ENERGY AND WORK 0 Q0.. Reason: The brakes n a car slow down the car by convertng ts knetc energy to thermal energy n the brake shoes through rcton. Cars have large knetc energes, and all o that energy s converted to thermal energy n the brake shoes, whch causes ther temperature to ncrease greatly. Thereore they must be made o materal that can tolerate very hgh temperatures wthout beng damaged. Assess: Ths s an example o an energy converson. All o the car s knetc energy s converted to thermal energy through rcton. To get an apprecaton o how much knetc energy s absorbed by the brake shoes, consder nstead the energy explct n stoppng the car by httng a statonary object nstead! Q0.. 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 o energy goes nto temporarly ncreasng the nal s knetc energy as t moves nto the object. Part o the energy goes nto permanently deormng the object to accept the nal. An apprecable porton o the ntal knetc energy o the hammer s converted to thermal energy through, or example, rcton between the nal and the object as the nal moves nto the object. Some gets drectly converted to knetc energy o the molecules that make up the nal (see secton.3 or an atomc vew o thermal energy and temperature) rom the collson between the nal and the hammer. Assess: I you ever try hammerng nals, the thermal energy generated s apprecable. Note that energy can be transormed drectly nto knetc energy o atoms or molecules that make up an object. As a smpler example, bangng a hammer on a sold object drectly wll ncrease the temperature o the both the hammer and the object. Q0.3. Reason: We must thnk o a process that ncreases an object s knetc energy wthout ncreasng any potental energy. Consder pullng an object across a level loor wth a constant orce. The orce does work on the object, whch wll ncrease the object s knetc energy. Snce the loor s level the gravtatonal potental energy does not change. The other orm o potental energy possble s that stored n a sprng, whch s also zero here. Assess: For there to be no potental energy change, the object n queston must reman at the same heght. Q0.4. Reason: Here we must ncrease potental energy wthout ncreasng knetc energy. Consder ltng an object at constant speed. Consder the object plus the earth as the system. The orce does work that ncreases the gravtatonal potental energy o the object, whle the knetc energy does not ncrease because the velocty o the object remans the same. Another possblty s the compresson o a sprng by an appled orce at constant velocty. Note that constant velocty s not necessary or the change n knetc energy between the begnnng and end o a process to be zero. Ltng an object or compressng a sprng n any way, as long as the ntal velocty s equal to the nal velocty at the end o the process leads to no change n net knetc energy. Any knetc energy ganed durng the process s lost when the object s brought the rest. Assess: Knetc energy does not change an object has the same velocty at the begnnng and end o a process. Copyrght 05 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst. 0-
0- Chapter 0 Q0.5. Reason: The system must convert knetc energy drectly to potental energy wth no external orce dong work. For gravtatonal potental energy we must change the heght o the object. One smple example would be rollng a ball up a hll. The ntal knetc energy s converted to gravtatonal potental energy as the ball ncreases ts heght. The ball loses knetc energy whle t gans potental energy. Another example s rollng a ball nto an uncompressed sprng on level ground. As the ball compresses the sprng, the system gans potental energy, whle losng knetc energy. Snce there are no orces external to the systems n these examples, no work s done on the systems by the envronment. Assess: As long as no orces external to the system are appled, work done on a system s zero. Q0.6. Reason: We need a process that converts knetc energy to work wthout any change n potental energy. Consder a block sldng on level ground, to whch s attached a cord you are holdng on to. As the block sldes, t exerts a orce on your hand by vrtue o ts knetc energy. As the block pulls your hand, t s dong work on you. The knetc energy o the block wll decrease as t contnues to exert the orce on your hand. Assess: To have a change n gravtatonal potental energy you must have a change n heght. Q0.7. Reason: Here we need to convert potental energy to knetc energy wthout any work done on the system. Consder droppng a ball rom a heght. The ball s gravtatonal energy s converted to the knetc energy o the ball as t alls. Another example would be releasng a ball at the end o a compressed sprng. The potental energy stored n the compressed sprng s converted to the knetc energy o the ball as the sprng stretches to ts equlbrum length. Snce no external orces act on a system, the work on the system s zero. Assess: Many examples n the problem secton wll nvolve just ths type o converson o potental energy to knetc energy. I no orces rom the envronment act on a system, the work done on the system s zero. Q0.8. Reason: We need a process that converts work totally nto thermal energy wthout any change n the knetc or potental energy. Consder movng a block o wood across a horzontal rough surace at constant speed. Because the surace s horzontal there s no change n potental energy, and because the speed s constant there s no change n knetc energy. All the work done n movng the block across the rough horzontal surace s transerred nto thermal energy. Assess: To have a change n gravtatonal energy you must have a change n heght and to have a change n knetc energy you must have a change n speed. Q0.9. Reason: We need a process that converts potental energy totally nto thermal energy wthout changng the knetc energy. Consder a wood block sldng down a rough nclned surace at a constant speed. The gravtatonal potental energy s decreasng and the knetc energy s constant. All the decrease n gravtatonal potental energy becomes an ncrease n thermal energy. Assess: Gravtatonal potental energy decreases because there s a change n the heght o the block. The knetc energy does not change because the speed o the block s constant. Q0.0. Reason: We need a process that converts knetc energy totally nto thermal energy wthout changng the gravtatonal potental energy. Consder a wood block sldng across a rough horzontal surace and slowng to a stop. The gravtatonal potental energy s not changng and all the knetc energy s beng transerred nto thermal energy. All the decrease n knetc energy becomes an ncrease n thermal energy. Assess: Gravtatonal potental energy does not change because there s a change n heght o the block. The knetc energy decreases as the block slows to a stop. Q0.. Reason: The energes nvolved here are knetc energy, gravtatonal potental energy, elastc potental energy, and thermal energy. For the system to be solated, we must not have any work beng done on the system and no heat beng transerred nto or out o the system. The ball s knetc and elastc energy s changng, so we should consder t part o the system. Snce ts gravtatonal potental energy s changng, we need to also consder the earth as part o the system. Thermal energy wll be generated n the ball and loor when the ball hts the loor, so we must consder both to be part o the system. Assess: In order to have an solated system no work can be done on the system and all orces must be nternal. Copyrght 05 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.
