end of the board where the 400 N acts gives (600 N)(2.00 m) (200 N)(1.00 m) (800 N)x= + and 1.25 mx = . The center of gravity of the motor is 0.75 m from the end of the board where the 600 N force is applied. EVALUATE: The motor is closest to the end of the board where the larger force is applied. 11.8. IDENTIFY: Apply the first and second conditions of equilibrium to the shelf. SET UP: The free-body diagram for the shelf is given in Figure 11.8. Take the axis at the left-hand end of the shelf and let counterclockwise torque be positive. The center of gravity of the uniform shelf is at its center. EXECUTE: (a) 0z? =? gives t s(0.200 m) (0.300 m) (0.400 m) 0w w T? ? + = . (25.0 N)(0.200 m) (50.0 N)(0.300 m) 50.0 N 0.400 m T += = 0yF =? gives 1 2 t s 0T T w w+ ? ? = and 1 25.0 NT = . The tension in the left-hand wire is 25.0 N and the tension in the right-hand wire is 50.0 N. Equilibrium and Elasticity 11-3 EVALUATE: We can verify that 0z? =? is zero for any axis, for example for an axis at the right-hand end of the shelf. Figure 11.8 11.9. IDENTIFY: Apply the conditions for equilibrium to the bar. Set each tension equal to its maximum value. SET UP: Let cable A be at the left-hand end. Take the axis to be at the left-hand end of the bar and x be the distance of the weight w from this end. The free-body diagram for the bar is given in Figure 11.9. EXECUTE: (a) 0yF =? gives bar 0A BT T w w+ ? ? = and bar 500.0 N 400.0 N 350.0 N 550 NA Bw T T w= + ? = + ? = . (b) 0z? =? gives bar(1.50 m) (0.750 m) 0BT wx w? ? = . bar(1.50 m) (0.750 m) (400.0 N)(1.50 m) (350 N)(0.750 m) 0.614 m 550 N BT wx w ? ?= = = . The weight should be placed 0.614 m from the left-hand end of the bar. EVALUATE: If the weight is moved to the left, AT exceeds 500.0 N and if it is moved to the right BT exceeds 400.0 N. Figure 11.9 11.10. IDENTIFY: Apply the first and second conditions for equilibrium to the ladder. SET UP: Let 2n be the upward normal force exerted by the ground and let 1n be the horizontal normal force exerted by the wall. The maximum possible static friction force that can be exerted by the ground is s 2n? . EXECUTE: (a) Since the wall is frictionless, the only vertical forces are the weights of the man and the ladder, and the normal force 2n . For the vertical forces to balance, 2 1 m 160 N 740 N 900 N,n w w= + = + = and the maximum frictional force is s 2 (0.40)(900N) 360N? n = = . (b) Note that the ladder makes contact with the wall at a height of 4.0 m above the ground. Balancing torques about the point of contact with the ground, 1(4.0 m) (1.5 m)(160 N) (1.0 m)(3 5)(740 N) 684 N m,n = + = ? so 1 171.0 Nn = . This horizontal force about must be balanced by the friction force, which must then be 170 N to two figures. (c) Setting the friction force, and hence 1n , equal to the maximum of 360 N and solving for the distance x along the ladder, (4.0 m)(360 N) (1.50 m)(160 N) (3 5)(740 N),x= + so 2.7 m.x = 11-4 Chapter 11 EVALUATE: The normal force exerted by the ground doesn?t change as the man climbs up the ladder. But the normal force exerted by the wall and the friction force exerted by the ground both increase as he moves up the ladder. 11.11. IDENTIFY: The system of the person and diving board is at rest so the two conditions of equilibrium apply. (a) SET UP: The free-body diagram for the diving board is given in Figure 11.11. Take the origin of coordinates at the left-hand end of the board (point A). 1F ! is the force applied at the support point and 2F ! is the force at the end that is held down. Figure 11.11 EXECUTE: 0A? =? gives 1(1.0 m) (500 N)(3.00 m) (280 N)(1.50 m) 0F+ ? ? = 1 (500 N)(3.00 m) (280 N)(1.50 m) 1920 N 1.00 m F += = (b) y yF ma=? 1 2 280 N 500 N 0F F? ? ? = 2 1 280 N 500 N 1920 N 280 N 500 N 1140 NF F= ? ? = ? ? = EVALUATE: We can check our answers by calculating the net torque about some point and checking that 0z? = for that point also. Net torque about the right-hand of the board: (1140 N)(3.00 m)+(280 N)(1.50 m) (1920 N)(2.00 m)? = 3420 N m 420 N m 3840 N m 0,? + ? ? ? = which checks. 11.12. IDENTIFY: Apply the first and second conditions of equilibrium to the beam. SET UP: The boy exerts a downward force on the beam that is equal to his weight. EXECUTE: (a) The graphs are given in Figure 11.12. (b) 6.25 m when 0,Ax F= = which is 1.25 m beyond point B. (c) Take torques about the right end. When the beam is just balanced, 0, so 900 N.A BF F= = The distance that point B must be from the right end is then (300 N)(4.50 m) 1.50 m. (900 N) = EVALUATE: When the beam is on the verge of tipping it starts to lift off the support A and the normal force AF exerted by the support goes to zero. Figure 11.12 Equilibrium and Elasticity 11-5 11.13. IDENTIFY: Apply the first and second conditions of equilibrium to the strut. (a) SET UP: The free-body diagram for the strut is given in Figure 11.13a. Take the origin of coordinates at the hinge (point A) and y+ upward. Let hF and vF be the horizontal and vertical components of the force F ! exerted on the strut by the pivot. The tension in the vertical cable is the weight w of the suspended object. The weight w of the strut can be taken to act at the center of the strut. Let L be the length of the strut. EXECUTE: y yF ma=? v 0F w w? ? = v 2F w= Figure 11.13a Sum torques about point A. The pivot force has zero moment arm for this axis and so doesn?t enter into the torque equation. 0A? = ( )sin30.0 ( / 2)cos30.0 ( cos30.0 ) 0TL w L w L° ? ° ? ° = sin30.0 (3 / 2)cos30.0 0T w° ? ° = 3 cos30.0 2.60 2sin30.0 w T w °= =° Then x xF ma=? implies h 0T F? = and h 2.60 .F w= We now have the components of F ! so can find its magnitude and direction (Figure 11.13b) 2 2 h vF F F= + 2 2(2.60 ) (2.00 )F w w= + 3.28F w= v h 2.00 tan 2.60 F w F w ? = = 37.6? = ° Figure 11.13b (b) SET UP: The free-body diagram for the strut is given in Figure 11.13c. Figure 11.13c The tension T has been replaced by its x and y components. The torque due to T equals the sum of the torques of its components, and the latter are easier to calculate. 11-6 Chapter 11 EXECUTE: 0 ( cos30.0 )( sin 45.0 ) ( sin30.0 )( cos45.0 )A T L T L? = + ° ° ? ° ° ?? (( / 2)cos45.0 ) ( cos45.0 ) 0w L w L° ? ° = The length L divides out of the equation. The equation can also be simplified by noting that sin 45.0 cos45.0 .° = ° Then (cos30.0 sin30.0 ) 3 / 2.T w° ? ° = 3 4.10 2(cos30.0 sin30.0 ) w T w= =° ? ° x xF ma=? h cos30.0 0F T? ° = h cos30.0 (4.10 )(cos30.0 ) 3.55F T w w= ° = ° = y yF ma=? v sin30.0 0F w w T? ? ? ° = v 2 (4.10 )sin30.0 4.05F w w w= + ° = From Figure 11.13d, 2 2 h vF F F= + 2 2(3.55 ) (4.05 ) 5.39F w w w= + = v h 4.05 tan 3.55 F w F w ? = = 48.8? = ° Figure 11.13d EVALUATE: In each case the force exerted by the pivot does not act along the strut. Consider the net torque about the upper end of the strut. If the pivot force acted along the strut, it would have zero torque about this point. The two forces acting at this point also have zero torque and there would be one nonzero torque, due to the weight of the strut. The net torque about this point would then not be zero, violating the second condition of equilibrium. 11.14. IDENTIFY: Apply the first and second conditions of equilibrium to the beam. SET UP: The free-body diagram for the beam is given in Figure 11.14. vH and hH are the vertical and horizontal components of the force exerted on the beam at the wall (by the hinge). Since the beam is uniform, its center of gravity is 2.00 m from each end. The angle ? has cos 0.800? = and sin 0.600? = . The tension T has been replaced by its x and y components. EXECUTE: (a) vH , hH and cosxT T ?= all produce zero torque. 0? =? gives load(2.00 m) (4.00 m) sin (4.00 m) 0w w T ?? ? + = and (150 N)(2.00 m) (300 N)(4.00 m) 625 N(4.00 m)(0.600)T += = . (b) 0xF =? gives h cos 0H T ?= = and h (625 N)(0.800) 500 NH = = . 0yF =? gives v load sin 0H w w T ?? ? + = and v load sin 150 N 300 N (625 N)(0.600) 75 NH w w T ?= + ? = + ? = . EVALUATE: For an axis at the right-hand end of the beam, only w and vH produce torque. The torque due to w is counterclockwise so the torque due to vH must be clockwise. To produce a counterclockwise torque, vH must be upward, in agreement with our result from 0yF =? . Figure 11.14 Equilibrium and Elasticity 11-7 11.15. IDENTIFY: Apply the first and second conditions of equilibrium to the door. SET UP: The free-body diagram for the door is given in Figure 11.15. Let 1H ! and 2H ! be the forces exerted by the upper and lower hinges. Take the origin of coordinates at the bottom hinge (point A) and y+ upward. EXECUTE: We are given that 1v 2v / 2 140 N.H H w= = = x xF ma=? 2h 1h 0H H? = 1h 2hH H= The horizontal components of the hinge forces are equal in magnitude and opposite in direction. Figure 11.15 Sum torques about point A. 1v ,H 2v ,H and 2hH all have zero moment arm and hence zero torque about an axis at this point. Thus 0A? =? gives 1h (1.00 m) (0.50 m) 0H w? = 1 1h 2 0.50 m (280 N) 140 N. 1.00 m H w? ?= ? =? ?? ? The horizontal component of each hinge force is 140 N. EVALUATE: The horizontal components of the force exerted by each hinge are the only horizontal forces so must be equal in magnitude and opposite in direction. With an axis at A, the torque due to the horizontal force exerted by the upper hinge must be counterclockwise to oppose the clockwise torque exerted by the weight of the door. So, the horizontal force exerted by the upper hinge must be to the left. You can also verify that the net torque is also zero if the axis is at the upper hinge. 11.16. IDENTIFY: Apply the conditions of equilibrium to the wheelbarrow plus its contents. The upward force applied by the person is 650 N. SET UP: The free-body diagram for the wheelbarrow is given in Figure 11.16. 650 NF = , wb 80.0 Nw = and w is the weight of the load placed in the wheelbarrow. EXECUTE: (a) 0z? =? with the axis at the center of gravity gives (0.50 m) (0.90 m) 0n F? = and 0.90 m 1170 N 0.50 m n F ? ?= =? ?? ? . 0yF =? gives wb 0F n w w+ ? ? = and wb 650 N 1170 N 80.0 N 1740 Nw F n w= + ? = + ? = . (b) The extra force is applied by the ground pushing up on the wheel. EVALUATE: You can verify that 0z? =? for any axis, for example for an axis where the wheel contacts the ground. Figure 11.16 11-8 Chapter 11 11.17. IDENTIFY: Apply the first and second conditions of equilibrium to Clea. SET UP: Consider the forces on Clea. The free-body diagram is given in Figure 11.17 EXECUTE: r 89 N,n = f 157 Nn = r fn n w+ = so 246 Nw = Figure 11.17 0,z? =? axis at rear feet Let x be the distance from the rear feet to the center of gravity. f (0.95 m) 0n xw? = 0.606 mx = from rear feet so 0.34 m from front feet. EVALUATE: The normal force at her front feet is greater than at her rear feet, so her center of gravity is closer to her front feet. 11.18. IDENTIFY: Apply the conditions for equilibrium to the crane. SET UP: The free-body diagram for the crane is sketched in Figure 11.18. hF and vF are the components of the force exerted by the axle. T ! pulls to the left so hF is to the right. T ! also pulls downward and the two weights are downward, so vF is upward. EXECUTE: (a) 0z? =? gives c b([13 m]sin 25 ([7.0 m]cos55 ) ([16.0 m]cos55 0T w w? ? =° ° ° . 4(11,000 N)([16.0 m]cos55 ) (15,000 N)([7.0 m]cos55 2.93 10 N (13.0 m)sin 25 T += = ×° °)° . (b) 0xF =? gives h cos30 0F T? =° and 4h 2.54 10 NF = × . 0yF =? gives v c bsin30 0F T w w? ? ? =° and 4v 4.06 10 NF = × . EVALUATE: 4 v 4 h 4.06 10 N tan 2.54 10 N F F ? ×= = × and 58? = ° . The force exerted by the axle is not directed along the crane. Figure 11.18 11.19. IDENTIFY: Apply the first and second conditions of equilibrium to the rod. SET UP: The force diagram for the rod is given in Figure 11.19. Figure 11.19 Equilibrium and Elasticity 11-9 EXECUTE: 0,z? =? axis at right end of rod, counterclockwise torque is positive 1(240 N)(1.50 m) (90 N)(0.50 m) ( sin30.0 )(3.00 m) 0T+ ? ° = 1 360 N m 45 N m 270 N 1.50 m T ? + ?= = x xF ma=? 2 1cos cos30 0T T? ? ° = and 2 cos 234 NT ? = y yF ma=? 1 2sin30 sin 240 N 90 N 0T T ?° + ? ? = 2 sin 330 N (270 N)sin30 195 NT ? = ? ° = Then 2 2 sin 195 N cos 234 N T T ? ? = gives tan 0.8333? = and 40? = ° And 2 195 N 303 N. sin 40 T = =° EVALUATE: The monkey is closer to the right rope than to the left one, so the tension is larger in the right rope. The horizontal components of the tensions must be equal in magnitude and opposite in direction. Since 2 1,T T> the rope on the right must be at a greater angle above the horizontal to have the same horizontal component as the tension in the other rope. 11.20. IDENTIFY: Apply the first and second conditions for equilibrium to the beam. SET UP: The free-body diagram for the beam is given in Figure 11.20. EXECUTE: The cable is given as perpendicular to the beam, so the tension is found by taking torques about the pivot point; (3.00 m) (1.00 kN)(2.00 m)cos25.0 (5.00 kN)(4.50 m)cos25.0T = ° + ° , and 7.40 kNT = . The vertical component of the force exerted on the beam by the pivot is the net weight minus the upward component of T , 6.00 kN cos25.0 0.17 kN.T? ° = The horizontal force is sin 25.0 3.13 kN.T ° = EVALUATE: The vertical component of the tension is nearly the same magnitude as the total weight of the object and the vertical component of the force exerted by the pivot is much less than its horizontal component. Figure 11.20 11.21. (a) IDENTIFY and SET UP: Use Eq.(10.3) to calculate the torque (magnitude and direction) for each force and add the torques as vectors. See Figure 11.21a. EXECUTE: 1 1 1 (8.00 N)(3.00 m)Fl? = = + 1 24.0 N m? = + ? 