Chapter 6 Work and Kinetic Energy

Transcription

1 Chapter 6 ork and Kinetic Energ Conceptual Probles True or alse: (a) I the net or work done on a particle was not zero, then its speed ust have changed. (b) I the net or work done on a particle was not zero, then its velocit ust have changed. (c) I the net or work done on a particle was not zero, then its direction o otion could not have changed. (d) No work is done b the orces acting on a particle i it reains at rest. (e) A orce that is alwas perpendicular to the velocit o a particle never does work on the particle. Deterine the Concept A orce does work on an object when its point o application oves through soe distance and there is a coponent o the orce along the line o otion. (a) True. The work done on a particle is equal to the change in its kinetic energ (the work-kinetic energ theore). (b) True. The work done on the particle changes the kinetic energ o the particle, which eans that its velocit ust change. (c) False. I the work done on the particle is negative, it could be decreasing the particle s speed and, eventuall, reversing its direction o otion. (d) True. The object could be at rest in one reerence rae and oving in another. I we consider onl the rae in which the object is at rest, then, because it ust undergo a displaceent in order or work to be done on it, we would conclude that the stateent is true. (e) True. A orce that is alwas perpendicular to the velocit o a particle changes the direction the particle is oving but does no work on the particle. You push a heav bo in a straight line along the top o a rough horizontal table. The bo starts at rest and ends at rest. Describe the work done on it (including sign) b each orce acting on it and the net work done on it. Deterine the Concept The noral and gravitational orces do zero work because the act at right angles to the bo s direction o otion. You do positive work on the bo and riction does negative work. Overall the net work is zero because its kinetic energ doesn t change (zero to zero). 59

2 53 Chapter 6 3 You are riding on a Ferris wheel that is rotating at constant speed. True or alse: During an raction o a revolution: (a) None o the orces acting on ou does work on ou. (b) The work done b all orces acting on ou is zero. (c) There is zero net orce on ou. (d) You are accelerating. (a) False. Both the orce eerted b the surace on which ou are sitting and the gravitational orce acting on ou do work on ou. (b) True. Because ou are rotating with a constant speed (ignoring the ver brie periods o acceleration at the beginning and end o the ride) and return, at the end o the ride, to the sae location ro which ou started the ride, our kinetic energ has not changed and so the net work done on ou is zero. (c) False. The net orce acting on ou is the su o the gravitational orce acting on ou and the orce eerted b the surace on which ou are sitting. (d) True. Because the direction ou are oving is continuall changing, our velocit is continuall changing and, hence, ou are eperiencing acceleration. 4 B what actor does the kinetic energ o a particle change i its speed is doubled but its ass is cut in hal? Deterine the Concept The kinetic energ o a particle is proportional to the square o its speed. Because K v, replacing v b v and b ields ( )( v) ( v ) K K'. Thus doubling the speed o a particle and halving its ass doubles its kinetic energ. 5 Give an eaple o a particle that has constant kinetic energ but is accelerating. Can a non-accelerating particle have a changing kinetic energ? I so, give an eaple. Deterine the Concept A particle oving along a circular path at constant speed has constant kinetic energ but is accelerating (because its velocit is continuall changing). No, because i the particle is not accelerating, the net orce acting on it ust be zero and, consequentl, its kinetic energ ust be constant. 6 A particle initiall has kinetic energ K. Later it is ound to be oving in the opposite direction with three ties its initial speed. hat is the kinetic energ now? (a) K, (b) 3K, (c) 3K, (d) 9K, (e) 9K Deterine the Concept The kinetic energ o a particle is proportional to the square o its speed and is alwas positive. Because K v, replacing v b 3v