Energy and Work 0-3 Q0.. Reason: The tenson orce s perpendcular to the drecton o moton o the mass. The work done by tenson s W = Td cosθ = Td cos90 = 0. No work s done by tenson. Assess: The work done by a orce depends not only on the orce and the dsplacement, but on the angle between them. When the orce s perpendcular to the dsplacement, the orce does no work at all. Q0.3. Reason: (a) The work done s W = Fd. Both partcles experence the same orce, so the greater work s done on the partcle that undergoes the greater dsplacement. Partcle A, whch s less massve than B, wll have the greater acceleraton and thus travel urther durng the s nterval. Thus more work s done on partcle A. (b) Impulse s FΔt. Both partcles experence the same orce F or the same tme nterval Δt = s. Thus the same mpulse s delvered to both partcles. (c) Both partcles receve the same mpulse, so the change n ther momenta s the same, that s, ma( v) A = mb( v ) B. But because m < m, t must be that A B ( v) A > ( v ) B. Ths result can also be ound rom knematcs, as n part (a). Assess: Work s the product o the orce and the dsplacement, whle mpulse s the product o the orce and the tme durng whch the orce acts. Q0.4. Reason: (a) The work done s W = Fd. Both partcles experence the same orce and move the same dstance, so the work done on them s the same. From the work-knetc energy theorem we know that the knetc energy o each puck has changed by the same amount. They both started rom rest, so they end up wth the same knetc energy. (b) The knetc energes are equal, but the speeds are not. va mb ma mv A A= mv B B = = = v m m So the speed o puck A s.4 tmes the speed o puck B. Assess: Ths makes ntutve sense. B A A Q0.5. Reason: Neglectng rctonal losses, the work you do on the jack s converted nto gravtatonal potental energy o the car as t s rased. The work you do s Fd, where F s the orce 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 Fw / = hd /. Assess: Because the orce F you can apply s so much less than the weght w o the car, h must be much less than d. Q0.6. Reason: The ball o mass m has an ntal potental energy o mgh and the ball o mass m has an ntal potental energy mgh. The ball o mass m has twce the ntal potental energy o the ball o mass m. There s no ntal knetc energy. Snce energy s conserved, as the balls ht the loor the ball o mass m stll has twce as much energy as the ball o mass m. As the balls ht the loor the energy s totally knetc. So the ball o mass m hts the loor wth twce the knetc energy o the ball o mass m. Assess: No energy was lost, rather t was changed rom potental to knetc energy. Q0.7. Reason: (a) I the car s to go twce as ast at the bottom, ts knetc energy, proportonal to v, wll be our tmes as great. You thus need to gve t our tmes as much gravtatonal potental energy at the top. Snce gravtatonal potental energy s lnearly proportonal to the heght h, you ll need to ncrease the heght o the track by a actor o our. (b) Usng consderatons o conservaton o energy, as n part (a), we see that the speed o the car at the bottom depends only on the heght o the track, not ts shape. Assess: Knetc energy s proportonal to the square o the velocty. Q0.8. Reason: (a) Because the elastc potental energy o a sprng s proportonal to the square o the dstance t s compressed, ts potental energy wll ncrease by a actor o our when ths dstance s doubled. (b) The ball now has our tmes the energy t had on the rst shot. When the ball s released, conservaton o energy tells us that the ball wll then have our tmes as much knetc energy. Knetc energy depends on the square o the speed, so the ball need travel only twce as ast to have our tmes the energy. Copyrght 05 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.
0-4 Chapter 0 Assess: Elastc potental energy and knetc energy both depend on the square o the dsplacement and speed, respectvely. So doublng the dsplacement leads to a doublng o the speed. Q0.9. Reason: Because both rocks are thrown rom 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. Assess: Although Chrs s rock was thrown angled upward so that t slows as t rst rses, t then speeds up as t begns to all, attanng the same speed as Sandy s as t passes the ntal heght. Sandy s rock wll ht the ground rst, but ts speed wll be no greater than Chrs s. Q0.0. Reason: By the tme the blocks reach the ground, they have transormed dentcal amounts o gravtatonal potental energy nto translatonal knetc energy o the blocks and rotatonal knetc energy o the cylnders. But the moment o nerta o a hollow cylnder s hgher than that o a sold cylnder o the same mass, so more o the energy o the system s n the orm o rotatonal knetc energy or the hollow cylnder than or the sold one. Ths leaves less energy n the orm o translatonal knetc energy or the hollow cylnder. But t s the translatonal knetc energy that determnes the speed o the block. So the block moves more slowly or the system wth the hollow cylnder, and so ts block reaches the ground last. Assess: The energy s shared between the rotatng cylnder and the allng block. The more energy the cylnder has, the less s avalable or the block. Q0.. Reason: As you land, the orce o the ground or pad does negatve work on your body, transerrng out the knetc energy you have just beore 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 orce F wll be much greater than t s wth the long stoppng dstance upon httng the pad. Assess: For a gven amount o work, the orce s large when the dsplacement s small. Q0.. Reason: I the crate sld down the ramp at a constant speed wthout Jason pushng back on t (as t could the angle were adjusted just rght) then, by conservaton o energy, the ncrease n thermal energy would exactly equal the loss n gravtatonal potental energy. But n act Jason s pushng on the crate n a drecton opposte ts moton. Thus he s dong negatve work on the crate, removng energy rom the system. So some o the loss o potental energy that would otherwse have gone nto ncreasng thermal energy s nstead removed rom the system, so the ncrease n thermal energy s actually less than the loss on potental energy. Assess: A numercal example wth made-up numbers mght help us vsualze ths. Suppose that as the crate sldes down the ramp t loses 00 J o gravtatonal potental energy, so that Δ Ug = 00 J. And suppose that the (negatve) work done by Jason s 0 J. Then the work-energy equaton, Δ Eth +Δ Ug = W s Δ Eth + ( 00 J) = 0 J, so that Δ Eth = 80 J, whch s less than the 00 J o potental energy loss. Q0.3. Reason: When the coaster s at the top U = mgy relatve to the ground. That amount o energy equals the knetc energy at the bottom. Halway down (or up) the potental energy s hal o what t was at the top, so the knetc energy must also be hal o what t s at the bottom. I v s the speed at the halway heght, then mv ( ') v' v' = = = ' = (30 m/s) = m/s () v mv v v So the correct choce s C. Assess: Even though the heght s hal the total, the speed s not hal o 30 m/s. Copyrght 05 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.