2 2 2 (8.00 N)( 3.00 m)F l l? = ? = ? + 2 24.0 N m (8.00 N)l? = ? ? ? Figure 11.21a 1 2 24.0 N m 24.0 N m (8.00 N) (8.00 N)z l l? ? ?= + = + ? ? ? ? = ?? Want l that makes 6.40 N mz? = ? ?? (net torque must be clockwise) (8.00 N) 6.40 N ml? = ? ? (6.40 N m)/8.00 N 0.800 ml = ? = 11-10 Chapter 11 (b) 2 1? ?> since 2F has a larger moment arm; the net torque is clockwise. (c) See Figure 11.21b. 1 1 1 (8.00 N)Fl l? = ? = ? 2 0? = since 2F ! is at the axis Figure 11.21b 6.40 N mz? = ? ?? gives (8.00 N) 6.40 N ml? = ? ? 0.800 m,l = same as in part (a). EVALUATE: The force couple gives the same magnitude of torque for the pivot at any point. 11.22. IDENTIFY: 0l FY A l ?= ? SET UP: 2 4 250.0 cm 50.0 10 mA ?= = × . EXECUTE: relaxed: 44 2 2(0.200 m)(25.0 N) 3.33 10 Pa(50.0 10 m )(3.0 10 m)Y ? ?= = ×× × maximum tension: 54 2 2 (0.200 m)(500 N) 6.67 10 Pa (50.0 10 m )(3.0 10 m) Y ? ?= = ×× × EVALUATE: The muscle tissue is much more difficult to stretch when it is under maximum tension. 11.23. IDENTIFY and SET UP: Apply Eq.(11.10) and solve for A and then use 2A r?= to get the radius and 2d r= to calculate the diameter. EXECUTE: 0 l F Y A l ?= ? so 0 l F A Y l ?= ? (A is the cross-section area of the wire) For steel, 112.0 10 PaY = × (Table 11.1) Thus 6 211 2 (2.00 m)(400 N) 1.6 10 m . (2.0 10 Pa)(0.25 10 m) A ??= = ×× × 2 ,A r?= so 6 2 4/ 1.6 10 m / 7.1 10 mr A ? ?? ?= = × = × 32 1.4 10 m 1.4 mmd r ?= = × = EVALUATE: Steel wire of this diameter doesn?t stretch much; 0/ 0.12%.l l? = 11.24. IDENTIFY: Apply Eq.(11.10). SET UP: From Table 11.1, for steel, 112.0 10 PaY = × and for copper, 111.1 10 PaY = × . 2 4 2( / 4) 1.77 10 mA d? ?= = × . 4000 NF? = for each rod. EXECUTE: (a) The strain is 0 l F l YA ? = . For steel 411 4 2 0 (4000N) 1.1 10 . (2.0 10 Pa)(1.77 10 m ) l l ? ? ? = = ×× × Similarly, the strain for copper is 42.1 10 .?× (b) Steel: 4 5(1.1 10 )(0.750 m) 8.3 10 m? ?× = × . Copper: 4 4(2.1 10 )(0.750 m) 1.6 10 m? ?× = × . EVALUATE: Copper has a smaller Y and therefore a greater elongation. 11.25. IDENTIFY: 0l FY A l ?= ? SET UP: 2 4 20.50 cm 0.50 10 mA ?= = × EXECUTE: 114 2 2(4.00 m)(5000 N) 2.0 10 Pa(0.50 10 m )(0.20 10 m)Y ? ?= = ×× × EVALUATE: Our result is the same as that given for steel in Table 11.1. 11.26. IDENTIFY: 0l FY A l ?= ? SET UP: 2 3 2 5 2(3.5 10 m) 3.85 10 mA r? ? ? ?= = × = × . The force applied to the end of the rope is the weight of the climber: 2(65.0 kg)(9.80 m/s ) 637 NF? = = . EXECUTE: 85 2(45.0 m)(637 N) 6.77 10 Pa(3.85 10 m )(1.10 m)Y ?= = ×× EVALUATE: Our result is a lot smaller than the values given in Table 11.1. An object made of rope material is much easier to stretch than if the object were made of metal. Equilibrium and Elasticity 11-11 11.27. IDENTIFY: Use the first condition of equilibrium to calculate the tensions 1T and 2T in the wires (Figure 11.27a). Then use Eq.(11.10) to calculate the strain and elongation of each wire. Figure 11.27a SET UP: The free-body diagram for 2m is given in Figure 11.27b. EXECUTE: y yF ma=? 2 2 0T m g? = 2 98.0 NT = Figure 11.27b SET UP: The free-body-diagram for 1m is given in Figure 11.27c EXECUTE: y yF ma=? 1 2 1 0T T m g? ? = 1 2 1T T m g= + 1 98.0 N 58.8 N 157 NT = + = Figure 11.27c (a) stress strain Y = so stressstrain F Y AY ?= = upper wire: 31 7 2 11 157 N strain 3.1 10 (2.5 10 m )(2.0 10 Pa) T AY ? ?= = = ×× × lower wire: 32 7 2 11 98 N strain 2.0 10 (2.5 10 m )(2.0 10 Pa) T AY ? ?= = = ×× × (b) 0strain /l l= ? so 0 (strain)l l? = upper wire: 3 3(0.50 m)(3.1 10 ) 1.6 10 m 1.6 mml ? ?? = × = × = lower wire: 3 3(0.50 m)(2.0 10 ) 1.0 10 m 1.0 mml ? ?? = × = × = EVALUATE: The tension is greater in the upper wire because it must support both objects. The wires have the same length and diameter, so the one with the greater tension has the greater strain and elongation. 11.28. IDENTIFY: Apply Eqs.(11.8), (11.9) and (11.10). SET UP: The cross-sectional area of the post is 2 2 2(0.125 m) 0.0491 mA r? ?= = = . The force applied to the end of the post is 2 4(8000 kg)(9.80 m/s ) 7.84 10 NF? = = × . The Young?s modulus of steel is 112.0 10 PaY = × . EXECUTE: (a) 4 6 2 7.84 10 N stress 1.60 10 Pa 0.0491 m F A ? ×= = = × (b) 6 6 11 stress 1.60 10 Pa strain 8.0 10 2.0 10 PaY ?×= = ? = ? ×× . The minus sign indicates that the length decreases. (c) 6 50 (strain) (2.50 m)( 8.0 10 ) 2.0 10 ml l ? ?? = = ? × = ? × EVALUATE: The fractional change in length of the post is very small. 11.29. IDENTIFY: F pA? = , so net ( )F p A= ? . SET UP: 51 atm 1.013 10 Pa= × . EXECUTE: 5 2 6(2.8 atm 1.0 atm)(1.013 10 Pa/atm)(50.0 m ) 9.1 10 N.? × = × EVALUATE: This is a very large net force. 11-12 Chapter 11 11.30. IDENTIFY: Apply Eq.(11.13). SET UP: 0V pV B ?? = ? . p? is positive when the pressure increases. EXECUTE: (a) The volume would increase slightly. (b) The volume change would be twice as great. (c) The volume change is inversely proportional to the bulk modulus for a given pressure change, so the volume change of the lead ingot would be four times that of the gold. EVALUATE: For lead, 104.1 10 PaB = × , so /p B? is very small and the fractional change in volume is very small. 11.31. IDENTIFY: /p F A= SET UP: 2 4 21 cm 1 10 m?= × EXECUTE: (a) 64 2250 N 3.33 10 Pa.0.75 10 m? = ×× (b) 6 4 2(3.33 10 Pa)(2)(200 10 m ) 133 kN.?× × = EVALUATE: The pressure in part (a) is over 30 times larger than normal atmospheric pressure. 11.32. IDENTIFY: Apply Eq.(11.13). Density /m V= . SET UP: At the surface the pressure is 51.0 10 Pa× , so 81.16 10 Pap? = × . 30 1.00 mV = . At the surface 31.00 m of water has mass 31.03 10 kg× . EXECUTE: (a) 0( )p VB V ?= ? ? gives 8 3 30 9 ( ) (1.16 10 Pa)(1.00 m ) 0.0527 m 2.2 10 Pa p V V B ? ×? = ? = ? = ?× (b) At this depth 31.03 10 kg× of seawater has volume 30 0.9473 mV V+ ? = . The density is 3 3 3 3 1.03 10 kg 1.09 10 kg/m 0.9473 m × = × . EVALUATE: The density is increased because the volume is compressed due to the increased pressure. 11.33. IDENTIFY and SET UP: Use Eqs.(11.13) and (11.14) to calculate B and k . EXECUTE: 6 3 9 3 0 (3.6 10 Pa)(600 cm ) 4.8 10 Pa / ( 0.45 cm ) p B V V ? ×= ? = ? = + ×? ? 9 10 11/ 1/ 4.8 10 Pa 2.1 10 Pak B ? ?= = × = × EVALUATE: k is the same as for glycerine (Table 11.2). 11.34. IDENTIFY: Apply Eq.(11.17). SET UP: 59.0 10 NF = ×" . 2(0.100 m)(0.500 10 m)A ?= × . 0.100 mh = . From Table 11.1, 107.5 10 PaS = × for steel. EXECUTE: (a) 5 || 2 2 10 (9 10 N) Shear strain 2.4 10 . [(0.100 m)(0.500 10 m)][7.5 10 Pa] F AS ? ? ×= = = ×× × (b) Using Eq.(11.16), 3(Shear strain) (0.024)(0.100 m) 2.4 10 mx h ?= ? = = × . EVALUATE: This very large force produces a small displacement; / 2.4%x h = . 11.35. IDENTIFY: The forces on the cube must balance. The deformation x is related to the force by F hS A x = " . F F=" since F is applied parallel to the upper face. SET UP: 2(0.0600 m)A = and 0.0600 mh = . Table 11.1 gives 104.4 10 PaS = × for copper and 100.6 10 Pa× for lead. EXECUTE: (a) Since the horizontal forces balance, the glue exerts a force F in the opposite direction. (b) 2 3 10 5(0.0600 m) (0.250 10 m)(4.4 10 Pa) 6.6 10 N 0.0600 m AxS F h ?× ×= = = × (c) 5 2 10 (6.6 10 N)(0.0600 m) 1.8 mm (0.0600 m) (0.6 10 Pa) Fh x AS ×= = =× EVALUATE: Lead has a smaller S than copper, so the lead cube has a greater deformation than the copper cube. 11.36. IDENTIFY and SET UP: Use Eq.(11.17). Same material implies same S EXECUTE: stress strain S = so /stressstrain F A S S = = " and same forces implies same .F" Equilibrium and Elasticity 11-13 For the smaller object, 1 1(strain) /F AS= " For the larger object, 2 2(strain) /F A S= " 2 1 1 1 2 2 (strain) (strain) F AS A A S F A ? ?? ?= =? ?? ?? ?? ?? ? " " Larger solid has triple each edge length, so 2 19 ,A A= and 2 1 (strain) 1 (strain) 9 = EVALUATE: The larger object has a smaller deformation. 11.37. IDENTIFY and SET UP: Use Eq.(11.8). EXECUTE: 72 3 290.8 NTensile stress 3.41 10 Pa(0.92 10 m) F F A r? ? ? ? ?= = = = ×× EVALUATE: A modest force produces a very large stress because the cross-sectional area is small. 11.38. IDENTIFY: The proportional limit and breaking stress are values of the stress, /F A? . Use Eq.(11.10) to calculate l? . SET UP: For steel, 1020 10 PaY = × . F w? = . EXECUTE: (a) 3 10 6 2 3(1.6 10 )(20 10 Pa)(5 10 m ) 1.60 10 N.w ? ?= × × × = × (b) 30 (1.6 10 )(4.0 m) 6.4 mm F l l A Y ??? ?? = = × =? ?? ? (c) 3 10 6 2 3(6.5 10 )(20 10 Pa)(5 10 m ) 6.5 10 N.? ?× × × = × EVALUATE: At the proportional limit, the fractional change in the length of the wire is 0.16%. 11.39. IDENTIFY: The elastic limit is a value of the stress, /F A? . Apply m=?F a! ! to the elevator in order to find the tension in the cable. SET UP: 8 813 (2.40 10 Pa) 0.80 10 PaFA ? = × = × . The free-body diagram for the elevator is given in Figure 11.39. F? is the tension in the cable. EXECUTE: 8 4 2 8 4(0.80 10 Pa) (3.00 10 m )(0.80 10 Pa) 2.40 10 NF A ?? = × = × × = × . y yF ma=? applied to the elevator gives F mg ma? ? = and 4 2 22.40 10 N 9.80 m/s 10.2 m/s 1200 kg F a g m ? ×= ? = ? = EVALUATE: The tension in the cable is about twice the weight of the elevator. Figure 11.39 11.40. IDENTIFY: The breaking stress of the wire is the value of /F A? at which the wire breaks. SET UP: From Table 11.3, the breaking stress of brass is 84.7 10 Pa× . The area A of the wire is related to its diameter by 2 / 4A d?= . EXECUTE: 7 2 8 350 N 7.45 10 m , so 4 0.97 mm. 4.7 10 Pa A d A ??= = × = = × EVALUATE: The maximum force a wire can withstand without breaking is proportional to the square of its diameter. 11-14 Chapter 11 11.41. IDENTIFY: Apply the conditions of equilibrium to the climber. For the minimum coefficient of friction the static friction force has the value s sf n?= . SET UP: The free-body diagram for the climber is given in Figure 11.41. sf and n are the vertical and horizontal components of the force exerted by the cliff face on the climber. The moment arm for the force T is (1.4 m)cos10° . EXECUTE: (a) 0z? =? gives (1.4 m)cos10 (1.1 m)cos35.0 0T w? =° ° . 2(1.1 m)cos35.0 (82.0 kg)(9.80 m/s ) 525 N (1.4 m)cos10 T = =°° (b) 0xF =? gives sin 25.0 222 Nn T= =° . 0yF =? gives s cos25 0f T w+ ? =° and 2 s (82.0 kg)(9.80 m/s ) (525 N)cos25 328 Nf = ? =° . (c) ss 328 N 1.48 222 N f n ? = = = EVALUATE: To achieve this large value of s? the climber must wear special rough-soled shoes. Figure 11.41 11.42. IDENTIFY: Apply 0z? =? to the bridge. SET UP: Let the axis of rotation be at the left end of the bridge and let counterclockwise torques be positive. EXECUTE: If Lancelot were at the end of the bridge, the tension in the cable would be (from taking torques about the hinge of the bridge) obtained from 2 2(12.0 N) (600 kg)(9.80 m s )(12.0 m) (200 kg)(9.80 m s )(6.0 m)T = + , so 6860 NT = . This exceeds the maximum tension that the cable can have, so Lancelot is going into the drink. To find the distance x Lancelot can ride, replace the 12.0 m multiplying Lancelot?s weight by x and the tension 3 max by 5.80 10 NT T = × and solve for x; 3 2 2 (5.80 10 N)(12.0 m) (200 kg)(9.80 m s )(6.0 m) 9.84 m. (600 kg)(9.80 m s ) x × ?= = EVALUATE: Before Lancelot goes onto the bridge, the tension in the supporting cable is 2(6.0 m)(200 kg)(9.80 m/s ) 9800 N 12.0 m T = = , well below the breaking strength of the cable. As he moves along the bridge, the increase in tension is proportional to x, the distance he has moved along the bridge. 11.43. IDENTIFY: For the airplane to remain in level flight, both 0 and 0y zF ?? = ? = . SET UP: The free-body diagram for the airplane is given in Figure 11.43. Let y+ be upward. EXECUTE: tail wing 0F W F? ? + = . Taking the counterclockwise direction as positive, and taking torques about the point where the tail force acts, wing(3.66 m)(6700 N) (3.36 m) 0.F? + = This gives wing 7300 N(up)F = and tail 7300 N 6700 N 600 N(down).F = ? = Equilibrium and Elasticity 11-15 EVALUATE: We assumed that the wing force was upward and the tail force was downward. When we solved for these forces we obtained positive values for them, which confirms that they do have these directions. Note that the rear stabilizer provides a downward force. It does not hold up the tail of the aircraft, but serves to counter the torque produced by the wing. Thus balance, along with weight, is a crucial factor in airplane loading. Figure 11.43 11.44. IDENTIFY: Apply the first and second conditions of equilibrium to the truck. SET UP: The weight on the front wheels is fn , the normal force exerted by the ground on the front wheels. The weight on the rear wheels is rn , the normal force exerted by the ground on the rear wheels. When the front wheels come off the ground, f 0n ? . The free-body diagram for the truck without the box is given in Figure 11.44a and with the box in Figure 11.44b. The center of gravity of the truck, without the box, is a distance x from the rear wheels. EXECUTE: 0yF =? in Fig.11.44a gives r f 8820 N 10,780 N 19,600 Nw n n= + = + = 0? =? in Fig.11.44a, with the axis at the rear wheels and counterclockwise torques positive, gives f (3.00 m) 0n wx? = and f (3.00 m) 10,780 N (3.00 m) 1.65 m19,600 N n x w ? ?= = =? ?? ? . (a) 0? =? in Fig.11.44b, with the axis at the rear wheels and counterclockwise torques positive, gives box f(1.00 m) (3.00 m) (1.65 m) 0w n w+ ? = . f (3600 N)(1.00 m) (19,600 N)(1.65 m) 9,580 N 3.00 m n ? += = 0yF =? gives r f boxn n w w+ = + and r 3600 N 19,600 N 9580 N 13,620 Nn = + ? = . There is 9,580 N on the front wheels and 13,620 N on the rear wheels. (b) f 0n ? . 0? =? gives box (1.00 m) (1.65 m) 0w w? = and 4box 1.65 3.23 10 Nw w= = × . EVALUATE: Placing the box on the tailgate in part (b) reduces the normal force exerted at the front wheels. Figure 11.44a, b 11.45. IDENTIFY: In each case, to achieve balance the center of gravity of the system must be at the fulcrum. Use Eq.