3 ields K' ( 3 ) 9( ) v v 9 K. ork and Energ 53 Hence tripling the speed o a particle increases its kinetic energ b a actor o 9 and (d) is correct. 7 [SSM] How does the work required to stretch a spring. c ro its unstressed length copare with the work required to stretch it. c ro its unstressed length? Deterine the Concept The work required to stretch or copress a spring a distance is given b k where k is the spring s stiness constant. Because, doubling the distance the spring is stretched will require our ties as uch work. 8 A spring is irst stretched. c ro its unstressed length. It is then stretched an additional. c. How does the work required or the second stretch copare to the work required or the irst stretch (give a ratio o second to irst)? Picture the Proble The work required to stretch or copress a spring a distance is given b k where k is the spring s stiness constant. Letting represent the length o the irst stretch, epress the work (the energ stored in the spring ater the stretch) done on the spring during the irst stretch: The work done in stretching the spring ro to is given b: Epress the ratio o to to obtain: k 4 3 k ( ) ( k ) k 3( k ) ( k ) k k 3 k k 9 The diension o power is (a) M L T, (b) M L /T, (c) M L /T, (d) M L /T 3. Deterine the Concept e can use the deinition o power as the scalar product o orce and velocit to epress the diension o power. Power is deined to be the product o orce and velocit: Epress the diension o orce: P F v M L T

4 53 Chapter 6 Epress the diension o velocit: L T Epress the diension o power in M L L M L ters o those o orce and velocit: 3 T T T ( d ) is correct. Show that the SI units o the orce constant o a spring can be written as kg s. Picture the Proble e can use the relationship o k. F k to establish the SI units Solve F k or k to obtain: F k Substitute the units o F/ and kg sipli to obtain: N s kg s True or alse: (a) The gravitational orce cannot do work on an object, because it is not a contact orce. (b) Static riction can never do work on an object. (c) As a negativel charged electron in an ato is pulled ro a positivel charged nucleus, the electric orce on the electron does work that has a positive value. (d) I a particle is oving along a circular path, the work being done on it is necessaril zero. (a) False. The deinition o work is not liited to displaceents caused b contact orces. Consider the work done b the gravitational orce on an object in reeall. (b) False. The orce that the earth eerts on our oot as ou walk (or run) is a static riction orce. Another eaple is the orce that the earth eerts on the tire ounted on a drive wheel o an autoobile. (c) False. The direction o the electric orce on the electron during its reoval ro the ato is opposite that o the displaceent o the electron. (d) False. I the particle is accelerating, there ust be a net orce acting on it and, hence, work is done on it. A hocke puck has an initial velocit in the + direction on a horizontal sheet o ice. Qualitativel sketch the orce-versus-position graph or the (constant) horizontal orce that would need to act on the puck to bring it to rest. Assue that the puck is located at when the orce begins to act. Show

5 ork and Energ 533 that the sign o the area under the curve agrees with the sign o the change in the puck s kinetic energ and interpret this in ters o the work kinetic energ theore. Deterine the Concept The graph o the orce F acting on the puck as a unction o its position is shown to the right. Note that, because the orce is negative, the area bounded b it and the ais is negative and, hence, the net work done b the orce is negative. In accordance with the work-kinetic energ theore, the change in kinetic energ is negative and the puck loses all o its initial kinetic energ. F 3 [SSM] True or alse: (a) The scalar product cannot have units. (b) I the scalar product o two nonzero vectors is zero, then the are parallel. (c) I the scalar product o two nonzero vectors is equal to the product o their agnitudes, then the two vectors are parallel. (d) As an object slides up an incline, the sign o the scalar product o the orce o gravit on it and its displaceent is negative. (a) False. ork is the scalar product o orce and displaceent. (b) False. Because A B AB cosθ, where θ is the angle between A and B, i the scalar product o the vectors is zero, then θ ust be 9 (or soe odd ultiple o 9 ) and the vectors are perpendicular. (c) True. Because A B AB cosθ, where θ is the angle between A and B, i the scalar product o the vectors is equal to the product o their agnitudes, then θ ust be and the vectors are parallel. (d) True. Because the angle between the gravitational orce and the displaceent o the object is greater than 9, its cosine is negative and, hence, the scalar product is negative. 4 (a) Must the scalar product o two perpendicular unit vectors alwas be zero? I not, give an eaple. (b) An object has a velocit v at soe instant. Interpret v v phsicall. (c) A ball rolls o a horizontal table. hat is the scalar product between its velocit and its acceleration the instant ater it leaves the table? Eplain. (d) In Part (c), what is the sign o the scalar product o its velocit and acceleration the instant beore it ipacts the loor?