Energy and Work 0-5 Q0.4. Reason: Work s dened by W = Fd when the orce s parallel to the dsplacement, as t s n ths case. Snce you and your rend each carry sutcases o the same mass up the same lghts o stars, you both exert the same orce on the sutcases over the same vertcal dstance and thereore do the same amount o work. Your rend takes longer than you to get up the lght o stars, so you expend a greater amount o power snce P= W/ Δt. The correct choce s C. Assess: For a gven amount o work, derng amounts o power are expended dependng on how quckly the work s done. Q0.5. Reason: Assumng the woman rases the weght at constant velocty, the orce she exerts must equal the weght o the object. Snce the mass o the object s 0 kg, ts weght s about w= mg = (0 kg)(9.80 m/s ) = 00 N. Snce she lts t m, the work done s W = Fd = (00 N)( m) = 400 J. She does ths work n 4 s, so the power she exerts s P = W/ Δ t = (400 J)/(4 s) = 00 W. The correct choce s A. Assess: Power s dened as the rate o dong work. Ths seems lke a reasonable amount o power or the woman to expend. Q0.6. Reason: Snce knetc energy s proportonal to the square o the velocty o an object, an object wth twce the velocty wll have our tmes the amount o knetc energy. In ths queston, all the knetc energy s converted to elastc potental energy n the sprng. The potental energy stored n a sprng s proportonal to the square o the compresson rom ts equlbrum poston. Snce we start wth our tmes the knetc energy, our tmes as much energy s stored n the sprng. But snce the energy stored n the sprng s proportonal to the square o the compresson, the compresson s only twce the compresson prevously, or (.0 cm) = 4.0 cm. The correct choce s C. Assess: Knetc energy s proportonal to the square o the velocty o an object and the potental energy o a sprng s proportonal to the square o the dsplacement rom the equlbrum poston. Q0.7 Reason: The potental energy at the top o the rst ramp s equal to the knetc energy at the bottom, so that mgh = (/) mv. By conservaton o energy, the energy when the block s movng up the second ramp at speed 0 v / 0 s equal to the ntal energy mgh, so we have mg h + m v 0 mg h + mv 4 ( 0 )= mgh. From the conservaton o energy appled to the rst ramp, we can rewrte ths as mgh + mgh = mgh or h = 3 h /4. The correct choce s C., 4 ( ) = mgh, whch we can rewrte as Assess: Because the block had energy mgh at the top o the ramp, t has energy mgh at all ponts along both ramps. Smlarly, ts energy s (/)mv at all ponts along both ramps. 0 Q0.8. Reason: As the ball alls, energy s conserved snce the only orce dong work s the orce o gravty. Snce gravty s conserved we may wrte ΔE = 0 or Δ K +Δ U g = 0 As the ball alls we have Δ K = mv / and Δ U g = mgh = mg( L Lcos30 ) = mgl ( cos30 ) Combnng these obtan mv / = mgl ( cos30 ) or v= gl ( cos30 ) = 3.6 m/s The correct response s C. Assess: The key to the problem s to realze that energy s conserved and then nd the change n knetc and potental energy. A speed o 3.6 m/s s reasonable. Copyrght 05 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.
0-6 Chapter 0 Problems P0.. Prepare: Equaton 0.6 gves the work done by a orce F on a partcle. The work s dened as W = Fd cos(θ), where d s the partcle s dsplacement. Snce there s a component o the ltng orce n the drecton o the dsplacement, we expect the work done n both parts o ths problem to be nonzero. Assume you lt the book steadly, so that the orce exerted on the book s constant. Solve: (a) Reer to the dagram. We are assumng the book does not accelerate, so the orce you exert on the book s exactly equal to the orce o gravty on the book. F hand on book = F gravty on book The total dsplacement o the book s.3 m 0.75 m =.55 m (keepng one extra sgncant gure or ths ntermedate result). The work done by gravty s then W gravty on book = wd cos(θ) = (.0 kg)(9.80 m/s )(.55 m)cos(80 ) = 30 J (b) The work done by hand s Whand on book = Fhand on bookd cos( θ). W hand on book = (.0 kg)(9.80 m/s )(.55 m)cos(0 ) =+30 J Assess: Note that the only derence s n the sgn o the answer. Ths s because the two orces are equal, but act n opposte drectons. The work done by gravty s negatve because gravty acts opposte to the dsplacement o the book. Your hand exerts a orce n the same drecton as the dsplacement, so t does postve work. We should expect the total work to be zero rom Equaton 0.4 snce energy s conserved n ths process. Reerrng to the results, we see that the work by your hand cancels the work done by gravty and the total work s zero as expected. P0.. Prepare: Note that not all the orces n ths problem are parallel to the dsplacement. Equaton 0.6 gves the work done by a constant orce whch s not parallel to the dsplacement: W = Fd cos(θ), where W s the work done by the orce F at an angle θ to the dsplacement d. Here the dsplacement s exactly downwards n the same drecton as w. We wll take all orces as havng our sgncant gures (as mpled by T = 95 N). Copyrght 05 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.