(11.3) to locate cmx , with im replaced by iw . SET UP: Let the origin be at the left-hand end of the rod and take the x+ axis to lie along the rod. Let 1 255 Nw = (the rod) so 1 1.00 mx = , let 2 225 Nw = so 2 2.00 mx = and let 3w W= . In part (a) 3 0.500 mx = and in part (b) 3 0.750 mx = . EXECUTE: (a) cm 1.25 mx = . 1 1 2 2 3 3cm 1 2 3 w x w x w x x w w w + += + + gives 1 2 cm 1 1 2 2 3 3 cm ( )w w x w x w x w x x + ? ?= ? and (480 N)(1.25 m) (255 N)(1.00 m) (225 N)(2.00 m) 140 N 0.500 m 1.25 m W ? ?= =? . (b) Now 3 140 Nw W= = and 3 0.750 mx = . cm (255 N)(1.00 m) (225 N)(2.00 m) (140 N)(0.750 m) 1.31 m 255 N 225 N 140 N x + += =+ + . W must be moved 1.31 m 1.25 m 6 cm? = to the right. EVALUATE: Moving W to the right means cmx for the system moves to the right. 11-16 Chapter 11 11.46. IDENTIFY: The center of gravity of the object must have the same x coordinate as the hook. Use Eq.(11.3) for cmx . The mass of a segment is proportional to its length. Define ? to be the mass per unit length, so i im l?= , where il is the length of a piece that has mass im . SET UP: Use coordinates with the origin at the right-hand edge of the object and x+ to the left. cmx L= . The mass of each piece can be taken at its center of gravity, which is at its geometrical center. Let 1 be the horizontal piece of length L, 2 be the vertical piece of length L and 3 be the horizontal piece with length x. EXECUTE: 1 1 2 2 3 3cm 1 2 3 m x m x m x x m m m + += + + gives ( / 2) ( / 2)L L x x L L L x ? ? ? ? ? += + + . ? divides out and the equation reduces to 2 22 3 0x xL L? ? = . 12 (2 4 )x L L= ± . x must be positive, so 3x L= . EVALUATE: cmx L= is equivalent to saying that the net torque is zero for an axis at the hook. 11.47. IDENTIFY: Apply the conditions of equilibrium to the horizontal beam. Since the two wires are symmetrically placed on either side of the middle of the sign, their tensions are equal and are each equal to w / 2 137 NT mg= = . SET UP: The free-body diagram for the beam is given in Figure 11.47. vF and hF are the horizontal and vertical forces exerted by the hinge on the sign. Since the cable is 2.00 m long and the beam is 1.50 m long, 1.50 m cos 2.00 m ? = and 41.4? = ° . The tension cT in the cable has been replaced by its horizontal and vertical components. EXECUTE: (a) 0z? =? gives c beam w w(sin 41.4 )(1.50 m) (0.750 m) (1.50 m) (0.60 m) 0T w T T? ? ? =° . 2 c (18.0 kg)(9.80 m/s )(0.750 m) (137 N)(1.50 m 0.60 m) 423 N (1.50 m)(sin 41.4 ) T + += =° . (b) 0yF =? gives v c beam wsin 41.4 2 0F T w T+ ? ? =° and 2 v w beam c2 sin 41.4 2(137 N) (18.0 kg)(9.80 m/s ) (423 N)(sin 41.4 ) 171 NF T w T= + ? = + ? =° ° . The hinge must be able to supply a vertical force of 171 N. EVALUATE: The force from the two wires could be replaced by the weight of the sign acting at a point 0.60 m to the left of the right-hand edge of the sign. Figure 11.47 11.48. IDENTIFY: Apply 0z? =? to the hammer. SET UP: Take the axis of rotation to be at point A. EXECUTE: The force 1F ! is directed along the length of the nail, and so has a moment arm of (0.800 m)sin 60° . The moment arm of 2F ! is 0.300 m, so 2 1 (0.0800 m)sin 60 (500 N)(0.231) 116 N. (0.300 m) F F °= = = EVALUATE: The force 2F that must be applied to the hammer handle is much less than the force that the hammer applies to the nail, because of the large difference in the lengths of the moment arms. 11.49. IDENTIFY: Apply the first and second conditions of equilibrium to the bar. SET UP: The free-body diagram for the bar is given in Figure 11.49. n is the normal force exerted on the bar by the surface. There is no friction force at this surface. hH and vH are the components of the force exerted on the Equilibrium and Elasticity 11-17 bar by the hinge. The components of the force of the bar on the hinge will be equal in magnitude and opposite in direction. EXECUTE: x xF ma=? h 120 NF H= = y yF ma=? v 0n H? = v ,H n= but we don?t know either of these forces. Figure 11.49 0B? =? gives (4.00 m) (3.00 m) 0F n? = 4 3(4.00 m/3.00 m) (120 N) 160 Nn F= = = and then v 160 NH = Force of bar on hinge: horizontal component 120 N, to right vertical component 160 N, upward EVALUATE: h v/ 120/160 3.00/ 4.00,H H = = so the force the hinge exerts on the bar is directed along the bar. n! and F ! have zero torque about point A, so the line of action of the hinge force H ! must pass through this point also if the net torque is to be zero. 11.50. IDENTIFY: Apply 0z? =? to the piece of art. SET UP: The free-body diagram for the piece of art is given in Figure 11.50. EXECUTE: 0z? =? gives (1.25 m) (1.02 m) 0BT w? = . 1.02 m(358 N) 292 N1.25 mBT ? ?= =? ?? ? . 0yF =? gives 0A BT T w+ ? = and 358 N 292 N 66 NA BT w T= ? = ? = . EVALUATE: If we consider the sum of torques about the center of gravity of the piece of art, AT has a larger moment arm than BT , and this is why A BT T< . Figure 11.50 11.51. IDENTIFY: Apply the conditions of equilibrium to the beam. SET UP: The free-body diagram for the beam is given in Figure 11.51. Let T ? and T ? be the tension in the two cables. Each tension has been replaced by its horizontal and vertical components. EXECUTE: (a) The center of gravity of the beam is a distance / 2L from each end and 0z? =? with the axis at the center of gravity of the beam gives sin ( / 2) sin ( / 2) 0T L T L? ?? ?? + = . sin sinT T? ?? ?= . 0xF =? gives cos cosT T? ?? ?= . Dividing the first equation by the second gives tan tan? ?= and ? ?= . Then the equations also say T T? ?= . (b) The center of gravity of the beam is a distance 3 / 4L from the left-hand end so a distance / 4L from the right- hand end. 0z? =? with the axis at the center of gravity of the beam gives sin (3 / 2) sin ( / 2) 0T L T L? ?? ?? + = and 3 sin sinT T? ?? ?= . 0xF =? gives cos cosT T? ?? ?= . Dividing the first equation by the second gives 3tan tan? ?= . 11-18 Chapter 11 EVALUATE: 3tan tan? ?= requires ? ?> . The cable closest to the center of gravity must be closer to the vertical direction. cos cos T T? ? ? ? ? ?= ? ?? ? and ? ?> means the tension is greater in the wire that is closest to the center of gravity. Figure 11.51 11.52. IDENTIFY: Apply the first and second conditions for equilibrium to the bridge. SET UP: Find torques about the hinge. Use L as the length of the bridge and T Band w w for the weights of the truck and the raised section of the bridge. Take y+ to be upward and x+ to be to the right. EXECUTE: (a) ( ) ( )3 1T B4 2sin70 cos30 cos30TL w L w L° = ° + ° , so ( ) 23 1T B 54 2 (9.80 m s )cos30 2.57 10 N. sin 70 m m T + °= = ×° (b) Horizontal: ( ) 5cos 70 30 1 97 10 NT .° ? ° = × (to the right). Vertical: 5 T B sin 40 2.46 10 Nw w T+ ? ° = × (upward). EVALUATE: If ? is the angle of the hinge force above the horizontal, 5 5 2.46 10 N tan 1.97 10 N ? ×= × and 51.3? = ° . The hinge force is not directed along the bridge. 11.53. IDENTIFY: Apply the conditions of equilibrium to the cylinder. SET UP: The free-body diagram for the cylinder is given in Figure 11.53. The center of gravity of the cylinder is at its geometrical center. The cylinder has radius R. EXECUTE: (a) T produces a clockwise torque about the center of gravity so there must be a friction force, that produces a counterclockwise torque about this axis. (b) Applying 0z? =? to an axis at the center of gravity gives 0TR fR? + = and T f= . 0z? =? applied to an axis at the point of contact between the cylinder and the ramp gives (2 ) sin 0T R MgR ?? + = . ( / 2)sinT Mg ?= . EVALUATE: We can show that 0xF =? and 0yF =? , for x and y axes parallel and perpendicular to the ramp, or for x and y axes that are horizontal and vertical. Figure 11.53 Equilibrium and Elasticity 11-19 11.54. IDENTIFY: Apply the first and second conditions of equilibrium to the ladder. SET UP: Take torques about the pivot. Let y+ be upward. EXECUTE: (a) The force VF that the ground exerts on the ladder is given to be vertical, so 0z? =? gives V (6.0 m)sin (250 N)(4.0 m)sin (750 N)(1.50 m)sinF ? ? ?= + , so V 354 N.F = (b) There are no other horizontal forces on the ladder, so the horizontal pivot force is zero. The vertical force that the pivot exerts on the ladder must be (750 N) (250 N) (354 N) 646 N,+ ? = up, so the ladder exerts a downward force of 646 N on the pivot. (c) The results in parts (a) and (b) are independent of ?. EVALUATE: All the forces on the ladder are vertical, so all the moment arms are vertical and are proportional to sin? . Therefore, sin? divides out of the torque equations and the results are independent of ? . 11.55. IDENTIFY: Apply the first and second conditions for equilibrium to the strut. SET UP: Denote the length of the strut by L . EXECUTE: (a) and .V mg w H T= + = To find the tension, take torques about the pivot point. 2 2 sin cos cos 3 3 6 L T L ? w L ? mg ?? ? ? ? ? ?= +? ? ? ? ? ?? ? ? ? ? ? and cot4 mg T w ?? ?= +? ?? ? . (b) Solving the above for w , and using the maximum tension for ,T 2 tan (700 N) tan55 0 (5.0 kg)(9.80 m s ) 951 N. 4 mg w T ? .= ? = ° ? = (c) Solving the expression obtained in part (a) for tan ? and letting 0, tan 0.700, so 4 00 . 4 mg? ? ? . T ? = = = ° EVALUATE: As the strut becomes closer to the horizontal, the moment arm for the horizontal tension force approaches zero and the tension approaches infinity. 11.56. IDENTIFY: Apply the first and second conditions of equilibrium to each rod. SET UP: Apply 0yF =? with y+ upward and apply 0? =? with the pivot at the point of suspension for each rod. EXECUTE: (a) The free-body diagram for each rod is given in Figure 11.56. (b) 0? =? for the lower rod: (6.0 N)(4.0 cm) (8.0 cm)Aw= and 3.0 NAw = . 0yF =? for the lower rod: 3 6.0 N 9.0 NAS w= + = 0? =? for the middle rod: 3(3.0 cm) (5.0 cm)Bw S= and 5.0 (9.0 N) 15.0 N3.0Bw ? ?= =? ?? ? . 0yF =? for the middle rod: 2 39.0 N 24.0 NS S= + = 0? =? for the upper rod: 2 (2.0 cm) (6.0 cm)CS w= and 2.0 (24.0 N) 8.0 N6.0Cw ? ?= =? ?? ? . 0yF =? for the upper rod: 1 2 32.0 NCS S w= + = . In summary, 3.0 NAw = , 15.0 NBw = , 8.0 NCw = . 1 32.0 NS = , 2 24.0 NS = , 3 9.0 NS = . (c) The center of gravity of the entire mobile must lie along a vertical line that passes through the point where 1S is located. EVALUATE: For the mobile as a whole the vertical forces must balance, so 1 6.0 NA B CS w w w= + + + . Figure 11.56 11-20 Chapter 11 11.57. IDENTIFY: Apply 0z? =? to the beam. SET UP: The free-body diagram for the beam is given in Figure 11.57. EXECUTE: 0, axis at hingez?? = , gives (6.0 m)(sin 40 ) (3.75 m)(cos30 ) 0T w° ? ° = and 7600 NT = . EVALUATE: The tension in the cable is less than the weight of the beam. sin 40T ° is the component of T that is perpendicular to the beam. Figure 11.57 11.58. IDENTIFY: Apply the first and second conditions of equilibrium to the drawbridge. SET UP: The free-body diagram for the drawbridge is given in Figure 11.58. vH and hH are the components of the force the hinge exerts on the bridge. EXECUTE: (a) 0z? =? with the axis at the hinge gives (7.0 m)(cos37 ) (3.5 m)(sin37 ) 0w T? + =° ° and 5cos37 (45,000 N)2 1.19 10 N sin37 tan37 T w= = = ×°° ° (b) 0xF =? gives 5h 1.19 10 NH T= = × . 0yF =? gives 4v 4.50 10 NH w= = × . 2 2 5 h v 1.27 10 NH H H= + = × . v h tan H H ? = and 20.7? = ° . The hinge force has magnitude 51.27 10 N × and is directed at 20.7° above the horizontal. EVALUATE: The hinge force is not directed along the bridge. If it were, it would have zero torque for an axis at the center of gravity of the bridge and for that axis the tension in the cable would produce a single, unbalanced torque. Figure 11.58 11.59. IDENTIFY: Apply the first and second conditions of equilibrium to the beam. SET UP: The free-body diagram for the beam is given in Figure 11.59. Figure 11.59 Equilibrium and Elasticity 11-21 EXECUTE: (a) 0,z? =? axis at lower end of beam Let the length of the beam be L. (sin 20 ) cos40 0 2 L T L mg ? ?° = ? ° =? ?? ? 1 2 cos40 2700 N sin 20 mg T °= =° (b) Take y+ upward. 0yF =? gives sin60 0n w T? + ° = so 73.6 Nn = 0xF =? gives s cos60 1372 Nf T= ° = s s ,f n?= ss 1372 N 1973.6 N f n ? = = = EVALUATE: The floor must be very rough for the beam not to slip. The friction force exerted by the floor is to the left because T has a component that pulls the beam to the right. 11.60. IDENTIFY: Apply 0z? =? to the beam. SET UP: The center of mass of the beam is 1.0 m from the suspension point. EXECUTE: (a) Taking torques about the suspension point, (4.00 m)sin30 (140.0 N)(1.00 m)sin30 (100 N)(2.00 m)sin30w + =° ° ° . The common factor of sin30° divides out, from which 15.0 N.w = (b) In this case, a common factor of sin 45° would be factored out, and the result would be the same. EVALUATE: All the forces are vertical, so the moments are all horizontal and all contain the factor sin? , where ? is the angle the beam makes with the horizontal. 11.61. IDENTIFY: Apply 0z? =? to the flagpole. SET UP: The free-body diagram for the flagpole is given in Figure 11.61. Let clockwise torques be positive. ? is the angle the cable makes with the horizontal pole. EXECUTE: (a) Taking torques about the hinged end of the pole (200 N)(2.50 m) (600 N)(5.00 m) (5.00 m) 0yT+ ? = . 700 NyT = . The x-component of the tension is then 2 2(1000 N) (700 N) 714 NxT = ? = . tan 5.00 m y x Th T ? = = . The height above the pole that the wire must be attached is 700(5.00 m) 4.90 m 714 = . (b) The y-component of the tension remains 700 N. Now 4.40 m tan 5.00 m ? = and 41.35? = ° , so 700 N 1060 N sin sin 41.35 yTT ?