( kg) (410 m/s) 0 m/s J. W mv mv m v v. 4 mv

Transcription

1 PHYS : Solution to Chapter 6 Home ork. RASONING a. The work done by the gravitational orce i given by quation 6. a = (F co θ). The gravitational orce point downward, oppoite to the upward vertical diplacement o.6 m. Thereore, the angle θ i 8º. b. The work done by the ecalator i done by the upward normal orce that the ecalator exert on the man. Since the man i moving at a contant velocity, he i in equilibrium, and the net orce acting on him mut be zero. Thi mean that the normal orce mut balance the man weight. Thu, the magnitude o the normal orce i F N = mg, and the work that the ecalator doe i alo given by quation 6.. However, ince the normal orce and the upward vertical diplacement point in the ame direction, the angle θ i º. a. According to quation 6., the work done by the gravitational orce i co co F mg kg 9.8 m/ co8.6 m 3.38 b. The work done by the ecalator i co co F F N kg 9.8 m/ co.6 m RASONING AND a. In both cae, the lit orce L i perpendicular to the diplacement o the plane, and, thereore, doe no work. A hown in the drawing in the text, when the plane i in the dive, there i a component o the weight that point in the direction o the diplacement o the plane. hen the plane i climbing, there i a component o the weight that point oppoite to the diplacement o the plane. Thu, ince the thrut T i the ame or both cae, the net orce in the direction o the diplacement i greater or the cae where the plane i diving. Since the diplacement i the ame in both cae, more net work i done during the dive. b. The work done during the dive i dive (T co 75), while the work done during the climb i climb (T co 5). Thereore, the dierence between the net work done during the dive and the climb i dive climb (T co 75) (T co 5) (co 75 co 5) = (5.9 N)(.7 3 m)(co 75 co 5) = RASONING AND The net work done on the car i T = F + + g + N T = F co. + co 8 mg in 5. + F N co 9 Rearranging thi reult give T F mg in = 5 N + kg 9.8 m/ in 5..7 N 9 m

2 7. RASONING The work done by the catapult catapult i one contribution to the work done by the net external orce that change the kinetic energy o the plane. The other contribution i the work done by the thrut orce o the plane engine thrut. According to the work-energy theorem (quation 6.3), the work done by the net external orce catapult + thrut i equal to the change in the kinetic energy. The change in the kinetic energy i the given kinetic energy o.5 7 at lit-o minu the initial kinetic energy, which i zero ince the plane tart at ret. The work done by the thrut orce can be determined rom quation 6. [ = (F co θ) ], ince the magnitude F o the thrut i.3 5 N and the magnitude o the diplacement i 87 m. e note that the angle θ between the thrut and the diplacement i º, becaue they have the ame direction. In ummary, we will calculate catapult rom catapult + thrut = K K. According to the work-energy theorem, we have catapult + thrut = K K Uing quation 6. and noting that K =, we can write the work energy theorem a ollow: catapult ork done by thrut Solving or catapult give K F co catapult F co K ork done by thrut N co 87 m.5 9. SSM RASONING The work done to launch either object can be ound rom quation 6.3, the work-energy theorem, K K mv mv. a. The work required to launch the hammer i mv mv m v v (7.3 kg) (9 m/) m/ 3. 3 b. Similarly, the work required to launch the bullet i m v v (.6 kg) ( m/) m/. 7. RASONING AND The net work done on the plane can be ound rom the work-energy theorem: mv mv m v v The tenion in the guideline provide the centripetal orce required to keep the plane moving in a circular path. Since the tenion in the guideline become our time greater, or mv r T = T mv r PHYS : Solution to Chapter 6 Home ork pg. /7

3 Solving or v give v vr r Subtitution o thi expreion into the work-energy theorem give Thereore, m v mv vr r r r m = (.9 kg)( m/) = 5. 6 m 3. RASONING At a height h above the ground, the gravitational potential energy o each clown i given by P mgh (quation 6.5). At a height h, uggle potential energy i, thereore, P = m gh, where m i uggle ma. Similarly, Bangle potential energy can be expreed in term o hi ma m B and height h B : P B = m B gh B. e will ue the act that thee two amount o potential energy are equal (P = P B ) to ind uggle ma m. Setting the gravitational potential energie o both clown equal, we olve or uggle ma and obtain m gh m g h B P P B B or m 86 kg.5 m 3.3 m mh B B h 65 kg 33. RASONING The work done by the weight o the baketball i given by quation 6. a F co, where F = mg i the magnitude o the weight, i the angle between the weight and the diplacement, and i the magnitude o the diplacement. The drawing how that the weight and diplacement are parallel, o that =. The potential energy o the baketball i given by quation 6.5 a P = mgh, where h i the height o the ball above the ground. a. The work done by the weight o the baketball i mg F co = mg (co )(h h ) = (.6 kg)(9.8 m/ )(6. m.5 m) = 7 b. The potential energy o the ball, relative to the ground, when it i releaed i P = mgh = (.6 kg)(9.8 m/ )(6. m) = 36 (6.5 c. The potential energy o the ball, relative to the ground, when it i caught i P = mgh = (.6 kg)(9.8 m/ )(.5 m) = 8.8 (6.5 PHYS : Solution to Chapter 6 Home ork pg. 3/7