Energy and Work 0-7 Solve: The angle between the orce w and the dsplacement s 0, so W w = wd cosθ = (500 N)(5 m)cos(0 ) =.5 kj The angle between the orce T and the dsplacement s 90 + 60 = 50. WT = T d cosθ = (830 N)(5 m)cos(50 ) = 7.9 kj The angle between the tenson T and the dsplacement s 90 + 45 = 35. W T = T d cosθ = (95 N)(5 m)cos(35 ) = 4.58 kj Assess: Note that the dsplacement d n all these cases s drected downwards and that t s always the angle between the orce and dsplacement used n the work equaton. For example, the angle between T and d s 50, not 60. P0.3. Prepare: Note that not all orces act n the same drecton as the dsplacement. We must use Equaton 0.6: W = Fd cos(θ) or each orce. W s the work done by a orce o magntude F on a partcle and d s the partcle s dsplacement. The crate s movng drectly to the rght. We assume all orces are gven to three sgncant gures, ncludng the 500 N orce snce the other two orces are gven to three sgncant gures. Solve: For the orce k, the dsplacement s exactly opposte the orce, so For the tenson T : For the tenson T : W k = kdcos(80 ) = (500 N)(3 m)( ) =.50 kj WT = Td cos(0 ) = (36 N)(3 m)(0.9397) = 0.99 kj WT = Tdcos(30 ) = (3 N)(3 m)(0.8660) = 0.579 kj Assess: Negatve work done by the orce o knetc rcton k means that.50 kj o energy has been transerred out o the crate and converted to heat. The other two orces have components along the dsplacement, and thereore do postve work to move the crate. P0.4. Prepare: We wll use the denton o work, Equaton 0.6 to calculate the work done. The sdewalk and escalators exert a normal orce on you, and may exert a orce to propel you orward. We wll assume that the escalator propels you at constant velocty, as the sdewalk does. Solve: (a) The escalator moves you across some dstance lke the sdewalk, but t also moves you upwards. See the ollowng dagram. Copyrght 05 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.
0-8 Chapter 0 The orce exerted on you by the escalator s the normal orce, 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 sgncant gures. Unlke the sdewalk case, there s a component o the dsplacement parallel to the normal orce. The angle between the orce and dsplacement s cos (4.5/7) = 50, so d cos(θ) = 4.5 m. Then W = Fd cosθ = (588 N)(4.5 m) =.6 kj (b) Reer to the gure. Here, the dsplacement s n the opposte drecton compared to part (b), so the angle between the orce and the dsplacement s now 80 50 = 30. So W = Fd cosθ = (588 N)(7.0 m)cos(30 ) =.6 kj Assess: In part (a), snce the orce has no component n the drecton o your dsplacement, the orce does no work. In part (b), there s a component o the orce along and n the same drecton as the dsplacement, so the orce does postve work. In part (c), the component o the orce along the dsplacement s n the opposte drecton to the dsplacement, so the orce does negatve work. P0.5. Prepare: Equaton 0.6 s the denton o work when the orce and dsplacement are not parallel, as s the case n ths problem. Solve: (a) The boy s standng stll n ths case, so the dsplacement s zero. W = Fd cos(θ) = (F)(0 m)cos(θ) = 0 J. The work done on the boy by the strng s exactly zero Joules. (b) The dsplacement s non-zero n ths case, so we expect the work done to be non-zero. Reer to the ollowng gure. The angle between the orce and the dsplacement s 80 30 = 50. The work s W = Fd cos(θ) = (4.5 N)( m)cos(50 ) = 43 J (c) The angle between the orce and dsplacement n ths case s 30 (Look at the gure and magne the drecton o the dsplacement vector s reversed). The work s W = Fd cos(θ) = (4.5 N)( m)cos(30 ) = 43 J Assess: For there to be work done, the dsplacement must not be zero. I there s no dsplacement there s no work done. Note that the answers to parts (b) and (c) have opposte sgns. Ths s because the dsplacement s exactly opposte n those cases or the same drecton o the orce. Copyrght 05 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.
Energy and Work 0-9 P0.6. Prepare: Equaton 0.6 s the denton o work when the orce and dsplacement are not parallel, as s the case n ths problem. Solve: Because Page pushes back the angle between the orce and the dsplacement s actually 60. W = Fd cos θ = (68 N)(3.5 m)cos60 = 0 N The amount o work Page dd s thereore 0 N. Assess: The cosne o the angle was negatve showng that the angle was greater than 90 degrees. P0.7. Prepare: The knetc energy or any object movng o mass m wth velocty v s gven n Equaton 0.8: K = mv. Solve: For the bullet, For the bowlng ball, K B = m v = B B (0.00 kg)(500 m/s) =.3 kj K BB = m v = BB BB (0 kg)(0 m/s) = 0.50 kj Thus, the bullet has the larger knetc energy. Assess: Knetc energy depends not only on mass but also on the square o the velocty. The prevous calculaton shows ths dependence. Although the mass o the bullet s 000 tmes smaller than the mass o the bowlng ball, ts speed s 50 tmes larger, whch leads to the bullet havng over twce the knetc energy o the bowlng ball. P0.8. Prepare: Use the denton o knetc energy, Equaton 0.8, to set up an equaton such that the knetc energy o the car s equal to that o the truck. Solve: For the knetc energy o the compact car and the knetc energy o the truck to be equal, mt 0 000 kg KC = KT mcvc = mtvt vc = vt = (5 km/h) = 0 km/h mc 000 kg To match the knetc energy o the truck, the car needs a velocty o 0 km/h (to two sgncant gures). Assess: Note that the smaller mass needs a greater velocty or ts knetc energy to be the same as that o the larger mass. Though the truck has 0 tmes the mass, the car only needs about our tmes the velocty o the truck to have the same knetc energy. Ths s because knetc energy s proportonal to the mass, but proportonal to the square o the velocty. P0.9. Prepare: In order to work ths problem, we need to know that the knetc energy o an object s gven by K = mv /. Solve: The problem may be solved n a qualtatve manner or n a quanttatve manner. Snce some students thnk one way and some the other, we wll use both methods. (a) Frst, n a qualtatve manner. Snce the knetc energy depends on the square o the speed, the knetc energy wll be doubled the speed s ncreases by a actor o. Ths s true because ( ) =. Then the new speed s (0 m/s) = 4 m/s. Copyrght 05 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.