= = =° , an increase of 60 N. EVALUATE: As the wire is fastened closer to the hinged end of the pole, the moment arm for T decreases and T must increase to produce the same torque about that end. Figure 11.61 11.62. IDENTIFY: Apply 0=?F! to each object, including the point where D, C and B are joined. Apply 0z? =? to the rod. SET UP: To find and ,C DT T use a coordinate system with axes parallel to the cords. EXECUTE: A and B are straightforward, the tensions being the weights suspended: 2(0.0360 kg)(9.80 m/s ) 0.353 NA? = = and 2(0.0240 kg 0.0360 kg)(9.80 m s ) 0.588 NBT = + = . Applying 0xF =? and 0yF =? to the point where the cords are joined, cos36.9 0.470 NC BT T= ° = and cos53.1 0.353 ND BT T= ° = . To find ,ET take torques about the point where string F is attached. 11-22 Chapter 11 2(1.00 m) sin36.9 (0.800 m) sin53.1 (0.200 m) (0 120 kg)(9 80 m s )(0 500 m)E D CT T T . . .= ° + ° + and 0.833 N.ET = FT may be found similarly, or from the fact that E FT T+ must be the total weight of the ornament. 2(0.180kg)(9.80m s ) 1.76 N, from which 0.931 N.FT= = EVALUATE: The vertical line through the spheres is closer to F than to E, so we expect F ET T> , and this is indeed the case. 11.63. IDENTIFY: Apply the equilibrium conditions to the plate. sinFr? ?= . SET UP: The free-body diagram for the plate is sketched in Figure 11.63. For the force T (tension in the cable), 2 2sin sinT Tr T h d? ? ?= = + . EXECUTE: (a) 0z? =? gives 2 2 sin 02dT h d W?+ ? = . T is least for 90? = ° , and in that case tan hd? = so 1tan h d ? ? ? ?= ? ?? ? . Then 2 22 d T W h d = + . (b) 0xF =? gives h 2 22 2 2 2sin 2( )2 Wd h Whd F T h dh d h d ? ? ?= = =? ? ++ +? ? . 0yF =? gives v cos 0F T W?+ ? = and 2 2 2 v 2 2 2 22 2 2 2 2 1 2( ) 2( )2 Wd d d h d F W W W h d h dh d h d ? ?? ? ? ? += ? = ? =? ?? ? ? ?+ ++ + ? ?? ?? ? EVALUATE: The angle ? that the net force exerted by the hinge makes with the horizontal is given by 2 2 2 2 2 2 v 2 2 h (2 ) 2( ) 2 tan 2( ) F W h d h d h d F h d Whd hd ? + + += = =+ . This force does not lie along the diagonal of the plate. Figure 11.63 11.64. IDENTIFY: Apply Eq.(11.10) and the relation 0 0/ /w w l l?? = ? ? that is given in the problem. SET UP: The steel rod in Example 11.5 has 40/ 9.0 10l l ?? = × . For nickel, 112.1 10 PaY = × . The width 0w is 0 4 /w A ?= . EXECUTE: (a) 4 4 20 ( ) (0.23)(9.0 10 ) 4(0.30 10 m ) 1.3 m.w ? l l w ??? ?? = ? ? = ? × × = (b) 1 l w F AY AY l ? w? ? ?= = and 11 2 2 3 6 2 (2.1 10 Pa) ( (2.0 10 m) ) 0.10 10 m 3.1 10 N 0.42 2.0 10 m ? F ? ? ? ? × × ×= = ×× . EVALUATE: For nickel and steel, 1? < and the fractional change in width is less than the fractional change in length. 11.65. IDENTIFY: Apply the equilibrium conditions to the crate. When the crate is on the verge of tipping it touches the floor only at its lower left-hand corner and the normal force acts at this point. The minimum coefficient of static friction is given by the equation s sf n?= . SET UP: The free-body diagram for the crate when it is ready to tip is given in Figure 11.65. EXECUTE: (a) 0z? =? gives (1.50 m)sin53.0 (1.10 m) 0P w? =° . 31.10 m 1.15 10 N[1.50 m][sin53.0 ]P w ? ?= = ×? ?? ?° (b) 0yF =? gives cos53.0 0n w P? ? =° . 3 3cos53.0 1250 N (1.15 10 N)cos53 1.94 10 Nn w P= + = + × = ×° ° (c) 0xF =? gives 3s sin53.0 (1.15 10 N)sin53.0 918 Nf P= = × =° ° . (d) ss 3 918 N 0.473 1.94 10 N f n ? = = =× Equilibrium and Elasticity 11-23 EVALUATE: The normal force is greater than the weight because P has a downward component. Figure 11.65 11.66. IDENTIFY: Apply 0z? =? to the meter stick. SET UP: The wall exerts an upward static friction force f and a horizontal normal force n on the stick. Denote the length of the stick by l. sf n?= . EXECUTE: (a) Taking torques about the right end of the stick, the friction force is half the weight of the stick, / 2f w= . Taking torques about the point where the cord is attached to the wall (the tension in the cord and the friction force exert no torque about this point), and noting that the moment arm of the normal force is tanl ? , tan / 2 Then, ( / ) tan 0.40, so arctan (0.40) 22 .n w f n ? ? ?= ? = < < = ° (b) Taking torques as in part (a), ( ) and n tan . 2 2 l l fl w w l x l ? w wx= + ? = + In terms of the coefficient of friction s ,? s / 2 ( ) 3 2tan tan ./ 2 2 f l l x l x ? ? n l x l x ? + ? ?> = =+ + Solving for x, s s 3tan 30 2 cm. 2 tan l ? ? x . ? ? ?> =+ (c) In the above expression, setting s10 cm and solving for givesx ?= s (3 20 ) tan 0.625.1 20 l ? ? l ?> =+ EVALUATE: For 15? = ° and without the block suspended from the stick, a value of s 0.268? ? is required to prevent slipping. Hanging the block from the stick increases the value of s? that is required. 11.67. IDENTIFY: Apply the first and second conditions of equilibrium to the crate. SET UP: The free-body diagram for the crate is given in Figure 11.67. (0.375 m)cos45wl = ° 2 (1.25 m)cos45l = ° Let 1F ! and 2F ! be the vertical forces exerted by you and your friend. Take the origin at the lower left-hand corner of the crate (point A). Figure 11.67 11-24 Chapter 11 EXECUTE: y yF ma=? gives 1 2 0F F w+ ? = 2 1 2 (200 kg)(9.80 m/s ) 1960 NF F w+ = = = 0A? =? gives 2 2 0wF l wl? = 2 2 0.375 mcos45 1960 N 590 N 1.25 mcos45 wlF w l ? ? °? ?= = =? ? ? ?°? ?? ? Then 1 2 1960 N 590 N 1370 N.F w F= ? = ? = EVALUATE: The person below (you) applies a force of 1370 N. The person above (your friend) applies a force of 590 N. It is better to be the person above. As the sketch shows, the moment arm for 1F ! is less than for 2 ,F ! so must have 1 2F F> to compensate. 11.68. IDENTIFY: Apply the first and second conditions for equilibrium to the forearm. SET UP: The free-body diagram is given in Figure 11.68a, and when holding the weight in Figure 11.68b. Let y+ be upward. EXECUTE: (a) Elbow 0?? = gives B(3.80 cm) (15.0 N)(15.0 cm)F = and B 59.2 NF = . (b) 0E?? = gives B(3.80 cm) (15.0 N)(15.0 cm) (80.0 N)(33.0 cm)F = + and B 754 NF = . The biceps force has a short lever arm, so it must be large to balance the torques. (c) 0yF? = gives E B 15.0 N 80.0 N 0F F? + ? ? = and E 754N 15.0 N 80.0 N 659 NF = ? ? = . EVALUATE: (d) The biceps muscle acts perpendicular to the forearm, so its lever arm stays the same, but those of the other two forces decrease as the arm is raised. Therefore the tension in the biceps muscle decreases. Figure 11.68a, b 11.69. IDENTIFY: Apply 0z? =? to the forearm. SET UP: The free-body diagram for the forearm is given in Fig. 11.10 in the textbook. EXECUTE: (a) 0, axis at elbowz?? = gives ( )sin 0wL T ? D? = . 2 2 2 2sin so h hD ? w Th D L h D= =+ + . max max 2 2 hD w T L h D = + . (b) 2 max max 2 22 2 1 ; the derivative is positive dw T h D dD h DL h D ? ?= ?? ?++ ? ? EVALUATE: (c) The result of part (b) shows that maxw increases when D increases, since the derivative is positive. maxw is larger for a chimp since D is larger. 11.70. IDENTIFY: Apply the first and second conditions for equilibrium to the table. SET UP: Label the legs as shown in Figure 11.70a. Legs A and C are 3.6 m apart. Let the weight be placed closest to legs C and D. By symmetry, A B= and C D= . Redraw the table as viewed from the AC side. The free-body diagram in this view is given in Figure 11.70b. EXECUTE: (about right end) 0z? =? gives ( ) ( )2 (3.6 m) 90.0 N (1.8 m) 1500 N (0.50 m)A = + and 130 NA B= = . 0yF =? gives 1590 NA B C D+ + + = . Using 130 NA B= = and C D= gives 670 NC D= = . By Newton?s third law of motion, the forces A, B, C, and D on the table are the same magnitude as the forces the table exerts on the floor. Equilibrium and Elasticity 11-25 EVALUATE: As expected, the legs closest to the 1500 N weight exert a greater force on the floor. Figure 11.70a, b 11.71. IDENTIFY: Apply 0z? =? first to the roof and then to one wall. (a) SET UP: Consider the forces on the roof; see Figure 11.71a. V and H are the vertical and horizontal forces each wall exerts on the roof. 20,000 Nw = is the total weight of the roof. 2V w= so / 2V w= Figure 11.71a Apply 0z? =? to one half of the roof, with the axis along the line where the two halves join. Let each half have length L. EXECUTE: ( / 2)( / 2)(cos35.0 ) sin35.0 cos35 0w L HL VL° + ° ? ° = L divides out, and use / 2V w= 1 4sin35.0 cos35.0H w° = ° 7140 N 4tan35.0 w H = =° EVALUATE: By Newton?s 3rd law, the roof exerts a horizontal, outward force on the wall. For torque about an axis at the lower end of the wall, at the ground, this force has a larger moment arm and hence larger torque the taller the walls. (b) SET UP: The force diagram for one wall is given in Figure 11.71b. Consider the torques on this wall. Figure 11.71b H is the horizontal force exerted by the roof, as considered in part (a). B is the horizontal force exerted by the buttress. Now the angle is 40 ,° so 5959 N 4tan 40 w H = =° EXECUTE: 0,z? =? axis at the ground (40 m) (30 m) 0H B? = and 7900 N.B = EVALUATE: The horizontal force exerted by the roof is larger as the roof becomes more horizontal, since for torques applied to the roof the moment arm for H decreases. The force B required from the buttress is less the higher up on the wall this force is applied. 11.72. IDENTIFY: Apply 0z? =? to the wheel. SET UP: Take torques about the upper corner of the curb. EXECUTE: The force F! acts at a perpendicular distance R h? and the weight acts at a perpendicular distance ( )22 22 .R R h Rh h? ? = ? Setting the torques equal for the minimum necessary force, 22 .Rh hF mg R h ?= ? 11-26 Chapter 11 (b) The torque due to gravity is the same, but the force F ! acts at a perpendicular distance 2 ,R h? so the minimum force is ( ) 2 /(2 ).mg Rh hv R h? ? EVALUATE: (c) Less force is required when the force is applied at the top of the wheel, since in this case F# has a larger moment arm. 11.73. IDENTIFY: Apply the first and second conditions of equilibrium to the gate. SET UP: The free-body diagram for the gate is given in Figure 11.73. Figure 11.73 Use coordinates with the origin at B. Let AH ! and BH ! be the forces exerted by the hinges at A and B. The problem states that AH ! has no horizontal component. Replace the tension T ! by its horizontal and vertical components. EXECUTE: (a) 0B? =? gives ( sin30.0 )(4.00 m) ( cos30.0 )(2.00 m) (2.00 m) 0T T w+ ° + ° ? = (2sin30.0 cos30.0 )T w° + ° = 500 N 268 N 2sin30.0 cos30.0 2sin30.0 cos30.0 w T = = =° + ° ° + ° (b) x xF ma=? says h cos30.0 0BH T? ° = h cos30.0 (268 N)cos30.0 232 NBH T= ° = ° = (c) y yF ma=? says v v sin30.0 0A BH H T w+ + ° ? = v v sin30.0 500 N (268 N)sin30.0 366 NA BH H w T+ = ? ° = ? ° = EVALUATE: T has a horizontal component to the left so hBH must be to the right, as these are the only two horizontal forces. Note that we cannot determine vAH and vBH separately, only their sum. 11.74. IDENTIFY: Use Eq.(11.3) to locate the x-coordinate of the center of gravity of the block combinations. SET UP: The center of mass and the center of gravity are the same point. For two identical blocks, the center of gravity is midway between the center of the two blocks. EXECUTE: (a) The center of gravity of top block can be as far out as the edge of the lower block. The center of gravity of this combination is then 3 4L to the left of the right edge of the upper block, so the overhang is 3 4.L (b) Take the two-block combination from part (a), and place it on top of the third block such that the overhang of 3 4L is from the right edge of the third block; that is, the center of gravity of the first two blocks is above the right edge of the third block. The center of mass of the three-block combination, measured from the right end of the bottom block, is 6L? and so the largest possible overhang is (3 4) ( 6) 11 12.L L L+ = Similarly, placing this three-block combination with its center of gravity over the right edge of the fourth block allows an extra overhang of 8,L for a total of 25 24.L (c) As the result of part (b) shows, with only four blocks, the overhang can be larger than the length of a single block. EVALUATE: The sequence of maximum overhangs is 18 24 L , 22 24 L , 25 24 L ,?. The increase of overhang when one more block is added is decreasing. Equilibrium and Elasticity 11-27 11.75. IDENTIFY: Apply the first and second conditions of equilibrium, first to both marbles considered as a composite object and then to the bottom marble. (a) SET UP: The forces on each marble are shown in Figure 11.75. EXECUTE: 2 1.47 NBF w= = sin / 2R R? = so 30? = ° 0,z? =? axis at P (2 cos ) 0CF R wR? ? = 0.424 N 2cos30C mg F = =° 0.424 NA CF F= = Figure 11.75 (b) Consider the forces on the bottom marble. The horizontal forces must sum to zero, so sinAF n ?= 0.848 N sin30 AFn = =° Could use instead that the vertical forces sum to zero cos 0BF mg n ?? ? = 0.848 N, cos30 BF mgn ?= =° which checks. EVALUATE: If we consider each marble separately, the line of action of every force passes through the center of the marble so there is clearly no torque about that point for each marble. We can use the results we obtained to show that 0xF =? and 0yF =? for the top marble. 11.76. IDENTIFY: Apply 0z? =? to the right-hand beam. SET UP: Use the hinge as the axis of rotation and take counterclockwise rotation as positive. If wireF is the tension in each wire and 200 Nw = is the weight of each beam, wire2 2 0F w? = and wireF w= . Let L be the length of each beam. EXECUTE: (a) 0z? =? gives wire csin cos sin 02 2 2 2 2L LF L F w? ? ?? ? = , where? is the angle between the beams and cF is the force exerted by the cross bar. The length drops out, and all other quantities except cF are known, so 1 wire 2 c wire1 2 sin( /2) sin( /2) (2 ) tan cos( /2) 2 F w F F w ? ? ? ? ?