Problem Set 7 Chapter 10: Energy & Work Questions: 2, 14, 15, 19, 21 Exercises & Problems: 5, 10, 21, 28, 45, 49, 58, 60, 68, 74

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1 Problem Set 7 Chapter 10: Energy & Work Questons:, 14, 15, 19, 1 Exercses & Problems: 5, 10, 1, 8, 45, 49, 58, 60, 68, 74 **Mdterm Exam Alert, Thursday, October 7 th ** Q10.: When you pound a nal wth a hammer, the nal gets qute warm. Descrbe the energy transformatons that lead to the addton of thermal energy n the nal. Q10.. Reason: When you ht a nal wth a hammer to pound t nto some object, many processes are at work. For example, some small amount of energy goes nto temporarly ncreasng the nal s knetc energy as t moves nto the object. Part of the energy goes nto permanently deformng the object to accept the nal. An apprecable porton of the ntal knetc energy of the hammer s converted to thermal energy through, for example, frcton between the nal and the object as the nal moves nto the object. Some gets drectly converted to knetc energy of the molecules that make up the nal (see secton 11.4 for an atomc vew of thermal energy and temperature) from the collson between the nal and the hammer. Assess: If you ever try hammerng nals, the thermal energy generated s apprecable. Note that energy can be transformed drectly nto knetc energy of atoms or molecules that make up an object. As a smpler example, bangng a hammer on a sold object drectly wll ncrease the temperature of the both the hammer and the object. Q10.14: The meanng of the word "work" s qute dfferent n physcs from ts everyday usage. Gve an example of an acton a person could do that "feels lke work" but that does not nvolve any work as we've defned t n ths chapter. Q Reason: If you hold a heavy object over your head, you do no work on t because ts dsplacement s zero. Or, f you walk whle carryng a heavy sutcase you do no work because the force you apply to the handle, drected upward, s perpendcular to the forward-drected dsplacement. Assess: Although you do no work n the physcs sense of the word, your muscles quckly tre because they must consume chemcal energy to reman contracted under a heavy load. It s n ths sense you feel that you are workng hard to hold the weght or carry the sutcase. Q10.15: To change a tre, you need to use a jack to rase one corner of your car. Whle dong so, you happen to notce that pushng the jack handle down 0 cm rases the car only 0. cm. Use energy concepts to explan why the handle must be moved so far to rase the car by such a small amount. Q Reason: Neglectng frctonal losses, the work you do on the jack s converted nto gravtatonal potental energy of the car as t s rased. The work you do s Fd, where F s the force you apply to the jack handle and d s the 0 cm dstance you move the handle. Ths work goes nto ncreasng the potental energy by an amount mgh wh, where w s the car s weght and h 0. cm s the change n the car s heght. So Fd wh so that F / w h / d. Assess: Because the force F you can apply s so much less than the weght w of the car, h must be much less than d. Q10.19: Sandy and Chrs stand on the edge of a clff and thrown dentcal mass rocks at the same speed. Sandy throws her rock horzontally whle Chrs throws hs upward at an angle of 45 to the horzontal. Are the rocks movng at the same speed when they ht the ground, or s one movng faster than the other? If one s movng faster, whch one? Explan. Q Reason: Because both rocks are thrown from the same heght, they have the same potental energy. And snce they are thrown wth the same speed, they have the same knetc energy. Thus both rocks have the same total energy. When they reach the ground, they wll have ths same total energy. Because they re both at the same heght at ground level, ther potental energy there s the same. Thus they must have the same knetc energy, and hence the same speed. 1