0-0 Chapter 0 Second, n a more quanttve manner. Use a subscrpt or the present case where the speed s 0 m/s and a subscrpt or the new case where the speed s such that the knetc energy s doubled. K = mv/ and K = mv / We want K = K Insertng expressons or K and K obtan mv mv = or v = v = 4 m/s (b) Frst, n a qualtatve manner, the speed s doubled and the knetc energy depends on the square o the speed, the knetc energy wll ncrease by a actor o our. Second, n a more quanttatve manner, use a subscrpt or the present case and a subscrpt or the new case where the speed s doubled. K = mv/ and K = mv / We want v = v. Insertng v nto K, we obtan mv m( v) mv K = = = 4 = 4K Ths expresson clearly shows that the knetc energy s ncreased by a actor o our when the speed s doubled. Assess: The key to the problem s to know that the knetc energy depends on the speed squared. Ater that, we can approach the problem n a qualtatve or a quanttatve manner. You may preer one method over the other, but you should be able to work n ether mode. P0.0. Prepare: Use the denton o knetc energy. Solve: K = mv = (60 kg)(33 m/s) = 33 kj P0.. Prepare: Use the denton o knetc energy. Solve: The man and the bullet have the same knetc energy. mv = mv v m m m b b 8.0 g (400 m/s) 4.0 m/s 80 kg b m = vb = = mm Assess: We expected the man to need much less speed than the bullet to have the same knetc energy. P0.. Prepare: We wll assume that all the work Sam does goes nto stoppng the boat. We can use conservaton o energy as expressed n Equaton 0.7 to calculate the work done rom the change n knetc energy. Solve: Reer to the beore and ater representaton o Sam stoppng a boat. Equaton 0.7 becomes mv mv = W Snce the boat s at rest at the end o the process, v = 0 m/s. Thereore, the nal knetc energy s zero. The work done on the boat s then W = mv = (00 kg)(. m/s) = 0.86 kj The amount o work s 0.86 kj. Copyrght 05 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.
Energy and Work 0- Assess: Note that the work done by Sam on the boat s negatve. Ths s because the orce Sam exerts on the boat must be opposte to the drecton o moton o the boat to slow t down. P0.3. Prepare: Use the law o conservaton o energy, Equaton 0.4, to nd the work done on the partcle. We wll assume there s no change n thermal energy o the ball. Solve: Consder the system to be the plastc ball. Snce there s no change n potental, thermal or chemcal energy o the ball, and there s no heat leavng or enterng the system, the conservaton o energy equaton becomes W =Δ K = mv mv = m( v v ) = (0.00 kg)[(30 m/s) ( 30 m/s) ] = 0 J Assess: Note that no work s done on the ball n reversng ts velocty. Ths s because negatve work s done n slowng the ball down to rest, and an equal amount o postve work s done n brngng the ball to the orgnal speed but n the opposte drecton. P0.4. Prepare: The denton o knetc energy or objects rotatng around a statonary axs s gven by Equaton 0.9. In Equaton 0.9, the rotatonal velocty should be n unts o radans/second. Solve: The turntable turns once every 4.0 s. So ts rotatonal velocty s rev π rad ω = =.57 rad/s 4.0 s rev Ths should be reported as.6 rad/s to two sgncant gures. We keep an extra sgncant gure or substtuton n the next step: Krot = I ω = (0.040 kg m )(.57 rad/s) = 0.049 J Assess: Ths s a reasonable result or such a low rotatonal velocty and moment o nerta. P0.5. Prepare: Energy s stored n the lywheel by vrtue o the moton o the partcles and s gven by Equaton 0.9. In ths equaton, unts or rotatonal velocty must be rad/s. Solve: Usng Equaton 0.9, rot rot, so = ω K K = I I ω We need to convert ω to proper unts, radans/s. Snce ω = 0 000 rev/mn and there are π rad/rev and 60 s/mn, rev mn π rad ω = 0 000 mn 60 s rev So. 6 ()(4.0 0 J) I = =.8 kg m rev mn π rad 0 000 mn 60 s rev Assess: The lywheel can store ths large amount o energy even though t has a low moment o nerta because o ts hgh rate o rotaton. P0.6. Prepare: As the heght o the hker changes so does hs potental energy. Gravtatonal potental energy s gven by Equaton 0.3. We only need to calculate the derence n potental energy. The reerence pont or measurng the hker s heght s arbtrary. Copyrght 05 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.
0- Chapter 0 We choose the orgn o the coordnate system chosen or ths problem at sea level so that the hker s poston n Death Valley s y 0 = 85.0 m. Solve: The hker s change n potental energy rom the bottom o Death Valley to the top o Mt. Whtney s Δ U = ( Ug) ( Ug) = mgy mgy= mg( y y) 6 = (65.0 kg)(9.80 m/s )[440 m ( 85.0 m)] =.87 0 J Assess: Ths s a large amount o energy, as expected. Note that ΔU s always ndependent o the orgn o the coordnate system. P0.7. Prepare: In part (a) we can smply use the denton o knetc energy n Equaton 0.8. We then use ths result n part (b) to nd the heght the car must be dropped rom to obtan the same knetc energy. The car s allng under the nluence o gravty. We can use conservaton o energy to calculate ts knetc energy as a result o the all. The sum o knetc and potental energy does not change as the car alls. Solve: (a) The knetc energy o the car s 5 5 KC = mcv C = (500 kg)(30 m/s) = 6.75 0 J 6.8 0 J We keep one addtonal sgncant gure here or use n part (b). (b) Reer to the dagram. Here we set K equal to K C n part (a) and place our coordnate system on the ground at = 0m. the car s potental energy ( U g) s zero, ts velocty s v, and ts knetc energy s K. At poston, so K = 0 J, and the only energy the car has s ( U g ) = mgy. Snce the sum + g moton, K + ( Ug) = K + ( U g). Ths means y At ths pont, y v = 0 m/s, K U s unchanged by Copyrght 05 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.