= = ? . Therefore 53(260 N) tan 130 N 2 F °= = (b) The crossbar is under compression, as can be seen by imagining the behavior of the two beams if the crossbar were removed. It is the crossbar that holds them apart. (c) The upward pull of the wire on each beam is balanced by the downward pull of gravity, due to the symmetry of the arrangement. The hinge therefore exerts no vertical force. It must, however, balance the outward push of the crossbar. The hinge exerts a force 130 N horizontally to the left for the right-hand beam and 130 N to the right for the left-hand beam. Again, it?s instructive to visualize what the beams would do if the hinge were removed. EVALUATE: The force exerted on each beam increases as ? increases and exceeds the weight of the beam for 90? ? ° . 11-28 Chapter 11 11.77. IDENTIFY: Apply the first and second conditions of equilibrium to the bale. (a) SET UP: Find the angle where the bale starts to tip. When it starts to tip only the lower left-hand corner of the bale makes contact with the conveyor belt. Therefore the line of action of the normal force n passes through the left-hand edge of the bale. Consider 0A? =? with point A at the lower left-hand corner. Then 0n? = and 0,f? = so it must be that 0mg? = also. This means that the line of action of the gravity must pass through point A. Thus the free-body diagram must be as shown in Figure 11.77a EXECUTE: 0.125 m tan 0.250 m ? = 27 ,? = ° angle where tips Figure 11.77a SET UP: At the angle where the bale is ready to slip down the incline sf has its maximum possible value, s s .f n?= The free-body diagram for the bale, with the origin of coordinates at the cg is given in Figure 11.77b EXECUTE: y yF ma=? cos 0n mg ?? = cosn mg ?= s s cosf mg? ?= s( f has maximum value when bale ready to slip) x xF ma=? s sin 0f mg ?? = s cos sin 0mg mg? ? ?? = stan? ?= s 0.60? = gives that 31? = ° Figure 11.77b 27? = ° to tip; 31? = ° to slip, so tips first (b) The magnitude of the friction force didn?t enter into the calculation of the tipping angle; still tips at 27 .? = ° For s 0.40? = tips at arctan(0.40) 22? = = ° Now the bale will start to slide down the incline before it tips. EVALUATE: With a smaller s? the slope angle ? where the bale slips is smaller. 11.78. IDENTIFY: Apply 0z? =? and 0xF =? to the bale. SET UP: Let x+ be horizontal to the right. Take the rotation axis to be at the forward edge of the bale, where it contacts the horizontal surface. When the bale just begins to tip, the only point of contact is this point and the normal force produces no torque. EXECUTE: (a) 2k k (0.35)(30.0 kg)(9.80 m s ) 103 NF f n mg? ?= = = = = (b) With respect to the forward edge of the bale, the lever arm of the weight is 0.250 m 0.125 m 2 = and the lever arm h of the applied force is then k 0.125 m1(0.125 m) (0.125 m) 0.36 m 0.35 mgh F ?= = = = . EVALUATE: As k? increases, F must increase and the bale tips at a smaller h. Equilibrium and Elasticity 11-29 11.79. IDENTIFY: Apply the first and second conditions of equilibrium to the door. (a) SET UP: The free-body diagram for the door is given in Figure 11.79. Figure 11.79 Take the origin of coordinates at the center of the door (at the cg). Let An k ,Af ,Bn and kBf be the normal and friction forces exerted on the door at each wheel. EXECUTE: y yF ma=? 0A Bn n w+ ? = 950 NA Bn n w+ = = x xF ma=? k k 0A Bf f F+ ? = k kA BF f f= + k k ,A Af n?= k k ,B Bf n?= so k k( ) (0.52)(950 N) 494 NA BF n n w? ?= + = = = 0B? =? ,Bn k ,Af and kBf all have zero moment arms and hence zero torque about this point. Thus (1.00 m) (2.00 m) ( ) 0Aw n F h+ ? ? = (1.00 m) ( ) (950 N)(1.00 m) (494 N)(1.60 m) 80 N 2.00 m 2.00 mA w F h n ? ?= = = And then 950 N 950 N 80 N 870 N.B An n= ? = ? = (b) SET UP: If h is too large the torque of F will cause wheel A to leave the track. When wheel A just starts to lift off the track An and kAf both go to zero. EXECUTE: The equations in part (a) still apply. 0A Bn n w+ ? = gives 950 NBn w= = Then k k 0.52(950 N) 494 NB Bf n?= = = k k 494 NA BF f f= + = (1.00 m) (2.00 m) ( ) 0Aw n F h+ ? ? = (1.00 m) (950 N)(1.00 m) 1.92 m 494 N w h F = = = EVALUATE: The result in part (b) is larger than the value of h in part (a). Increasing h increases the clockwise torque about B due to F and therefore decreases the clockwise torque that An must apply. 11.80. IDENTIFY: Apply the first and second conditions for equilibrium to the boom. SET UP: Take the rotation axis at the left end of the boom. EXECUTE: (a) The magnitude of the torque exerted by the cable must equal the magnitude of the torque due to the weight of the boom. The torque exerted by the cable about the left end is sinTL ? . For any angle ,? sin (180 ) sin ,? ?° ? = so the tension T will be the same for either angle. The horizontal component of the force that the pivot exerts on the boom will be cos or cos(180 ) cosT T T? ? ?° ? = ? . (b) From the result of part (a), T is proportional to 1 sin? and this becomes infinite as 0 or 180 .? ?? ? ° (c) The tension is a minimum when sin? is a maximum, or 90 ,? = ° a vertical cable. (d) There are no other horizontal forces, so for the boom to be in equilibrium, the pivot exerts zero horizontal force on the boom. 11-30 Chapter 11 EVALUATE: As the cable approaches the horizontal direction, its moment arm for the axis at the pivot approaches zero, so T must go to infinity in order for the torque due to the cable to continue to equal the gravity torque. 11.81. IDENTIFY: Apply the first and second conditions of equilibrium to the pole. (a) SET UP: The free-body diagram for the pole is given in Figure 11.81. n and f are the vertical and horizontal components of the force the ground exerts on the pole. x xF ma=? 0f = The force exerted by the ground has no horizontal component. Figure 11.81 EXECUTE: 0A? =? (7.0 m)cos (4.5 m)cos 0T mg? ?+ ? = (4.5 m/7.0 m) (4.5/ 7.0)(5700 N) 3700 NT mg= = = 0yF =? 0n T mg+ ? = 5700 N 3700 N 2000 Nn mg T= ? = ? = The force exerted by the ground is vertical (upward) and has magnitude 2000 N. EVALUATE: We can verify that 0z? =? for an axis at the cg of the pole. T n> since T acts at a point closer to the cg and therefore has a smaller moment arm for this axis than n does. (b) In the 0A? =? equation the angle ? divided out. All forces on the pole are vertical and their moment arms are all proportional to cos .? 11.82. IDENTIFY: Apply the equilibrium conditions to the pole. The horizontal component of the tension in the wire is 22.0 N. SET UP: The free-body diagram for the pole is given in Figure 11.82. The tension in the cord equals the weight W. vF and hF are the components of the force exerted by the hinge. If either of these forces is actually in the opposite direction to what we have assumed, we will get a negative value when we solve for it. EXECUTE: (a) sin37.0 22.0 NT =° so 36.6 NT = . 0z? =? gives ( sin37.0 )(1.75 m) (1.35 m) 0T W? =° . (22.0 N)(1.75 m) 28.5 N 1.35 m W = = . (b) 0yF =? gives v cos37.0 0F T W? ? =° and v (36.6 N)cos37.0 55.0 N 84.2 NF = + =° . 0xF =? gives hsin37.0 0W T F? ? =° and h 28.5 N 22.0 N 6.5 NF = ? = . The magnitude of the hinge force is 2 2 h v 84.5 NF F F= + = . EVALUATE: If we consider torques about an axis at the top of the plate, we see that hF must be to the left in order for its torque to oppose the torque produced by the force W. Figure 11.82 Equilibrium and Elasticity 11-31 11.83. IDENTIFY: Apply 0z? =? to the slab. SET UP: The free-body diagram is given in Figure 11.83a. 3.75 mtan 1.75 m ? = so 65.0? = ° . 20.0 90? ?+ + =° ° so 5.0? = ° . The distance from the axis to the center of the block is 2 23.75 m 1.75 m 2.07 m 2 2 ? ? ? ?+ =? ? ? ?? ? ? ? . EXECUTE: (a) (2.07 m)sin5.0 (3.75 m)sin52.0 0w T? =° ° . 0.061T w= . Each worker must exert a force of 0.012w , where w is the weight of the slab. (b) As ? increases, the moment arm for w decreases and the moment arm for T increases, so the worker needs to exert less force. (c) 0T ? when w passes through the support point. This situation is sketched in Figure 11.83b. (1.75 m) / 2 tan (3.75 m) / 2 ? = and 25.0? = ° . If ? exceeds this value the gravity torque causes the slab to tip over. EVALUATE: The moment arm for T is much greater than the moment arm for w, so the force the workers apply is much less than the weight of the slab. Figure 11.83a, b 11.84. IDENTIFY: For a spring, F kx= . 0F lY A l ?= ? . SET UP: F F W? = = and l x? = . For copper, 1011 10 PaY = × . EXECUTE: (a) 0 0 YA YA F l x l l ? ? ? ?= ? =? ? ? ?? ? ? ? . This in the form of F kx= , with 0 YA k l = . (b) 10 4 2 5 0 (11 10 Pa) (6.455 10 m) 1.9 10 N/m 0.750 m YA k l ? ?× ×= = = × (c) 5 3(1.9 10 N/m)(1.25 10 m) 240 NW kx ?= = × × = EVALUATE: For the wire the force constant is very large, much larger than for a typical spring. 11.85. IDENTIFY: Apply Newton?s 2nd law to the mass to find the tension in the wire. Then apply Eq.(11.10) to the wire to find the elongation this tensile force produces. (a) SET UP: Calculate the tension in the wire as the mass passes through the lowest point. The free-body diagram for the mass is given in Figure 11.85a. The mass moves in an arc of a circle with radius 0.50 m.R = It has acceleration rada ! directed in toward the center of the circle, so at this point rada ! is upward. Figure 11.85a EXECUTE: y yF ma=? 2T mg mR ?? = so that 2( ).T m g R ?= + 11-32 Chapter 11 But ? must be in rad/s: (120 rev/min)(2 rad/1 rev)(1 min/60 s) 12.57 rad/s.? ?= = Then 2 2(12.0 kg)(9.80 m/s (0.50 m)(12.57 rad/s) ) 1066 N.T = + = Now calculate the elongation l? of the wire that this tensile force produces: 0 F l Y A l ?= ? so 0 10 4 2 (1066 N)(0.50 m) 0.54 cm. (7.0 10 Pa)(0.014 10 m ) F l l YA ? ?? = = =× × (b) SET UP: The acceleration rada! is directed in towards the center of the circular path, and at this point in the motion this direction is downward. The free-body diagram is given in Figure 11.85b. EXECUTE: y yF ma=? 2mg T mR ?+ = 2( )T m R g?= ? Figure 11.85b 2 2(12.0 kg)((0.50 m)(12.57 rad/s) 9.80 m/s ) 830 NT = ? = 0 10 4 2 (830 N)(0.50 m) 0.42 cm. (7.0 10 Pa)(0.014 10 m ) F l l YA ? ?? = = =× × EVALUATE: At the lowest point T and w are in opposite directions and at the highest point they are in the same direction, so T is greater at the lowest point and the elongation is greatest there. The elongation is at most 1% of the length. 11.86. IDENTIFY: 0 YA F l l? ? ?= ?? ?? ? so the slope of the graph in part (a) depends on Young?s modulus. SET UP: F? is the total load, 20 N plus the added load. EXECUTE: (a) The graph is given in Figure 11.86. (b) The slope is 42 60 N 2.0 10 N/m (3.32 3.02) 10 m? = ×? × . 4 4 110 2 3 2 3.50 m (2.0 10 N/m)= (2.0 10 N/m) 1.8 10 Pa [0.35 10 m] l Y r? ? ? ? ?? ?= × × = ×? ?? ? ×? ? ? ? (c) The stress is /F A? . The total load at the proportional limit is 60 N 20 N 80 N+ = . 8 3 2 80 N stress 2.1 10 Pa (0.35 10 m)? ?= = ×× EVALUATE: The value of Y we calculated is close to the value for iron, nickel and steel in Table 11.1. Figure 11.86 Equilibrium and Elasticity 11-33 11.87. IDENTIFY: Use the second condition of equilibrium to relate the tension in the two wires to the distance w is from the left end. Use Eqs.(11.8) and (11.10) to relate the tension in each wire to its stress and strain. (a) SET UP: stress / ,F A?= so equal stress implies /T A same for each wire. 2 2/ 2.00 mm / 4.00 mmA BT T= so 2.00B AT T= The question is where along the rod to hang the weight in order to produce this relation between the tensions in the two wires. Let the weight be suspended at point C, a distance x to the right of wire A. The free-body diagram for the rod is given in Figure 11.87. EXECUTE: 0C? =? (1.05 m ) 0B AT x T x+ ? ? = Figure 11.87 But 2.00B AT T= so 2.00 (1.05 m ) 0A AT x T x? ? = 2.10 m 2.00x x? = and 2.10 m/3.00 0.70 mx = = (measured from A). (b) SET UP: stress/strainY = gives that strain stress / / .Y F AY?= = EXECUTE: Equal strain thus implies 2 11 2 11(2.00 mm )(1.80 10 Pa) (4.00 mm )(1.20 10 Pa) A BT T=× × 4.00 1.20 1.333 . 2.00 1.80B A A T T T? ?? ?= =? ?? ?? ?? ? The 0C? =? equation still gives (1.05 m ) 0.B AT x T x? ? = But now 1.333B AT T= so (1.333 )(1.05 m ) 0A AT x T x? ? = 1.40 m 2.33x= and 1.40 m/2.33 0.60 mx = = (measured from A). EVALUATE: Wire B has twice the diameter so it takes twice the tension to produce the same stress. For equal stress the moment arm for BT (0.35 m) is half that for AT (0.70 m), since the torques must be equal. The smaller Y for B partially compensates for the larger area in determining the strain and for equal strain the moment arms are closer to being equal. 11.88. IDENTIFY: Apply Eq.(11.10) and calculate l? . SET UP: When the ride is at rest the tension F? in the rod is the weight 1900 N of the car and occupants. When the ride is operating, the tension F? in the rod is obtained by applying m=?F a! ! to a car and its occupants. The free-body diagram is shown in Figure 11.88. The car travels in a circle of radius sinr l ?= , where l is the length of the rod and ? is the angle the rod makes with the vertical. For steel, 112.0 10 PaY = × . 8.00 rev/min 0.838 rad/s? = = . EXECUTE: (a) 40 11 4 2(15.0 m)(1900 N) 1.78 10 m 0.18 mm(2.0 10 Pa)(8.00 10 m ) l F l YA ?? ?? = = = × =× × (b) x xF ma=? gives 2 2sin sinF mr ml? ? ??? = = and 2 2 3 2 1900 N (15.0 m)(0.838 rad/s) 2.04 10 N 9.80 m/s F ml?? ? ?= = = ×? ?? ? . 32.04 10 N (0.18 mm) 0.19 mm 1900 N l ? ?×? = =? ?? ? 11-34 Chapter 11 EVALUATE: y yF ma=? gives cosF mg?? = and cos /mg F? ?= . As ? increases F? increases and cos? becomes small. Smaller cos? means ? increases, so the rods move toward the horizontal as ? increases. Figure 11.88 11.89. IDENTIFY and SET UP: The tension is the same at all points along the composite rod. Apply Eqs.