4 d. The change in the ball gravitational potential energy i P = P P = = 7 e ee that the change in the gravitational potential energy i equal to 7 =, where i the work done by the weight o the ball (ee part a). 36. RASONING The girl gravitational potential energy P mgh (quation 6.5) increae a he rie. Thi increae come at the expene o her kinetic energy K mv (quation 6.), which decreae a he rie. Becaue air reitance i negligible, all o the kinetic energy he loe i tranormed into potential energy, o that her total mechanical energy K P remain contant. e will ue thi principle to determine the ditance he rie during thi interval. The conervation principle give K P K P or P P K K or mg h h K K Solving thi expreion or h h, which i the ditance he rie, we obtain h K K mg h 35 kg9.8 m/.67 m 5. SSM RASONING Friction and air reitance are being ignored. The normal orce rom the lide i perpendicular to the motion, o it doe no work. Thu, no net work i done by nonconervative orce, and the principle o conervation o mechanical energy applie. have Applying the principle o conervation o mechanical energy to the wimmer at the top and the bottom o the lide, we mv mgh mv mgh I we let h be the height o the bottom o the lide above the water, h h, and h H. Since the wimmer tart rom ret, v m/, and the above expreion become Solving or H, we obtain v gh gh H h v g Beore we can calculate H, we mut ind v and h. Since the velocity in the horizontal direction i contant, v x t 5. m. m/.5 The vertical diplacement o the wimmer ater leaving the lide i, rom quation 3.5b (with down being negative), y a yt ( 9.8 m/ )(.5 ).3 m PHYS : Solution to Chapter 6 Home ork pg. /7

5 Thereore, h =.3 m. Uing thee value o v and h in the above expreion or H, we ind H h v g.3 m (. m/) (9.8 m/ ) 6.33 m 56. RASONING According to quation 6.6, the work nc done by nonconervative orce, uch a kinetic riction and air reitance, i equal to the object change in kinetic energy, mv mv plu the change in it potential energy, nc mgh mgh : mv mv mgh mgh (6.6) Thi relation will be ued in both part (a) and (b) o the problem. a. Since nonconervative orce are abent, the work done by them i zero, o nc =. In thi cae, we can algebraically rearrange quation 6.6 to how that the initial total mechanical energy (kinetic energy plu potential energy) at the top i equal to the inal total mechanical energy at the bottom: mv mgh mv mgh Solving or the inal peed v o the keleton and rider give v v g h h The initial peed o the rider i v = m/, and the drop in height i hh m, o the inal peed at the bottom o the run i v v g h h m/ 9.8 m/ m 5. m/ b. The work nc done by the nonconervative orce ollow directly rom quation 6.6: nc m v v mg h h 86. kg 35.8 m/ m/ 86. kg 9.8 m/ m 3.5 Note that the dierence in height, height h. h h m, i a negative number becaue the inal height h i le than the initial 58. RASONING AND According to the work-energy theorem a given in quation 6.8, we have nc mv mgh mv mgh The metal piece tart at ret and i at ret jut a it barely trike the bell, o that v v m/. In addition, h h and h m, while nc.5 M v become, where M and v are the ma and peed o the hammer. Thu, the work-energy theorem PHYS : Solution to Chapter 6 Home ork pg. 5/7.5 M v mgh

6 Solving or the peed o the hammer, we ind v mgh.5m (. kg)(9.8 m/ )(5. m).5 (9. kg).7 m/ 63. SSM RASONING AND One kilowatt houri the amount o work or energy generated when one kilowatt o power i upplied or a time o one hour. From quation 6.a, we know that P t. Uing the act that k=. 3 / and that h = 36, we have. kh = (. 3 /)( h)= (. 3 /)(36 )= Change in energy 65. RASONING The average power i given by quation 6.b ( P ). The time i given. Since the road i level, there Time i no change in the gravitational potential energy, and the change in energy reer only to the kinetic energy. According to quation 6., the kinetic energy i mv. The peed v i given, but the ma m i not. However, we can obtain the ma rom the given weight, ince the weight i mg. a. Uing quation 6.b and 6., we ind that the average power i Change in energy K K mv mv P Time Time Time Since the car tart rom ret, v = m/, and ince the weight i = mg, the ma i m = /g. Thereore, the average power i mv mv v P () Time g Time Uing the given value or the weight, inal peed, and the time, we ind that 3 9. N. m/ v P g Time 9.8 m/ ( hp) b. From quation () it ollow that. N. m/ v P g Time 9.8 m/ (68 hp) PHYS : Solution to Chapter 6 Home ork pg. 6/7

7 66. RASONING The work done i given by quation 6. a F co, where F i the magnitude o the orce that you exert on the rowing bar, i the magnitude o the bar diplacement, and i the angle between the orce and the diplacement. Solving thi equation or F give F co (6.) The work i alo equal to the average power P multiplied by the time t, or Pt (6.a) Subtituting quation (6.a) into (6.) give Pt F co co The angle i, ince the direction o the pulling orce i the ame a the diplacement o the rowing bar. Thu, the magnitude o the orce exerted on the handle i Pt 8.5 F. N co co. m 67. RASONING The average power developed by the cheetah i equal to the work done by the cheetah divided by the elaped time (quation 6.a). The work, on the other hand, can be related to the change in the cheetah kinetic energy by the work-energy theorem, quation 6.3. a. The average power i P t (6.a) where i the work done by the cheetah. Thi work i related to the change in the cheetah kinetic energy by quation 6.3, mv mv, o the average power i P t mv mv t kg7 m / kg m /.. b. The power, in unit o horepower (hp), i hp P. 3 hp 75.7 PHYS : Solution to Chapter 6 Home ork pg. 7/7