Energy and Work 0-3 K + mgy = K + mgy K + 0 = K + mgy 5 ( K K) (6.75 0 J 0J) y = = = 46 m mg (500 kg)(9.80 m/s ) (c) To check ths result depends on the car s mass, rewrte the result o part (b) leavng m as a varable, and check t cancels out. ( K K) mv mv ( v v ) y = = = mg mg g Snce m cancels out, the dstance does not depend upon the mass. Assess: A car travelng at 30 m/s s travelng at 08 km/h or 67 m/h. At that speed, t has the same amount o energy as rom beng dropped 46 m, whch s 5 t, or rom the top o an approxmately 9 story buldng! P0.8. Prepare: We need to know how to nd knetc energy ( K = mv /) and gravtatonal potental energy ( U = mgh ) and be aware that we want the potental energy to change by an amount equal to the knetc energy. g Solve: Equate the potental energy to the knetc energy mv = mgh And solve or h to obtan h = v g = ( m/s) (9.8 m/s ) = 6. m Assess: Ths s as hgh as a two-story buldng. P0.9. Prepare: In ths case the bke racer wll ncrease hs gravtatonal potental energy by an amount ΔU g = mgh, where h s the change n hs vertcal poston. Solve: The change n the bke racer s gravtaton potental energy as he executes the clmb s ΔU g = mgh = mgd sn4.3 =63 kj Assess: The bke racer would ncrease hs gravtatonal potental energy ths same about by clmbng up a lght o stars that ncreased hs vertcal poston by 90 m. Ths allows one to conclude that star clmbng s excellent exercse. P0.0. Prepare: The wreckng ball wll ncrease ts gravtatonal potental energy by an amount ΔU g = mgh, where h s the change n the vertcal poston. Solve: The ncrease n gravtatonal potental energy o the wreckng ball s determned by Δ U = mgh = mg( L Lcos5 ) = mgl ( cos5 ) = 4 kj g Assess: Ths s a reasonable change n gravtatonal potental energy or such a massve object. P0.. Prepare: Assume an deal sprng that obeys Hooke s law. Equaton 0.5 gves the energy stored n a sprng. The elastc potental energy o a sprng s dened as U s = kx, where x s the magntude o the stretchng or compresson relatve to the unstretched or uncompressed length. ΔU s = 0 when the sprng s at ts equlbrum length and x = 0. Solve: We have U = 00 J and k = 000 N/m. Solvng or x: s x= U / k = (00 J)/(000 N/m) = 0.63 m s Assess: In the equaton or the elastc potental energy stored n a sprng, t s always the dstance o the stretchng o compresson relatve to the unstretched or equlbrum length. P0.. Prepare: Assume an deal sprng that obeys Hooke s law. Equaton 0.5 gves the potental energy stored n an deal sprng. The elastc potental energy o a sprng s dened as Us = kx, where x s the magntude o the stretchng or compresson relatve to the unstretched or uncompressed length. Copyrght 05 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.
0-4 Chapter 0 Solve: We have x = 0 cm = 0.0 m and k = 500 N/m. Ths means Us = kx = (500 N/m)(0.0 m) = 0 J Assess: Snce x s squared, s sprng s n ts equlbrum poston. U s postve or a sprng that s ether compressed or stretched. U s s zero when the P0.3. Prepare: We wll assume the knee extensor tendon behaves accordng to Hooke s Law and stretches n a straght lne. The elastc energy stored n a sprng s gven by Equaton 0.5, U = kx Solve: For athletes, For non-athletes, U s,athlete s. (33 000 N/m)(0.04 m) kx 7.7 J = = = (33 000 N/m)(0.033 m) kx 8.0 J Us,non-athlete = = = The derence n energy stored between athletes and non-athletes s thereore 9.7 J. Assess: Notce the energy stored by athletes s over.5 tmes the energy stored by non-athletes. P0.4. Prepare: Snce the loor s horzontal, there s no change n gravtatonal potental energy. I Marssa drags the bag across the loor at a constant speed, there s no change n knetc energy and hence no net work done on the bag. I ths s the case, then WMarssa + W rcton = 0 and ΔE thermal = W Marssa = W rcton Solve: The thermal energy created by Marssa may be determned by: Δ E = W = W = F dcosθ = μ Ndcos(80 ) = μ mgd = 4.7 0 J thermal Marssa rcton rcton knetc knetc Assess: I Marssa accomplshed ths n 30 s (whch s a reasonable tme) the power delvered would lght a 50 watt lght bulb. Ths s a reasonable amount. P0.5. Prepare: Snce the gravtatonal potental energy and the knetc energy o the car do not change, all the work Mark does on the car goes nto thermal energy. Solve: The thermal energy created n the tres and the road may be determned by: Δ Et = WMark = FMarkdcos0 = (0 N)(50 m)cos0 = 6.5 kj Assess: All the work Mark does n pushng the car, becomes thermal energy o the tres and road. Snce Mark s pushng n the drecton the car s movng, the angle between the drecton o F and d s 50 P0.6. Prepare: In ths case all the work done on the crate by gravty goes nto thermal energy. The work done by gravty depends on the angle between the orce o gravty and the dsplacement o the crate, whch s 55 degrees. Solve: The thermal energy created may be determned by: Δ Et = Wgravty = FgravtyLcos55 = (900 N)( m)cos 55 = 6. kj Assess: Ths seems lke a lot o thermal energy, however the crate s heavy (about 00 lbs) and the ramp s long so we should expect a large number. We could also solve ths problem as ollows: Δ E = W = Δ U = ( mgh) = F Lsn35 = F Lcos55 Th gravty gravty gravty gravty P0.7. Reason: The orce o gravty and the orce o rcton are dong work on the chld. Snce the chld sldes at a constant speed, the net work (whch s the change n knetc energy) s zero. Ths allows us to wrte: Wg + W =Δ K = 0 or W = W g Knowng how the work done by gravty s related to the change n gravtatonal potental energy and how the work done by rcton s related to the orce o rcton, we can determne the orce o rcton. Solve: Wrtng expressons or the work done by rcton obtan W = W g = ( ΔU g ) =ΔU g = Mgh and W = F Lcos80 = F L Combnng these and solvng or the orce o rcton obtan F = Mgh/ L= (5 kg)(9.8 m/s )(3.0 m)/(7.0 m) =.0 0 N Copyrght 05 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.