(11.8) and (11.10) to relate the elongations, stresses, and strains for each rod in the compound. EXECUTE: Each piece of the composite rod is subjected to a tensile force of 44.00 10 N.× (a) 0 F l Y A l ?= ? so 0F ll YA ?? = b nl l? = ? gives that 0,b 0,n b b n n F l F l Y A Y A ? ?= (b for brass and n for nickel); 0,nl L= But the F? is the same for both, so n n 0,n 0,b b b Y A l l Y A = 10 2 10 2 21 10 Pa 1.00 cm (1.40 m) 1.63 m 9.0 10 Pa 2.00 cm L ? ?? ?×= =? ?? ?×? ?? ? (b) stress / /F A T A?= = brass: 4 4 2 8stress / (4.00 10 N)/(2.00 10 m ) 2.00 10 PaT A ?= = × × = × nickel: 4 4 2 8stress / (4.00 10 N)/(1.00 10 m ) 4.00 10 PaT A ?= = × × = × (c) stress/strainY = and strain stress/Y= brass: 8 10 3strain (2.00 10 Pa)/(9.0 10 Pa) 2.22 10?= × × = × nickel: 8 10 3strain (4.00 10 Pa)/(21 10 Pa) 1.90 10?= × × = × EVALUATE: Larger Y means less l? and smaller A means greater ,l? so the two effects largely cancel and the lengths don?t differ greatly. Equal l? and nearly equal l means the strains are nearly the same. But equal tensions and A differing by a factor of 2 means the stresses differ by a factor of 2. 11.90. IDENTIFY: Apply 0 F l Y A l ? ? ??= ? ?? ? . The height from which he jumps determines his speed at the ground. The acceleration as he stops depends on the force exerted on his legs by the ground. SET UP: In considering his motion take y+ downward. Assume constant acceleration as he is stopped by the floor. EXECUTE: (a) 4 2 9 4 0 (3.0 10 m )(14 10 Pa)(0.010) 4.2 10 N l F YA l ? ? ? ??= = × × = ×? ?? ? (b) As he is stopped by the ground, the net force on him is netF F mg?= ? , where F? is the force exerted on him by the ground. From part (a), 4 42(4.2 10 N) 8.4 10 NF? = × = × and 4 2 48.4 10 N (70 kg)(9.80 m/s ) 8.33 10 NF = × ? = × . netF ma= gives 3 21.19 10 m/sa = × . 3 21.19 10 m/sya = ? × since the acceleration is upward. 0y y yv v a t= + gives 3 2 0 ( 1.19 10 m/s )(0.030 s) 35.7 m/sy yv a t= ? = ? × = . His speed at the ground therefore is 35.7 m/sv = . This speed is related to his initial height h above the floor by 212 mv mgh= and 2 2 2 (35.7 m/s) 65 m 2 2(9.80 m/s ) v h g = = = . Equilibrium and Elasticity 11-35 EVALUATE: Our estimate is based solely on compressive stress; other injuries are likely at a much lower height. 11.91. IDENTIFY and SET UP: 0 / Y F l A l?= ? (Eq.11.10 holds since the problem states that the stress is proportional to the strain.) Thus 0 / .l F l AY?? = Use proportionality to see how changing the wire properties affects .l? EXECUTE: (a) Change 0l but F? (same floodlamp), A (same diameter wire), and Y (same material) all stay the same. 0 constant, l F l AY ?? = = so 1 2 01 02 l l l l ? ?=? ? 2 1 02 01 1( / ) 2 2(0.18 mm) 0.36 mml l l l l? = ? = ? = = (b) 2 214( / 2) ,A d d? ?= = so 021 4 F l l d Y? ?? = ,F? 0 ,l Y all stay the same, so ( )2 10 4( ) / constantl d F l Y??? = = 2 2 1 1 2 2( ) ( )l d l d? = ? 2 2 2 1 1 2( / ) (0.18 mm)(1/2) 0.045 mml l d d? = ? = = (c) ,F? 0 ,l A all stay the same so 0 / constantlY F l A?? = = 1 1 2 2l Y l Y? = ? 10 10 2 1 1 2( / ) (0.18 mm)(20 10 Pa/11 10 Pa) 0.33 mml l Y Y? = ? = × × = EVALUATE: Greater l means greater ,l? greater diameter means less ,l? and smaller Y means greater .l? 11.92. IDENTIFY: Apply Eq.(11.13) and calculate V? . SET UP: The pressure increase is /w A , where w is the weight of the bricks and A is the area 2r? of the piston. EXECUTE: 2 5 2 (1420 kg)(9.80 m/s ) 1.97 10 Pa (0.150 m) p ?? = = × 0 V p B V ?? = ? gives 5 0 8 ( ) (1.97 10 Pa)(250 L) 0.0542 L 9.09 10 Pa p V V B ? ×? = ? = ? = ?× EVALUATE: The fractional change in volume is only 0.022%, so this attempt is not worth the effort. 11.93. IDENTIFY and SET UP: Apply Eqs.(11.8) and (11.15). The tensile stress depends on the component of F! perpendicular to the plane and the shear stress depends on the component of F ! parallel to the plane. The forces are shown in Figure 11.93a Figure 11.93a (a) EXECUTE: The components of F are shown in Figure 11.93b. The area of the diagonal face is / cos .A ? Figure 11.93b 2cos tensile stress cos /( / cos ) . ( / cos ) F F F A A A ?? ?? ?= = = (b) sin cos sin 2 shear stress sin /( / cos ) ( / cos ) 2 F F F F A A A A ? ? ?? ??= = = = " (using a trig identity). EVALUATE: (c) From the result of part (a) the tensile stress is a maximum for cos 1,? = so 0 .? = ° (d) From the result of part (b) the shear stress is a maximum for sin 2 1,? = so for 2 90? = ° and thus 45? = ° 11-36 Chapter 11 11.94. IDENTIFY: Apply the first and second conditions of equilibrium to the rod. Then apply Eq.(11.10) to relate the compressive force on the rod to its change in length. SET UP: For copper, 111.1 10 PaY = × . EXECUTE: (a) Taking torques about the pivot, the tension T in the cable is related to the weight by 0 0(sin ) 2, so .2sin mg T l mgl T? ?= = The horizontal component of the force that the cable exerts on the rod, and hence the horizontal component of the force that the pivot exerts on the rod, is cot 2 mg ? and the stress is cot . 2 mg A ? (b) 0 0 cot . 2 l F mgl l AY AY ?? = = l? corresponds to a decrease in length. (c) In terms of the density and length, 0( ) ,m A l?= so the stress is 0( 2)cotl g? ? and the change in length is 2 0( 2 )cotl g Y? ? . (d) Using the numerical values, the stress is 51.4 10× Pa and the change in length is 62.2 10 m.?× (e) The stress is proportional to the length and the change in length is proportional to the square of the length, and so the quantities change by factors of 2 and 4. EVALUATE: The compressive force and therefore the decrease in length increase as ? decreases and the cable becomes more nearly horizontal. 11.95. IDENTIFY: Apply the first and second conditions for equilibrium to the bookcase. SET UP: When the bookcase is on the verge of tipping, it contacts the floor only at its lower left-hand edge and the normal force acts at this point. When the bookcase is on the verge of slipping, the static friction force has its largest possible value, s n? . EXECUTE: (a) Taking torques about the left edge of the left leg, the bookcase would tip when (1500 )(0.90 m) 750 (1.80 m) F ?= = ? and would slip when s( )(1500 ) 600 ,F ?= ? = ? so the bookcase slides before tipping. (b) If F is vertical, there will be no net horizontal force and the bookcase could not slide. Again taking torques about the left edge of the left leg, the force necessary to tip the case is (1500 )(0.90 m) 13.5 kN (0.10 m) ? = . (c) To slide, the friction force is s ( cos ),f w F? ?= + and setting this equal to sinF ? and solving for F gives s ssin cos w F ? ? ? ?= ? (to slide). To tip, the condition is that the normal force exerted by the right leg is zero, and taking torques about the left edge of the left leg, sin (1.80 m) cos (0.10 m) (0.90 m),F F w? ?+ = and solving for F gives (1 9)cos 2sin w F ? ?= + (to tip). Setting the two expressions equal to each other gives s s((1 9)cos 2sin ) sin cos? ? ? ? ? ?+ = ? and solving for? gives s s (10 9) arctan 66 . (1 2 ) ?? ? ? ?= = °? ??? ? EVALUATE: The result in (c) depends not only on the numerical value of s? but also on the width and height of the bookcase. 11.96. IDENTIFY: Apply 0z? =? to the post, for various choices of the location of the rotation axis. SET UP: When the post is on the verge of slipping, sf has its largest possible value, s sf n?= . EXECUTE: (a) Taking torques about the point where the rope is fastened to the ground, the lever arm of the applied force is / 2h and the lever arm of both the weight and the normal force is tan ,h ? and so ( ) tan . 2 hF n w h ?= ? Taking torques about the upper point (where the rope is attached to the post), . 2 hfh F= Using sf n?? and solving for F, 1 1 s 1 1 1 1 2 2(400 N) 400 N tan 0.30 tan36.9 F w ? ? ? ?? ? ? ?? ? = ? =? ? ? ?°? ?? ? . (b) The above relations between , and becomeF n f 3 2 ( ) tan , , 5 5 F h n w h f F?= ? = and eliminating f and n and solving for F gives 1 s 2 5 3 5 , tan F w ? ? ?? ?? ?? ?? ? and substitution of numerical values gives 750 N to two figures. Equilibrium and Elasticity 11-37 (c) If the force is applied a distance y above the ground, the above relations become ( ) tan , ( ) ,Fy n w h F h y fh?= ? ? = which become, on eliminating and ,n f ( ) ( ) s 1 / / . tan y h y h w F ? ? ? ??? ?? ?? ? As the term in square brackets approaches zero, the necessary force becomes unboundedly large. The limiting value of y is found by setting the term in square brackets equal to zero. Solving for y gives tan tan36.9 0.71. tan 0.30 tan36.9s y h ? ? ? °= = =+ + ° EVALUATE: For the post to slip, for an axis at the top of the post the torque due to F must balance the torque due to the friction force. As the point of application of F approaches the top of the post, its moment arm for this axis approaches zero. 11.97. IDENTIFY: Apply 0z? =? to the girder. SET UP: Assume that the center of gravity of the loaded girder is at 2,L and that the cable is attached a distance x to the right of the pivot. The sine of the angle between the lever arm and the cable is then 2 2(( 2) )h h L x+ ? . EXECUTE: The tension is obtained from balancing torques about the pivot; 2 2 2, (( 2) ) hx T wL h L x ? ?? ? =? ?+ ?? ? where w is the total load . The minimum tension will occur when the term in square brackets is a maximum; differentiating and setting the derivative equal to zero gives a maximum, and hence a minimum tension, at 2min ( ) ( 2).x h L L= + However, if min , which occurs if 2,x L h L> > the cable must be attached at L, the farthest point to the right. EVALUATE: Note that minx is greater than / 2L but approaches / 2L as 0h ? . The tension is a minimum when the cable is attached somewhere on the right-hand half of the girder. 11.98. IDENTIFY: Apply the equilibrium conditions to the ladder combination and also to each ladder. SET UP: The geometry of the 3-4-5 right triangle simplifies some of the intermediate algebra. Denote the forces on the ends of the ladders by and L RF F (left and right). The contact forces at the ground will be vertical, since the floor is assumed to be frictionless. EXECUTE: (a) Taking torques about the right end, (5.00 m) (480 N)(3.40 m) (360 N)(0.90 m)LF = + , so 391 NLF = . RF may be found in a similar manner, or from 840 N 449 N.R LF F= ? = (b) The tension in the rope may be found by finding the torque on each ladder, using the point A as the origin. The lever arm of the rope is 1.50 m. For the left ladder, (1.50 m) (3.20 m) (480 N)(1.60 m), so 322.1 NLT F T= ? = (322 N to three figures). As a check, using the torques on the right ladder, (1.50 m) (1.80 m) (360 N)(0.90 m)RT F= ? gives the same result. (c) The horizontal component of the force at A must be equal to the tension found in part (b). The vertical force must be equal in magnitude to the difference between the weight of each ladder and the force on the bottom of each ladder, 480 N 391 N 449 N 360 N 89 N.? = ? = The magnitude of the force at A is then 2 2(322.1 N) (89 N) 334 N.+ = (d) The easiest way to do this is to see that the added load will be distributed at the floor in such a way that (0.36)(800 N) 679 N, and (0.64)(800 N) 961 N.L L R RF F F F? ?= + = = + = Using these forces in the form for the tension found in part (b) gives (3.20 m) (480 N)(1.60 m) (1.80 m) (360 N)(0.90 m) 937 N (1.50 m) (1.50 m) L RF FT ? ?? ?= = = . EVALUATE: The presence of the painter increases the tension in the rope, even though his weight is vertical and the tension force is horizontal. 11.99. IDENTIFY: Apply Eq.(11.14) to each material, the oil and the sodium. For each material, /p F A? = . SET UP: The total volume change, totV? , is related to the distance the piston moves by totV Ax? = . EXECUTE: The change in the volume of the oil is O Ok v p? and the change in the volume of the sodium is s s .k v p? Setting the total volume change equal to Ax (x is positive) and using ,p F A? = O O s s( )( ),Ax k V k V F A= + and solving for sk gives 2 s O O s 1A x k k V F V ? ?= ? ?? ?? ? 11-38 Chapter 11 EVALUATE: Neglecting the volume change of the oil corresponds to setting O 0k = , and in that case 2 s s A x k FV = . In either case, x is larger when sk is larger. 11.100. IDENTIFY: Write ( )p V? or ( )pV ?? in terms of p? and V? and use the fact that p V or pV ? is constant. SET UP: B is given by Eq.(11.13). EXECUTE: (a) For constant temperature ( )0T? = , ( ) ( ) ( ) 0pV p V p V? = ? + ? = and ( ) ( )p V B p V ?= ? =? . (b) In this situation, ( ) 1( ) 0, ( ) 0,? Vp V ?p V V p ?p V ? ? ?? + ? = ? + = and ( ) .p VB ?p V ?= ? =? EVALUATE: We will see later that 1? > , so B is larger in part (b). 11.101. IDENTIFY: Apply Eq.(11.10) to calculate l? . SET UP: For steel, 112.0 10 PaY = × . EXECUTE: (a) From Eq.(11.10), 2 4 10 7 2 (4.50 kg)(9.80 m/s )(1.50 m) 6.62 10 m, or 0.66 mm (20 10 Pa)(5.00 10 m ) l ??? = = ×× × to two figures. (b) 2 2(4.50 kg)(9.80 m s )(0.0500 10 m) 0.022 J.?× = (c) The magnitude F will vary with distance; the average force is 0(0.0250 cm ) 16.7 N,YA l = and so the work done by the applied force is 2 3(16.7 N)(0.0500 10 m) 8.35 10 J.? ?× = × (d) The average force the wire exerts is (450 kg) 16.7 N 60.8 N.g + = The work done is negative, and equal to 2 2(60.8 N)(0.0500 10 m) 3.04 10 J.? ?? × = ? × (e) Eq.(11.10) is in the form of Hooke?s law, with 0 YA k l = . 21el 2U kx= , so 2 21el 2 12 ( )U k x x? = ? . 4 1 6.62 10 mx ?= × and 3 42 10.500 10 m 11.62 10 mx x? ?= × + = × . The change in elastic potential energy is ( ) 10 7 2 4 2 4 2 2(20 10 Pa)(5.00 10 m ) ((11.62 10 m) (6.62 10 m) ) 3.04 10 J, 2 1.50 m ? ? ? ?