2 Assess: Although Chrs s rock was thrown angled upward so that t slows as t frst rses, t then speeds up as t begns to fall, attanng the same speed as Sandy s as t passes the ntal heght. Sandy s rock wll ht the ground frst, but ts speed wll be no greater than Chrs s. Q10.1: You are much more lkely to be njured f you fall and your head strkes the ground than f your head strkes a gymnastcs pad. Use energy and work concepts to explan why ths s so. Q10.1. Reason: As you land, the force of the ground or pad does negatve work on your body, transferrng out the knetc energy you have just before mpact. Ths work s Fd, where d s the dstance over whch your body stops. Wth the short stoppng dstance nvolved upon httng the ground, the force F wll be much greater than t s wth the long stoppng dstance upon httng the pad. Assess: For a gven amount of work, the force s large when the dsplacement s small. P10.5: a. At the arport, you rde a "movng sdewalk" that carres you horzontally for 5 m at 0.70 m/s. Assumng that you were movng at 0.70 m/s before steppng onto the movng sdewalk and contnue at 0.70 m/s afterward, how much work does the movng sdewalk do on you? Your mass s 60 kg. b. An escalator carres you from one level to the next n the arport termnal. The upper level s 4.5 m above the lower level, and the length of the escalator s 7.0 m. How much work does the up escalator do on you when you rde t from the lower level to the upper level? c. How much work does the down escalator do on you when you rde t from the upper level to the lower level? P10.5. Prepare: We wll use the defnton of work, Equaton 10.6 to calculate the work done. The sdewalk and escalators exert a normal force on you, and may exert a force to propel you forward. We wll assume that the escalator propels you at constant velocty, as the sdewalk does. Solve: (a) Snce you get on the sdewalk movng at 0.70 m/s, and you contnue at 0.70 m/s afterwards, there s no acceleraton and therefore no force on you n the horzontal drecton. See the followng dagram. The work s then W Fd cos ( F)( d)cos(90 ) ( F)( d)(0) 0.0 J. The work done by the sdewalk s exactly zero Joules. (b) The escalator moves you across some dstance lke the sdewalk, but t also moves you upwards. See the followng dagram. The force exerted on you by the escalator s the normal force, whch s equal to your weght. n w mg (60 kg)(9.80 m/s ) 588 N whch should be reported as 590 N to two sgnfcant fgures. Unlke the sdewalk case, there s a component of the dsplacement parallel to the normal force. The angle 1 between the force and dsplacement s cos (4.5/7) 50, so d cos( ) 4.5 m. Then (c) Refer to the fgure. W Fd cos (588 N)(4.5 m).6 kj

3 Here, the dsplacement s n the opposte drecton compared to part (b), so the angle between the force and the dsplacement s now So W Fd cos (588 N)(7.0 m)cos(130 ).6 kj Assess: In part (a), snce the force has no component n the drecton of your dsplacement, the force does no work. In part (b), there s a component of the force along and n the same drecton as the dsplacement, so the force does postve work. In part (c), the component of the force along the dsplacement s n the opposte drecton to the dsplacement, so the force does negatve work. P10.10: Sam's job at the amusement part s to slow down and brng to a stop the boats n the log rde. If a boat and ts rders have a mass of 100 kg and the boat drfts n at 1. m/s, how much work does Sam do to stop t? P Prepare: We wll assume that all the work Sam does goes nto stoppng the boat. We can use conservaton of energy as expressed n Equaton 10.7 to calculate the work done from the change n knetc energy. Solve: Refer to the before and after representaton of Sam stoppng a boat. Equaton 10.7 becomes 1 1 mvf mv W Snce the boat s at rest at the end of the process, vf 0 m/s. Therefore, the fnal knetc energy s zero. The work done on the boat s then W 1 mv 1 (100 kg)(1. m/s) 0.86 kj Assess: Note that the work done by Sam on the boat s negatve. Ths s because the force Sam exerts on the boat must be opposte to the drecton of moton of the boat to slow t down. P10.1: The elastc energy stored n your tendons can contrbute up to 35% of your energy needs when runnng. Sports scentsts have studed the change n length of the knee extensor tendon n sprnters and nonathletes. The fnd (on average) that the sprnters' tendons stretch 41 mm, whle nonathletes' stretch only 33 mm. The sprng constant for the tendon s the same for both groups, 33 N/mm. What s the dfference n maxmum stored energy between the sprnters and the nonathletes? P10.1. Prepare: We wll assume the knee extensor tendon behaves accordng to Hooke s Law and stretches 1 n a straght lne. The elastc energy stored n a sprng s gven by equaton 10.15, Us kx. Solve: For athletes, For non-athletes, U s,athlete 1 1 (33,000 N/m)(0.041 m) kx 7.7 J 1 1 (33,000 N/m)(0.033 m) kx 18.0 J Us,non-athlete The dfference n energy stored between athletes and non-athletes s therefore 9.7 J. Assess: Notce the energy stored by athletes s over 1.5 tmes the energy stored by non-athletes. 3