Energy and Work 0-5 The mnus remnds us that the orce o rcton opposes the moton o the object, so the magntude s 00 N. Assess: A 00 N orce o rcton or a chld sldng down a playground sld s a reasonable number. P0.8. Prepare: Ths s a case o ree all, so the sum o the knetc and gravtatonal potental energy does not change as the ball rses and alls. The gure shows a ball s beore-and-ater pctoral representaton or the three stuatons n parts (a), (b) and (c). Solve: The quantty K + U g s the same durng ree all: K + U = +. g K U g At the top o ts trajectory where the ball turns around the ball s velocty s 0 m/s. We have (a) mv + mgy = mv0 + mgy0 y = ( v v ) / g = [(0 m/s) (0 m/s) ]/( 9.80 m/s ) = 5. m 0 5. m s thereore the maxmum heght o the ball above the wndow. Ths s 5 m above the ground. (b) mv + mgy = mv0 + mgy 0 Snce y = y 0 = 0, we get or the magntudes v = v 0 = 0 m/s. (c) mv3 + mgy3 = mv0 + mgy0 v3 + gy3 = v0 + gy0 v3 = v0 + g( y0 y 3) v 3 = (0m/s) + (9.80m/s )[0 m ( 0m)] Takng the square root, the magntude o v 3 s equal to m/s. Assess: Note that the ball s speed as t passes the wndow on ts way down s the same as the speed wth whch t was tossed up, but n the opposte drecton. P0.9. Prepare: The only orce actng on the ball durng ts trp s gravty. The sum o the knetc and gravtatonal potental energy or the ball, consdered as a partcle, does not change durng ts moton. Use Equaton 0.4. Note that at the top o ts trajectory when the ball turns around, the velocty o the ball s zero. Assume there s no rcton. Copyrght 05 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.
0-6 Chapter 0 The gure shows the ball s beore-and-ater pctoral representaton or the two stuatons descrbed n parts (a) and (b). Solve: Snce energy s conserved, the quantty K + U s the same durng the entre trp. Thus, K + U = K + U. g g g (a) mv + mgy = mv0 + mgy0 v0 = v + g( y y0) v0 = (0 m/s) + (9.80 m/s )(0 m.5 m) = 67 m /s v0 = 3 m/s (b) mv + mgy = mv0 + mgy0 v = v0 + g( y0 y) v = 67 m /s + (9.80 m/s )(.5 m 0 m) v = 4 m/s Assess: An ncrease n speed rom 3 m/s to 4 m/s as the ball alls through a dstance o.5 m s reasonable. Also, note that mass does not appear n the calculatons that nvolve ree all snce both gravtatonal potental energy and knetc energy are proportonal to mass. The mass cancels out n the equatons. P0.30. Prepare: Snce the ramp s rctonless, the sum o the puck s knetc and gravtatonal potental energy does not change durng ts sldng moton. Use Equaton 0.4 or the conservaton o energy. Solve: The quantty K + U g s the same at the top o the ramp as t was at the bottom. The energy conservaton equaton K +U g = K +U g s mv + mgy = mv + mgy v = v + g( y y) v = (0 m/s) + (9.80 m/s )(.03 m 0 m) = 0.m /s v = 4.5 m/s Assess: An ntal push wth a speed o 4.5 m/s 0 mph to cover a dstance o 3.0 m up a 0 ramp seems reasonable. Note that a ramp o any angle to the same nal heght would lead to the same nal velocty or the puck. Note that the mass cancels out n the equaton snce both knetc energy and gravtatonal potental energy are proportonal to mass. Copyrght 05 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.
Energy and Work 0-7 P0.3. Prepare: The ollowng gure shows a beore-and-ater pctoral representaton o the rollng car. The car starts at rest rom the top o the hll snce t slps out o gear. Snce we are gnorng rcton, energy s conserved. The total energy o the car at the top o the hll s equal to total energy o the car at any other pont. Solve: The energy conservaton equaton then becomes K + ( Ug) = K + ( U g ) or mv + mgy = mv + mgy The car starts rom rest, so v = 0 m/s, whch gves K = 0 J. Takng the bottom o the hll as the reerence pont or gravtatonal potental, y = 0 m and so U = 0 J. The energy conservaton equaton becomes mv mgy = Cancelng m and solvng or v, v = gy = = ()(9.80 m/s )(50 m) 3 m/s Assess: Note that the problem does not gve the shape o the hll, so the acceleraton o the car s not necessarly constant. Constant acceleraton knematcs can t be used to nd the car s nal speed. However, energy s conserved no matter what the shape o the hll. Note that the mass o the car s not needed. Snce knetc energy and gravtatonal potental energy are both proportonal to mass, the mass cancels out n the equaton. The nal speed o the car, ater travelng to the bottom o the 50 m hll s 3 m/s whch s nearly 70 m/h! P0.3. Prepare: Assume there s zero rollng rcton snce rcton s not mentoned n the problem. The sum o the knetc and gravtatonal potental energy, thereore, does not change durng the car s moton. Consder the other sde o the hll to be the zero or gravtatonal potental energy. Copyrght 05 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.
0-8 Chapter 0 Solve: (a) The ntal energy o the car s 4 K0 + U (500 kg)(0 m/s) 7.5 0 J g0 = mv0 + mgy 0 = = The energy o the car at the top o the hll s 4 K+ Ug = K+ mgy = K+ (500kg)(9.80m/s )(5.0m) = K + 7.4 0 J I the car just wants to make t to the top, then K = 0. That s, the car has no velocty at the top o the hll. In other 4 words, a mnmum energy o 7.4 0 J s needed to get to the top. Snce ths energy s less than the avalable 4 energy o 7.5 0 J, the car can make t to the top. (b) The conservaton o energy equaton K0 + Ug0 = K + U g s 4 4 7.5 0 J = mv + mgy 7.5 0 J = (500 kg) v + (500 kg)(9.80 m/s )( 5.0 m) v = 4 m/s Assess: A hgher speed on the other sde o the hll s reasonable because the car has ncreased ts knetc energy and lowered ts potental energy compared to ts startng values. Note that the shape o the hll s rrelevant because gravtatonal potental energy depends only on heght. P0.33. Prepare: Consder the sprng as an deal sprng that obeys Hooke s law. We wll also assume zero rollng rcton durng the compresson o the sprng, so that mechancal energy s conserved. At the maxmum compresson o the sprng, 60 cm, the velocty o the cart wll be zero. The gure shows a beore-and-ater pctoral representaton. The beore stuaton s when the cart hts the sprng n ts equlbrum poston. We put the orgn o our coordnate system at ths equlbrum poston o the ree end o the sprng. Ths gve x = xe = 0 and x = 60 cm. Solve: The conservaton o energy equaton K + Us = K+ U s s mv + kx = mv + kx Usng v = 0 m/s, x = 0.60 m, and x = 0 m gves: k 50 N/m kx = mv v = x = (0.60 m) = 3.0 m/s m 0kg Assess: Elastc potental energy s always measured rom the unstretched or uncompressed length o the sprng. Copyrght 05 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.