× × × ? × = × the negative of the result of part (d). EVALUATE: The tensile force in the wire is conservative and obeys the relation W U= ?? . 12-1 G RAVITATION 12.1. IDENTIFY and SET UP: Use the law of gravitation, Eq.(12.1), to determine g.F EXECUTE: S MS on M 2 SM (S sun, M moon); m m F G r = = E ME on M 2 EM (E earth) m m F G r = = 22 EMS on M S M S EM 2 E on M SM E M E SM rF m m m r G F r Gm m m r ? ?? ? ? ?= =? ?? ? ? ?? ?? ? ? ?? ?? ? EM ,r the radius of the moon?s orbit around the earth is given in Appendix F as 83.84 10 m.× The moon is much closer to the earth than it is to the sun, so take the distance SMr of the moon from the sun to be SE ,r the radius of the earth?s orbit around the sun. 230 8 S on M 24 11 E on M 1.99 10 kg 3.84 10 m 2.18. 5.98 10 kg 1.50 10 m F F ? ?? ?× ×= =? ?? ?× ×? ?? ? EVALUATE: The force exerted by the sun is larger than the force exerted by the earth. The moon?s motion is a combination of orbiting the sun and orbiting the earth. 12.2. IDENTIFY: The gravity force between spherically symmetric spheres is 1 2g 2Gm mF r= , where r is the separation between their centers. SET UP: 11 2 26.67 10 N m /kgG ?= × ? . The moment arm for the torque due to each force is 0.150 m. EXECUTE: (a) For each pair of spheres, 11 2 2 7 g 2 (6.67 10 N m /kg )(1.10 kg)(25.0 kg) 1.27 10 N (0.120 m) F ? ?× ?= = × . From Figure 12.4 in the textbook we see that the forces for each pair are in opposite directions, so net 0F = . (b) The net torque is 7 8net g2 2(1.27 10 N)(0.150 m) 3.81 10 N mF l? ? ?= = × = × ? . (c) The torque is very small and the apparatus must be very sensitive. The torque could be increased by increasing the mass of the spheres or by decreasing their separation. EVALUATE: The quartz fiber must twist through a measurable angle when a small torque is applied to it. 12.3. IDENTIFY: The force exerted on the particle by the earth is w mg= , where m is the mass of the particle. The force exerted by the 100 kg ball is 1 2g 2 Gm m F r = , where r is the distance of the particle from the center of the ball. SET UP: 11 2 26.67 10 N m /kgG ?= × ? , 29.80 m/sg = . EXECUTE: gF w= gives ball2Gmm mgr = and 11 2 2 5ball 2 (6.67 10 N m /kg )(100 kg) 2.61 10 m 0.0261 mm 9.80 m/s Gm r g ? ?× ?= = = × = . It is not feasible to do this; a 100 kg ball would have a radius much larger than 0.0261 mm. EVALUATE: The gravitational force between ordinary objects is very small. The gravitational force exerted by the earth on objects near its surface is large enough to be important because the mass of the earth is very large. 12.4. IDENTIFY: Apply Eq.(12.2), generalized to any pair of spherically symmetric objects. SET UP: The separation of the centers of the spheres is 2R. EXECUTE: The magnitude of the gravitational attraction is 2 2 2 2/(2 ) /4 .GM R GM R= EVALUATE: Eq.(12.2) applies to any pair of spherically symmetric objects; one of the objects doesn't have to be the earth. 12 12-2 Chapter 12 12.5. IDENTIFY: Use Eq.(12.1) to calculate gF exerted by the earth and by the sun and add these forces as vectors. (a) SET UP: The forces and distances are shown in Figure 12.5. Let EF ! and SF ! be the gravitational forces exerted on the spaceship by the earth and by the sun. Figure 12.5 EXECUTE: The distance from the earth to the sun is 111.50 10 m.r = × Let the ship be a distance x from the earth; it is then a distance r x? from the sun. E SF F= says that 2 2E S/ /( )Gmm x Gmm r x= ? ( )22E/ /m x ms r x= ? and 2 2 S E( ) ( / )r x x m m? = S E/r x x m m? = and S E(1 / )r x m m= + 11 8 30 24 S E 1.50 10 m 2.59 10 m 1 / 1+ 1.99 10 kg/5.97 10 kg r x m m ×= = = ×+ × × (from center of earth) (b) EVALUATE: At the instant when the spaceship passes through this point its acceleration is zero. Since S Em m" this equal-force point is much closer to the earth than to the sun. 12.6. IDENTIFY: Apply Eq.(12.1) to calculate the magnitude of the gravitational force exerted by each sphere. Each force is attractive. The net force is the vector sum of the individual forces. SET UP: Let + x be to the right. EXECUTE: (a) ( )( ) ( )( ) ( ) ( ) 11 2 2 11 g 2 2 5.00 kg 10.0 kg 6.673 10 N m /kg 0.100 kg 2.32 10 0.400 m 0.600 m xF ? ?? ?= × ? ? + = ? × ?? ?? ?? ? , with the minus sign indicating a net force to the left. (b) No, the force found in part (a) is the net force due to the other two spheres. EVALUATE: The force from the 5.00 kg sphere is greater than for the 10.0 kg sphere even though its mass is less, because r is smaller for this mass. 12.7. IDENTIFY: The force exerted by the moon is the gravitational force, Mg 2Gm mF r= . The force exerted on the person by the earth is w mg= . SET UP: The mass of the moon is 22M 7.35 10 kgm = × . 11 2 26.67 10 N m /kgG ?= × ? . EXECUTE: (a) 22 11 2 2 3 moon g 8 2 (7.35 10 kg)(70 kg) (6.67 10 N m /kg ) 2.4 10 N (3.78 10 m) F F ? ? ×= = × ? = ×× . (b) 2earth (70 kg)(9.80 m/s ) 690 NF w= = = . 6moon earth/ 3.5 10F F ?= × . EVALUATE: The force exerted by the earth is much greater than the force exerted by the moon. The mass of the moon is less than the mass of the earth and the center of the earth is much closer to the person than is the center of the moon. 12.8. IDENTIFY: Use Eq.(12.2) to find the force each point mass exerts on the particle, find the net force, and use Newton?s second law to calculate the acceleration. SET UP: Each force is attractive. The particle (mass m) is a distance 1 0.200 mr = from 1 8.00 kgm = and therefore a distance 2 0.300 mr = from 2 15.0 kgm = . Let + x be toward the 15.0 kg mass. EXECUTE: 11 2 2 811 2 2 1 (8.00 kg) (6.67 10 N m /kg ) (1.334 10 N/kg) (0.200 m) Gm m m F m r ? ?= = × ? = × , in the x? -direction. 11 2 2 82 2 2 2 2 (15.0 kg) (6.67 10 N m /kg ) (1.112 10 N/kg) (0.300 m) Gm m m F m r ? ?= = × ? = × , in the x+ -direction. The net force is 8 8 9 1 2 ( 1.334 10 N/kg 1.112 10 N/kg) ( 2.2 10 N/kg)x x xF F F m m ? ? ?= + = ? × + × = ? × . 9 22.2 10 m/sxx Fa m ?= = ? × . The acceleration is 9 22.2 10 m/s?× , toward the 8.00 kg mass. EVALUATE: The smaller mass exerts the greater force, because the particle is closer to the smaller mass. Gravitation 12-3 12.9. IDENTIFY: Apply Eq.(12.1) to calculate the magnitude of each gravitational force. Each force is attractive. SET UP: The masses are 22M 7.35 10 kgm = × , 30S 1.99 10 kgm = × and 24E 5.97 10 kgm = × . Denote the earth-sun separation as 1r and the earth-moon separation as 2r . EXECUTE: (a) ( ) 20S EM 2 2 1 2 2 6.30 10 , ( ) m m Gm r r r ? ?+ = × ?? ?+? ? toward the sun. (b) The earth-moon distance is sufficiently small compared to the earth-sun distance (r2 << r1) that the vector from the earth to the moon can be taken to be perpendicular to the vector from the sun to the moon. The gravitational forces are then 20M S2 1 4.34 10 Gm m r = × ? and 20M E2 2 1.99 10 Gm m r = × ? , and so the force has magnitude 204.77 10 × ? and is directed 24.6° from the direction toward the sun. (c) ( ) ( ) 20S E M 2 2 21 2 2.37 10 , m m Gm rr r ? ?? = × ?? ??? ?? ? toward the sun. EVALUATE: The net force is very different in each of these three positions, even though the magnitudes of the forces from the sun and earth change very little. 12.10. IDENTIFY: Apply Eq.(12.1) to calculate the magnitude of each gravitational force. Each force is attractive. SET UP: The forces on one of the masses are sketched in Figure 12.10. The figure shows that the vector sum of the three forces is toward the center of the square. EXECUTE: A B A DonA B D 2 2 AB AD cos 45 2 cos 45 F 2 Gm m Gm m F F r r °= ° + = + . 11 2 2 2 11 2 2 2 3 onA 2 2 2(6.67 10 N m /kg )(800 kg) cos 45 (6.67 10 N m /kg )(800 kg) 8.2 10 N (0.10 m) (0.10 m) F ? ? ?× ? ° × ?= + = × toward the center of the square. EVALUATE: We have assumed each mass can be treated as a uniform sphere. Each mass must have an unusually large density in order to have mass 800 kg and still fit into a square of side length 10.0 cm. Figure 12.10 12.11. IDENTIFY: Use Eq.(12.2) to calculate the gravitational force each particle exerts on the third mass. The equilibrium is stable when for a displacement from equilibrium the net force is directed toward the equilibrium position and it is unstable when the net force is directed away from the equilibrium position. SET UP: For the net force to be zero, the two forces on M must be in opposite directions. This is the case only when M is on the line connecting the two particles and between them. The free-body diagram for M is given in Figure 12.11. 1 3m m= and 2m m= . If M is a distance x from 1m , it is a distance 1.00 m x? from 2m . EXECUTE: (a) 1 2 2 23 0(1.00 m )x x x mm mM F F F G G x x = + = ? + =? . 2 23(1.00 m )x x? = . 1.00 m / 3x x? = ± . Since M is between the two particles, x must be less than 1.00 m and 1.00 m 0.634 m 1 1/ 3 x = =+ . M must be placed at a point that is 0.634 m from the particle of mass 3m and 0.366 m from the particle of mass m. (b) (i) If M is displaced slightly to the right in Figure 12.11, the attractive force from m is larger than the force from 3m and the net force is to the right. If M is displaced slightly to the left in Figure 12.11, the attractive force from 3m is larger than the force from m and the net force is to the left. In each case the net force is away from equilibrium and the equilibrium is unstable. (ii) If M is displaced a very small distance along the y axis in Figure 12.11, the net force is directed opposite to the direction of the displacement and therefore the equilibrium is stable. 12-4 Chapter 12 EVALUATE: The point where the net force on M is zero is closer to the smaller mass. Figure 12.11 12.12. IDENTIFY: The force 1F ! exerted by m on M and the force 2F ! exerted by 2m on M are each given by Eq.(12.2) and the net force is the vector sum of these two forces. SET UP: Each force is attractive. The forces on M in each region are sketched in Figure 12.12a. Let M be at coordinate x on the x-axis. EXECUTE: (a) For the net force to be zero, 1F ! and 2F ! must be in opposite directions and this is the case only for 0 x L< < . 1 2 0F + F = ! ! then requires 1 2F F= . 2 2(2 )( ) GmM G m M x L x = ? . 2 22 ( )x L x= ? and 2L x x? = ± . x must be less than L, so 0.414 1 2 L x L= =+ . (b) For 0x < , 0xF > . 0xF ? as x ? ?? and xF ? +? as 0x ? . For x L> , 0xF < . 0xF ? as x ? ? and xF ? ?? as x L? . For 0 0.414x L< < , 0xF < and xF increases from ?? to 0 as x goes from 0 to 0.414L. For 0.414L x L< < , 0xF > and xF increases from 0 to +? as x goes from 0.414L to L. The graph of xF versus x is sketched in Figure 12.12b. EVALUATE: Any real object is not exactly a point so it is not possible to have both m and M exactly at 0x = or 2m and M both exactly at x L= . But the magnitude of the gravitational force between two objects approaches infinity as the objects get very close together. Figure 12.12 Gravitation 12-5 12.13. IDENTIFY: Use Eq.(12.1) to find the force exerted by each large sphere. Add these forces as vectors to get the net force and then use Newton?s 2nd law to calculate the acceleration. SET UP: The forces are shown in Figure 12.13. sin 0.80? = cos 0.60? = Take the origin of coordinate at point P. Figure 12.13 EXECUTE: 112 2(0.26 kg)(0.010 kg) 1.735 10 N(0.100 m) A A m m F G G r ?= = = × 11 2 1.735 10 N B B m m F G r ?= = × 11 11sin (1.735 10 N)(0.80) 1.39 10 NAx AF F ? ? ?= ? = ? × = ? × 11 11cos (1.735 10 N)(0.60) 1.04 10 NAy AF F ? ? ?= ? = + × = + × 11sin 1.39 10 NBx BF F ? ?= + = + × 11cos 1.04 10 NBy BF F ? ?= + = + × x xF ma=? gives Ax Bx xF F ma+ = 0 xma= so 0xa = y yF ma=? gives Ay By yF F ma+ = 112(1.04 10 N) (0.010 kg) ya ?× = 9 22.1 10 m/s ,ya ?= × directed downward midway between A and B EVALUATE: For ordinary size objects the gravitational force is very small, so the initial acceleration is very small. By symmetry there is no x-component of net force and the y-component is in the direction of the two large spheres, since they attract the small sphere. 12.14. IDENTIFY: Apply Eq.(12.4) to Pluto. SET UP: Pluto has mass 221.5 10 kgm = × and radius 61.15 10 mR = × . EXECUTE: Equation (12.4) gives ( )( )( ) 11 2 2 22 2 2 26 6.763 10 N m /kg 1.5 10 kg 0.757 m /s . 1.15 10 m g ?× ? ×= = × EVALUATE: g at the surface of Pluto is much less than g at the surface of Earth. Eq.(12.4) applies to any spherically symmetric object. 12.15. IDENTIFY: Eg 2mmF G r= , so E g 2 m a G r = , where r is the distance of the object from the center of the earth. SET UP: Er h R= + , where h is the distance of the object above the surface of the earth and 6E 6.38 10 mR = × is the radius of the earth. EXECUTE: To decrease the acceleration due to gravity by one-tenth, the distance from the center of the earth must be increased by a factor of 10, and so the distance above the surface of the earth is ( ) 7E10 1 1.38 10 m.R? = × EVALUATE: This height is about twice the radius of the earth. 12.16. IDENTIFY: Apply Eq.(12.4) to the earth and to Venus. w mg= . SET UP: 2E2 E 9.80 m/s Gm g R = = . V E0.815m m= and V E0.949R R= . E E 75.0 Nw mg= = . EXECUTE: (a) V E EV E2 2 2 V E E (0.815 ) 0.905 0.905 (0.949 ) Gm G m Gm g g R R R = = = = . (b) V V E0.905 (0.905)(75.0 N) 67.9 Nw mg mg= = = = . 12-6 Chapter 12 EVALUATE: The mass of the rock is independent of its location but its weight equals the gravitational force on it and that depends on its location. 12.17. (a) IDENTIFY and SET UP: Apply Eq.(12.4) to the earth and to Titania. The acceleration due to gravity at the surface of Titania is given by 2T T T/ ,g Gm R= where Tm is its mass and TR is its radius. For the earth, 2E E E/ .g Gm R= EXECUTE: For Titania, T E/1700m m= and T E/8,R R= so T E ET E2 2 2 T E E ( /1700) 64 0.0377 . ( /8) 1700 Gm G m Gm g g R R R ? ?