4 P10.8: What mnmum speed does a 100 g puck need to make t to the top of a frctonless ramp that s 3.0 m long and nclned at 0? P10.8. Prepare: Snce the ramp s frctonless, the sum of the puck s knetc and gravtatonal potental energy does not change durng ts sldng moton. Use Equaton 10.4 for the conservaton of energy. Solve: The quantty K + U g s the same at the top of the ramp as t was at the bottom. The energy conservaton equaton Kf Ugf K Ug s 1 1 mvf mgyf mv mgy v vf g( yf y) v (0 m/s) (9.80 m/s )(1.03 m 0 m) 0.1m /s v 4.5 m/s Assess: An ntal push wth a speed of 4.5 m/s 10 mph to cover a dstance of 3.0 m up a 0 ramp seems reasonable. Note that a ramp of any angle to the same fnal heght would lead to the same fnal velocty for the puck. Note that the mass cancels out n the equaton snce both knetc energy and gravtatonal potental energy are proportonal to mass. P10.45: In the wnter sport of curlng, players gve a 0 kg stone a push across a sheet of ce. A curler accelerates a stone to a speed of 3.0 m/s over a tme of.0 s. a. How much force does the curler exert on the stone? b. What average power does the curler use to brng the stone up to speed? P Prepare: Neglectng frcton, the only horzontal force on the stone s the force exerted by the curler. The work done by ths force wll be transferred entrely nto the stone s knetc energy. We can use the conservaton of energy equaton to calculate ths work. We wll assume the ce s frctonless, that the acceleraton of the stone s constant, and that the stone starts from rest. Solve: (a) Refer to the dagram. We can fnd the force exerted on the stone by the curler from the acceleraton of the stone. Snce v 0 m/s, The force on the stone s gven by vf v 3.0 m/s a 1.5 m/s t.0 s F ma (0 kg)(1.5 m/s) 30 N (b) Snce Eth E U U 0 J, n ths case, the law of conservaton of energy (10.3) reads chem g s Snce v 0 m/s 1 1 W K K mv mv f f 4

5 The average power s gven by Equaton 10., 1 1 (0 kg)(3.0 m/s) W mv 90 J f W 90 J P 45 W t.0 s Assess: Note that the amount of muscle power needed to quckly accelerate ths relatvely heavy stone would not even fully lght a 60 W lght bulb. P10.49: A 55 kg skateboarder wants to just make t to the upper edge of a "half-ppe" wth a radus of 3.0 m, as shown n the fgure. What speed v does he need at the bottom f he s to coast all the way up? a. Frst do the calculaton treatng the skateboarder and board as a pont partcle, wth the entre mass nearly n contact wth the half-ppe. b. More realstcally, the mass of the skateboarder n a deep crouch mght be thought of as concentrated 0.75 m from the half-ppe. Assumng he remans n that poston all the way up, what v s needed to reach the upper edge? P Prepare: Assumng that the track offers no rollng frcton, the sum of the skateboarder s knetc and gravtatonal potental energy does not change durng hs rollng moton. The vertcal dsplacement of the skateboarder s equal to the radus of the track. Solve: (a) The quantty K + U g s the same at the upper edge of the quarter-ppe track as t was at the bottom. The energy conservaton equaton Kf Ugf K Ug s 1 1 mvf mgyf mv mgy v vf g( yf y) v (0 m/s) (9.80 m/s )(3.0 m 0 m) 58.8 m/s v 7.7 m/s (b) If the skateboarder s n a low crouch, hs heght above ground at the begnnng of the trp changes to 0.75 m. Hs heght above ground at the top of the ppe remans the same snce he s horzontal at that pont. Followng the same procedure as for part (a), 1 1 mvf mgyf mv mgy v vf g( yf y) v (0 m/s) (9.80 m/s )(3.0 m 0.75 m) 44.1 m/s v 6.6 m/s Assess: Note that we dd not need to know the skateboarder s mass, as s the case wth free-fall moton. Note that the shape of the track s rrelevant. P10.58: The maxmum energy a bone can absorb wthout breakng s surprsngly small. For a healthy human of mass 60 kg, expermental data show that the leg bones can absorb about 00 J. a. From what maxmum heght could a person jump and land rgdly uprght on both feet wthout breakng hs legs? b. People jump from much greater heghts than ths; explan how ths s possble. 5