Energy and Work 0-9 P0.34. Prepare: Model the jet plane as a partcle, and the sprng as an deal that obeys Hooke s law. We wll also assume zero rollng rcton durng the stretchng o the sprng, so that mechancal energy s conserved. The gure shows a beore-and-ater pctoral representaton. The beore stuaton occurs just as the jet plane lands on the arcrat carrer and the sprng s n ts equlbrum poston. We put the orgn o our coordnate system at the rght ree end o the sprng. Ths gves x = x e = 0 m. Snce the sprng stretches 30 m to stop the plane, x = 30 m. Solve: The conservaton o energy equaton K + Us = K+ U s or the sprng-jet plane system s mv + kx = mv + kx Usng v = 0 m/s, x = xe = 0 m, and x = 30 m yelds k 60 000 N/m kx = mv v = ( x) = (30m) = 60 m/s m 5000kg Assess: A landng speed o 60 m/s or 30 mph s reasonable. P0.35. Prepare: Ths s a case o ree all, so the sum o the knetc and gravtatonal potental energy does not change as the rock s thrown. Assume there s no rcton. The drecton the rock s thrown s not known. The coordnate system s put on the ground or ths system, so that y = 6 m. The rock s nal velocty v must be at least 5.0 m/s to dslodge the Frsbee. Solve: The energy conservaton equaton or the rock K + Ug = K + U g s mv + mgy = mv + mgy Ths equaton nvolves only the velocty magntudes and not the angle at whch the rock s to be thrown to dslodge the Frsbee. Ths equaton s true or all angles that wll take the rock to the Frsbee 6 m above the ground and movng wth a speed o 5.0 m/s. Copyrght 05 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.
0-0 Chapter 0 Usng the prevous equaton we get v + gy = v + gy v = v + g( y y ) = (5.0 m/s) + (9.80 m/s )(6 m.0 m) = 7 m/s Assess: Knetc energy s dened as K = mv and s a scalar quantty. Scalar quanttes do not have drectonal propertes. Note also that the mass o the rock s not needed. P0.36. Prepare: We wll use conservaton o energy, Equaton 0.4, to calculate the ncrease n thermal energy. Assume the ntal velocty o the reman to be zero. Solve: Consder the gure. Usng ground level as the reerence or gravtatonal potental energy, the conservaton o energy equaton becomes Δ K +Δ Ug +Δ Eth = 0 Δ Eth = ( Δ K +ΔU g) The change n hs gravtatonal potental energy s Δ Ug = mgy mgy = 0 (80 kg)(9.80 m/s )(4. m) = 3.3 kj The change n hs knetc energy s Δ K = K K = 0 J (80 kg)(. m/s) = 0.9 kj So Δ E th = (90 J 3300 J) =+ 3. kj. Assess: Note that most o the gravtatonal potental energy s converted to thermal energy. P0.37. Prepare: The thermal energy o the slde and the chld s pants changes durng the slde. I we consder the system to be the chld and slde, total energy s conserved durng the slde. The energy transormatons durng the slde are governed by the conservaton o energy equaton, Equaton 0.4. Solve: (a) The chld s knetc and gravtatonal potental energy wll be changng durng the slde. There s no heat enterng or leavng the system, and no external work done on the chld. There s a possble change n the thermal Copyrght 05 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.
Energy and Work 0- energy o the slde and seat o the chld s pants. Use the ground as reerence or calculatng gravtatonal potental energy. K = K0 = mv0 = 0 J U = Ug0 = mgy0 = (0 kg)(9.80 m/s )(3.0 m) = 590 J W = 0 J K = K = mv = (0 kg)(.0 m/s) = 40 J U = Ug = mgy = 0 J At the top o the slde, the chld has gravtatonal potental energy o 590 J. Ths energy s transormed partly nto the knetc energy o the chld at the bottom o the slde. Note that the nal knetc energy o the chld s only 40 J, much less than the ntal gravtatonal potental energy o 590 J. The remander s the change n thermal energy o the chld s pants and the slde. (b) The energy conservaton equaton becomes Δ K +Δ Ug +Δ E th = 0. Wth Δ U g = 590 J and Δ K = 40 J, the change n the thermal energy o the slde and o the chld s pants s then 590 J 40 J = 550 J. Assess: Note that most o the gravtatonal potental energy s converted to thermal energy, and only a small amount s avalable to be converted to knetc energy. P0.38. Prepare: The only orce dong work on the puck s rcton. Knowng that the change n knetc energy o the puck s equal to the work done on the puck, we can determne how ar the puck wll slde beore comng to rest. Solve: The change n knetc energy o the puck s Δ K = K = K mv / The work done on the puck by the orce o rcton s W = Fd cos80 = μnd = μmgd Knowng that W =ΔK obtan μ mgd = mv / or d = v / μg = (5.0 m/s) /(0.05)(9.8 m/s ) = 6 m Assess: Ths s a reasonable dstance or a puck travelng 5 m/s to slde across the ce beore comng to rest. P0.39. Prepare: Call the system the tube, rder, and slope (but not the rope or rope puller). The snow s not rctonless. Assume the tow rope s parallel to the slope. Use the work-energy equaton and W = Fd. Solve: Snce there are no sprngs or chemcal reactons nvolved the work-energy equaton s W =Δ K +Δ Ug +ΔE th We are told the towng takes place at a constant speed so Δ K = 0. Solve or the change n thermal energy. Δ Eth = W Δ Ug = Fd mgδ y= (340 N)(0 m) (80 kg)(9.8 m/s )(30 m) = 7 kj Assess: W s the work done on the system by the rope. P0.40. Prepare: Call the system the bke, rder. Use the work-energy equaton and W = Fd. Assume the cyclst and ar do not heat up: Δ Eth = 0. Solve: Snce there are no sprngs or chemcal reactons nvolved the work-energy equaton s K + ( Ug) +Δ Eth = K + ( Ug) + W mv + 0+ 0= mv + mgh FDL FL D ( N)(450 m) v = v + gh = ( m/s) + (9.8 m/s )(30 m) m 70 kg v = 4 m/s Assess: 4 m/s s qute ast (54 mph) or a cyclst, but the drag was small. Copyrght 05 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.