= = = =? ?? ? Since 2E 9.80 m/s ,g = 2 2T (0.0377)(9.80 m/s ) 0.37 m/s .g = = EVALUATE: g on Titania is much smaller than on earth. The smaller mass reduces g and is a greater effect than the smaller radius, which increases g . (b) IDENTIFY and SET UP: Use density mass/volume.= Assume Titania is a sphere. EXECUTE: From Section 12.2 we know that the average density of the earth is 35500 kg/m . For Titania 3 3T T E3 34 4 T E3 3 /1700 512 512 (5500 kg/m ) 1700 kg/m ( /8) 1700 1700 Em m R R ? ?? ?= = = = = EVALUATE: The average density of Titania is about a factor of 3 smaller than for earth. We can write Eq.(12.4) for Titania as 4T T T3 .g GR? ?= T Eg g< both because T E? ?< and T E .R R< 12.18. IDENTIFY: Apply Eq.(12.4) to Rhea. SET UP: /m V? = . The volume of a sphere is 343V R?= . EXECUTE: 2 212.44 10 kggRM G = = × and ( ) 3 33 1.30 10 kg/m .4 /3 M R ? ?= = × EVALUATE: The average density of Rhea is about one-fourth that of the earth. 12.19. IDENTIFY: Apply Eq.(12.2) to the astronaut. SET UP: 24E 5.97 10 kgm = × and 6E 6.38 10 mR = × . EXECUTE: Eg 2mmF G r= . 3 E600 10 mr R= × + so g 610 NF = . At the surface of the earth, g 735 N.w m= = The gravity force is not zero in orbit. The satellite and the astronaut have the same acceleration so the astronaut?s apparent weight is zero. EVALUATE: In Eq.(12.2), r is the distance of the object from the center of the earth. 12.20. IDENTIFY: nn 2 n m g G R = , where the subscript n refers to the neutron star. w mg= . SET UP: 3n 10.0 10 mR = × . 30n 1.99 10 kgm = × . Your mass is 2675 N 68.9 kg.9.80 m/s w m g = = = EXECUTE: 30 11 2 2 12 2 n 3 2 1.99 10 kg (6.673 10 N m /kg ) 1.33 10 m/s (10.0 10 m) g ? ×= × ? = ×× Your weight on the neutron star would be 12 2 13n n (68.9 kg)(1.33 10 m/s ) 9.16 10 Nw mg= = × = × . EVALUATE: Since nR is much less than the radius of the sun, the gravitational force exerted by the neutron star on an object at its surface is immense. 12.21. IDENTIFY and SET UP: Use the measured gravitational force to calculate the gravitational constant G , using Eq.(12.1). Then use Eq.(12.4) to calculate the mass of the earth: EXECUTE: 1 2g 2mmF G r= so 2 10 2 g 11 2 2 3 1 2 (8.00 10 N)(0.0100 m) 6.667 10 N m /kg . (0.400 kg)(3.00 10 kg) F r G m m ? ? ? ×= = = × ?× E 2 E Gm g R = gives 2 6 2 2 24E E 11 2 2 (6.38 10 m) (9.80 m/s ) 5.98 10 kg. 6.667 10 N m /kg R g m G ? ×= = = ×× ? EVALUATE: Our result agrees with the value given in Appendix F. 12.22. IDENTIFY: Use Eq.(12.4) to calculate g for Europa. The acceleration of a particle moving in a circular path is 2 rada r?= . SET UP: In 2rada r?= , ? must be in rad/s. For Europa, 61.569 10 m.R = × Gravitation 12-7 EXECUTE: 11 2 2 22 2 2 6 2 (6.67 10 N m /kg )(4.8 10 kg) 1.30 m/s (1.569 10 m) Gm g R ?× ? ×= = =× . radg a= gives 21.30 m/s 60 s 1 rev (0.553 rad/s) 5.28 rpm 4.25 m 1 min 2 rad g r ? ? ? ?? ?= = = =? ?? ?? ?? ? . EVALUATE: The radius of Europa is about one-fourth that of the earth and its mass is about one-hundredth that of earth, so g on Europa is much less than g on earth. The lander would have some spatial extent so different points on it would be different distances from the rotation axis and rada would have different values. For the ? we calculated, rada g= at a point that is precisely 4.25 m from the rotation axis. 12.23. IDENTIFY and SET UP: Example 12.5 gives the escape speed as 1 2 / ,v GM R= where M and R are the mass and radius of the astronomical object. EXECUTE: 11 2 2 121 2(6.673 10 N m /kg )(3.6 10 kg)/700 m 0.83 m/s.v ?= × ? × = EVALUATE: At this speed a person can walk 100 m in 120 s; easily achieved for the average person. We can write the escape speed as 241 3 ,v GR??= where ? is the average density of Dactyl. Its radius is much smaller than earth?s and its density is about the same, so the escape speed is much less on Dactyl than on earth. 12.24. IDENTIFY: In part (a) use the expression for the escape speed that is derived in Example 12.5. In part (b) apply conservation of energy. SET UP: 34.5 10 mR = × . In part (b) let point 1 be at the surface of the comet. EXECUTE: (a) The escape speed is 2GMv R = so 2 3 2 13 11 2 2 (4.5 10 m)(1.0 m/s) 3.37 10 kg 2 2(6.67 10 N m /kg ) Rv M G ? ×= = = ×× ? . (b) (i) 211 12K mv= . 2 10.100K K= . 1 GMmU R= ? ; 2 GMm U r = ? . 1 1 2 2K U K U+ = + gives 2 21 1 1 12 2(0.100)( ) GMm GMm mv mv R r ? = ? . Solving for r gives 2 2 1 3 11 2 2 13 1 1 0.450 1 0.450(1.0 m/s) 4.5 10 m (6.67 10 N m /kg )(3.37 10 kg) v r R GM ? = ? = ?× × ? × and 45 kmr = . (ii) The debris never loses all of its initial kinetic energy, but 2 0K ? as r ?? . The farther the debris are from the comet?s center, the smaller is their kinetic energy. EVALUATE: The debris will have lost 90.0% of their initial kinetic energy when they are at a distance from the comet?s center of about ten times the radius of the comet. 12.25. IDENTIFY: The escape speed, from the results of Example 12.5, is 2 / .GM R SET UP: For Mars, 236.42 10 kgM = × and 63.40 10 mR = × . For Jupiter, 271.90 10 kgM = × and 76.91 10 mR = × . EXECUTE: (a) 11 2 2 23 6 32(6.673 10 N m /kg )(6.42 10 kg)/(3.40 10 m) 5.02 10 m/s.v ?= × ? × × = × (b) 11 2 2 27 7 42(6.673 10 N m /kg (1.90 10 kg)/(6.91 10 m) 6.06 10 m/s.v ?= × ? × × = × (c) Both the kinetic energy and the gravitational potential energy are proportional to the mass of the spacecraft. EVALUATE: Example 12.5 calculates the escape speed for earth to be 41.12 10 m/s× . This is larger than our result for Mars and less than our result for Jupiter. 12.26. IDENTIFY: The kinetic energy is 212K mv= and the potential energy is GMmU r= ? SET UP: The mass of the earth is 24E 5.97 10 kgM = × . EXECUTE: (a) 3 2 912 (629 kg)(3.33 10 m/s) 3.49 10 JK = × = × (b) 11 2 2 24 7E 9 (6.673 10 N m /kg )(5.97 10 kg)(629 kg) 8.73 10 J 2.87 10 m GM m U r ?× ? ×= ? = ? = ? ×× . EVALUATE: The total energy K U+ is positive. 12-8 Chapter 12 12.27. IDENTIFY: Apply Newton?s 2nd law to the motion of the satellite and obtain and equation that relates the orbital speed v to the orbital radius r. SET UP: The distances are shown in Figure 12.27a. The radius of the orbit is E .r h R= + 5 6 67.80 10 m 6.38 10 m 7.16 10 m.r = × + × = × Figure 12.27a The free-body diagram for the satellite is given in Figure 12.27b. (a) EXECUTE: y yF ma=? g radF ma= 2 2 Emm vG m r r = Figure 12.27b 11 2 2 24 3E 6 (6.673 10 N m /kg )(5.97 10 kg) 7.46 10 m/s 7.16 10 m Gm v r ?× ? ×= = = ×× (b) 6 3 2 2 (7.16 10 m) 6030 s 1.68 h. 7.46 10 m/s r T v ? ? ×= = = =× EVALUATE: Note that Er h R= + is the radius of the orbit, measured from the center of the earth. For this satellite r is greater than for the satellite in Example 12.6, so its orbital speed is less. 12.28. IDENTIFY: The time to complete one orbit is the period T , given by Eq.(12.12). The speed v of the satellite is given by 2 r v T ?= . SET UP: If h is the height of the orbit above the earth?s surface, the radius of the orbit is Er h R= + . 6 E 6.38 10 mR = × and 24E 5.97 10 kgm = × . EXECUTE: (a) 3 / 2 5 6 3 / 2 3 11 2 2 24 E 2 2 (7.05 10 m 6.38 10 m) 5.94 10 s 99.0 min (6.67 10 N m /kg )(5.97 10 kg) r T Gm ? ? ? × + ×= = = × =× ? × (b) 5 6 3 3 2 (7.05 10 m 6.38 10 m) 7.49 10 m/s 7.49 km/s 5.94 10 s v ? × + ×= = × =× EVALUATE: The satellite in Example 12.6 is at a lower altitude and therefore has a smaller orbit radius than the satellite in this problem. Therefore, the satellite in this problem has a larger period and a smaller orbital speed. But a large percentage change in h corresponds to a small percentage change in r and the values of T and v for the two satellites do not differ very much. 12.29. IDENTIFY: Apply m=?F a! ! to the motion of the earth around the sun. SET UP: For the earth, 7365.3 days 3.156 10 sT = = × and 111.50 10 m.r = × 2 .rT v ?= EXECUTE: 11 4 7 2 2 (1.50 10 m) 2.99 10 s. 3.156 10 s r v T ? ? ×= = = ×× g radF ma= gives 2 E S E2 . m m v G m r r = 2 4 2 11 30 S 11 2 2 (2.99 10 s) (1.50 10 m) 2.01 10 kg 6.673 10 N m /kg v r m G ? × ×= = = ×× ? EVALUATE: Appendix F gives 30S 1.99 10 kgm = × , in good agreement with our calculation. 12.30. IDENTIFY: We can calculate the orbital period T from the number of revolutions per day. Then the period and the orbit radius are related by Eq.(12.12). SET UP: 24E 5.97 10 kgm = × and 6E 6.38 10 mR = × . The height h of the orbit above the surface of the earth is related to the orbit radius r by Er h R= + . 41 day 8.64 10 s= × . Gravitation 12-9 EXECUTE: The satellite moves 15.65 revolutions in 48.64 10 s× , so the time for 1.00 revolution is 4 38.64 10 s 5.52 10 s 15.65 T ×= = × . 3 / 2 E 2 r T Gm ?= gives 1/ 3 1/ 32 11 2 2 24 3 2 E 2 2 [6.67 10 N m /kg ][5.97 10 kg][5.52 10 s] 4 4 Gm T r ? ? ?? ? ? ?× ? × ×= =? ? ? ?? ? ? ? . 66.75 10 mr = × and 5 E 3.7 10 m 370 kmh r R= ? = × = . EVALUATE: The period of this satellite is slightly larger than the period for the satellite in Example 12.6 and the altitude of this satellite is therefore somewhat greater. 12.31. IDENTIFY: Apply m=?F a! ! to the motion of the baseball. 2 rv T?= . SET UP: 3D 6 10 mr = × . EXECUTE: (a) g radF ma= gives 2 D 2 D D m m v G m r r = . 11 2 2 15 D 3 D (6.673 10 N m /kg )(2.0 10 kg) 4.7 m/s 6 10 m Gm v r ?× ? ×= = =× 4.7 m/s 11 mph= , which is easy to achieve. (b) 32 2 (6 10 m) 8020 s 134 min 4.7 m/s r T v ? ? ×= = = = . The game would last a long time. EVALUATE: The speed v is relative to the center of Deimos. The baseball would already have some speed before we throw it, because of the rotational motion of Deimos. 12.32. IDENTIFY: 2 rT v ?= and g radF ma= . SET UP: The sun has mass 30S 1.99 10 kgm = × . The radius of Mercury?s orbit is 105.79 10 m× , so the radius of Vulcan?s orbit is 103.86 10 m× . EXECUTE: g radF ma= gives 2 S 2 m m v G m r r = and 2 SGmv r = . 3 / 2 10 3/2 6 11 2 2 30 S S 2 2 (3.86 10 m) 2 4.13 10 s 47.8 days (6.673 10 N m /kg )(1.99 10 kg) r r T r Gm Gm ? ?? ? ×= = = = × =× ? × EVALUATE: The orbital period of Mercury is 88.0 d, so we could calculate T for Vulcan as 3 / 2(88.0 d)(2 /3) 47.9 daysT = = . 12.33. IDENTIFY: The orbital speed is given by /v Gm r= , where m is the mass of the star. The orbital period is given by 2 r T v ?= . SET UP: The sun has mass 30S 1.99 10 kgm = × . The orbit radius of the earth is 111.50 10 m× . EXECUTE: (a) / .v Gm r= 11 2 2 30 11 4(6.673 10 N m /kg )(0.85 1.99 10 kg)/((1.50 10 m)(0.11)) 8.27 10 m/s.v ?= × ? × × × = × (b) 62 / 1.25 10 s 14.5 daysr v? = × = (about two weeks). EVALUATE: The orbital period is less than the 88 day orbital period of Mercury; this planet is orbiting very close to its star, compared to the orbital radius of Mercury. 12.34. IDENTIFY: The period of each satellite is given by Eq.(12.12). Set up a ratio involving T and r. SET UP: 3 / 2 p 2 r T Gm ?= gives 3/ 2 p 2 constant T r Gm ?= = , so 1 23 / 2 3 / 2 1 2 T T r r = . EXECUTE: 3 / 2 3 / 2 2 2 1 1 48,000 km (6.39 days) 24.5 days 19,600 km r T T r ? ? ? ?= = =? ? ? ?? ?? ? . For the other satellite, 3 / 2 2 64,000 km (6.39 days) 37.7 days 19,600 km T ? ?= =? ?? ? . EVALUATE: T increases when r increases. 12.35. IDENTIFY: In part (b) apply the results from part (a). SET UP: For Pluto, 0.248e = and 125.92 10 ma = × . For Neptune, 0.010e = and 124.50 10 ma = × . The orbital period for Pluto is 247.9 yT = . 12-10 Chapter 12 EXECUTE: (a) The result follows directly from Figure 12.19 in the textbook. (b) The closest distance for Pluto is 12 12(1 0.248)(5.92 10 m) 4.45 10 m? × = × . The greatest distance for Neptune is 12 12(1 0.010)(4.50 10 m) 4.55 10 m+ × = × . (c) The time is the orbital period of Pluto, 248 yT = . EVALUATE: Pluto's closest distance calculated in part (a) is 12 80.10 10 m 1.0 10 km× = × , so Pluto is about 100 million km closer to the sun than Neptune, as is stated in the problem. The eccentricity of Neptune's orbit is small, so its distance from the sun is approximately constant. 12.36. IDENTIFY: 3 / 2 star 2 r T Gm ?= , where starm is the mass of the star. 2 rv T ?= . SET UP: 53.09 days 2.67 10 s= × . The orbit radius of Mercury is 105.79 10 m× . The mass of our sun is 301.99 10 kg× . EXECUTE: (a) 52.67 10 sT = × . 10 9(5.79 10 m)/9 6.43 10 mr = × = × . 3 / 2 star 2 r T Gm ?= gives 2 3 2 9 3 30 star 2 5 2 11 2 2 4 4 (6.43 10 m) 2.21 10 kg (2.67 10 s) (6.67 10 N m /kg ) r m T G ? ? ? ×= = = ×× × ? . star sun 1.11 m m = , so star sun1.11m m= . (b) 9 5 5 2 2 (6.43 10 m) 1.51 10 m/s 2.67 10 s r v T ? ? ×= = = ×× EVALUATE: The orbital period of Mercury is 88.0 d. The period for this planet is much less primarily because the orbit radius is much less and also because the mass of the star is greater than the mass of our sun. 12.37. (a) IDENTIFY: If the orbit is circular, Newton?s 2nd law requires a particular relation between its orbit radius and orbital speed. SET UP: The gravitational force exedrted on the spacecraft by the sun is 2g S H/ ,F Gm m r= where Sm is the mass of the sun and Hm is the mass of the Helios B spacecraft. For a circular orbit, 2rad /a v r= and 2H / .F m v r=? If we neglect all forces on the spacecraft except for the force exerted by the sun, 2g H / ,F F m v r= =? so 2 2S H H/ /Gm m r m v r= EXECUTE: 11 2 2 30 9S/ (6.673 10 N m /kg )(1.99 10 kg)/43 10 mv Gm r ?= = × ? × × 45.6 10 m/s 56 km/s= × = EVALUATE: The actual speed is 71 km/s, so the orbit cannot be circular. (b) IDENTIFY and SET UP: The orbit is a circle or an ellipse if it is closed, a parabola or hyperbola if open. The orbit is closed if the total energy (kinetic potential)+ is negative, so that the object cannot reach .r ? ? EXECUTE: For Helios B, 2 3 2 9 2 21 1 H H H2 2 (71 10 m/s) (2.52 10 m /s )K m v m m= = × = × 11 2 2 30 S H H/ ( (6.673 10 N m /kg )(1.99 10 kg)/(43U Gm m r m ?= ? = ? × ? × × 9 9