6 Hnt: Thnk about how people land when they jump from greater heghts. P Prepare: We wll take the system to be the person plus the earth. When a person drops from a certan heght, the ntal potental energy s transformed to knetc energy. When the person hts the ground, f they land rgdly uprght, we assume that all of ths energy s transformed nto elastc potental energy of the compressed leg bones. The maxmum energy that can be absorbed by the leg bones s 00 J; ths lmts the maxmum heght. Solve: (a) The ntal potental energy can be at most 00 J, so the heght h of the jump s lmted by mgh 00 J For m 60 kg, ths lmts the heght to h 00 J mg 00 J 60 kg 9.8 m/s 0.34 m (b) If some of the energy s transformed to other forms than elastc energy n the bones, the ntal heght can be greater. If a person flexes her legs on landng, some energy s transformed to thermal energy. Ths allows for a greater ntal heght. Assess: There are other tssues n the body wth elastc propertes that wll absorb energy as well, so ths lmt s qute conservatve. P10.60: The 5.0-m-long rope n the fgure hangs vertcally from a tree rght at the edge of a ravne. A woman wants to use the rope to swng to the other sde of the ravne. She runs as fast as she can, grabs the rope, and swngs out over the ravne. a. As she swngs, what energy converson s takng place?. b. When she's drectly over the far edge of the ravne, how much hgher s she than when she started? c. Gven your answers to parts a and b, how fast much she be runnng when she grabs the rope n order to swng all the way across the ravne? P Prepare: Snce there s no frcton, total mechancal energy s conserved. Solve: (a) As she swngs, her heght above the clff ncreases snce the rope doesn t stretch. Her ntal knetc energy s beng converted to gravtatonal potental energy durng the swng. (b) Refer to the dagram. When she s drectly above the opposte sde of the ravne, she has moved 3.0 m horzontally whle the rope s also swngng her upwards. Usng the Pythagorean theorem, we can fnd the dstance between the branch and her new heght. l (5.0 m) (3.0 m) 4.0 m. Therefore, she s 5.0 m 4.0 m = 1.0 m above the clff. (c) To make t to the other sde of the ravne, she must have enough knetc energy to be converted to the equvalent gravtatonal potental energy of her addtonal 1.0 m of heght. The mnmum ntal velocty she wll need wll be when she just makes t to the other sde of the ravne wth no knetc energy (all her ntal knetc energy beng converted to potental energy of her heght above the clff). Usng the clff as the reference for gravtatonal potental energy, the conservaton of energy equaton reads v K U K U 1 g 1 g 1 mv mgy 1 1 gy (9.80 m/s )(1.0 m) 4.4 m/s Assess: We calculated for the case where all her ntal knetc energy s converted to gravtatonal potental energy at the other sde of the ravne. Note she could start wth a greater ntal knetc energy, whch would also 6

7 get her to the other sde of the ravne. In ths case, when she s above the other sde of the ravne, she wll have some addtonal knetc energy nstead of just makng t to the other sde. P10.68: A 70 kg human sprnter can accelerate from rest to 10 m/s n 3.0 s. Durng the same nterval, a 30 kg greyhound can accelerate from rest to 0 m/s. Compute (a) the change n knetc energy and (b) the average power output for each. P Prepare: We can fnd the knetc energes drectly from the runner s masses and fnal speeds. The power s the rate at whch each runner s nternal chemcal energy s converted nto knetc energy, so the average power s K / t. Solve: (a) We have (70 kg)(10 m/s) 3500 J, 1 KS msv S whle KG mgvg (30 kg)(0 m/s) 6000 J (here, S stands for sprnter and G for greyhound). (b) We have P K S S / t (3500 J)/(3.0 s) 100 W, and P K G G / t (6000 J)/(3.0 s) 000 W. Assess: Although the greyhound has less than half the mass of the human, ts fnal speed s twce as great. And snce knetc energy depends on the square of the speed, the hgher speed of the greyhound means that ts knetc energy s greater than the human s. Because 1 hp = 760 W, a human can output about hp and a greyhound about 3 hp n very short bursts. P10.74: The human heart has to pump the average adult's 6.0 L of blood through the body every mnute. The heart must do work to overcome frctonal forces that resst the blood flow. The average blood pressure s N/m. a. Compute the work done movng the 6.0 L of blood completely through the body, assumng the blood pressure always takes t average value. b. What power output must the heart have to do ths task once a mnute? Hnt: When the heart contracts, t apples force to the blood. Pressure s just force/area, so we can wrte work = (pressure)(area)(dstance). But (area)(dstance) s just the blood volume passng through the heart. P Prepare: The heart provdes the pressure to move blood through the body and therefore does work on the blood. We assume all the work goes nto pushng the blood through the body Solve: (a) Usng the hnt, W PAd PV (1.310 N/m )(6.010 m ) 78 J (n ths equaton P represents pressure, not power). W 78 J (b) P 1.3 W. t 60 s Assess: Ths seems lke a reasonable answer, as t s a small fracton of the power requred for most human actvtes. 7