6 534 Chapter 6 Deterine the Concept (a) No. An unit vectors that are not perpendicular to each other (as are î, ĵ, and kˆ ) will have a scalar product dierent ro zero. (b) v v v cos v is the object s speed. (c) Zero, because at the instant the ball leaves the horizontal table, v and a( g) are perpendicular. (d) Positive, because the inal velocit o the ball has a downward coponent in the direction o the acceleration. 5 You lit a package verticall upward a distance L in tie Δt. You then lit a second package that has twice the ass o the irst package verticall upward the sae distance while providing the sae power as required or the irst package. How uch tie does liting the second package take (answer in ters o Δt)? Picture the Proble Power is the rate at which work is done. Epress the power ou eert in liting the package one eter in Δt seconds: P Δt Δt Epress the power ou develop in liting a package o twice the ass one eter in t seconds: P Δt Δt Because ou eert the sae power in liting both packages: t t Δt Δt Δ Δ 6 There are lasers that output ore than. G o power. A tpical large odern electric generation plant tpicall produces. G o electrical power Does this ean the laser outputs a huge aount o energ? Eplain. Hint: These high-power lasers are pulsed on and o, so the are not outputting power or ver long tie intervals. Deterine the Concept No, the power a onl last or a short tie interval. 7 [SSM] You are driving a car that accelerates ro rest on a level road without spinning its wheels. Use the center-o-ass work translationalkinetic-energ relation and ree-bod diagras to clearl eplain which orce (or orces) is (are) directl responsible or the gain in translational kinetic energ o

7 ork and Energ 535 both ou and the car. Hint: The relation reers to eternal orces onl, so the car s engine is not the answer. Pick our sste correctl or each case. Deterine the Concept The car shown in the ree-bod diagra is accelerating in the positive direction, as are ou (shown to the right). The net eternal orce (neglecting air resistance) acting on the car (and on ou) is the static riction orce s eerted b the road and acting on the car tires. The positive center o ass work this riction orce does is translated into a gain o kinetic energ. F r g F r n r s F r n, on ou b the seat r r F on F r g, on ou b seat back ou s Estiation and Approiation 8 (a) Estiate the work done on ou b gravit as ou take an elevator ro the ground loor to the top o the Epire State Building, a building stories high. (b) Estiate the aount o work the noral orce o the loor did on ou. Hint: The answer is not zero. (c) Estiate the average power o the orce o gravit. Picture the Proble According to the work-kinetic energ theore, the work done on ou b gravit and the noral orce eerted b the loor equals our change in kinetic energ. (a) The deinition o work is: Assuing a coordinate sste in which the upward direction is the + direction, F and ds are given b: s F ds s F g ˆj and ds d ˆj () Substitute or F and ds h h in equation () to obtain: g ˆ j d ˆ b gravit j g d where h is the height o the Epire State building.

8 536 Chapter 6 Evaluating this integral ields: Assue that our ass is 7 kg and that the height o the Epire State building is 3 ( loors). Then: b gravit b gravit gh ( 7 kg)( 9.8 /s )( 3 ) J J (b) Appl the work-kinetic energ theore to obtain: + b gravit b loor ΔK or, because our change in kinetic energ is zero, + b gravit b loor Solve or b loor to obtain: b loor b gravit Substituting or 5 b gravit ields: b loor (.6 J). 5 J (c) I it takes in to ride the elevator to the top loor, then the average power o the orce o gravit is: P Δ Δt.6 6 s 5 av J 3.4 k Rearks: Soe books sa the noral orce cannot do work. Having solved this proble, what do ou think about that stateent? 9 The nearest stars, apart ro the Sun, are light-ears awa ro Earth. I we are to investigate these stars, our space ships will have to travel at an appreciable raction o the speed o light. (a) You are in charge o estiating the energ required to accelerate a,-kg capsule ro rest to percent o the speed o light in one ear. hat is the iniu aount o energ that is required? Note that at velocities approaching the speed o light, the kinetic energ orula v is not correct. However, it gives a value that is within % o the correct value or speeds up to % o the speed o light. (b) Copare our estiate to the aount o energ that the United States uses in a ear (about 5 J). (c) Estiate the iniu average power required o the propulsion sste. Picture the Proble e can ind the kinetic energ K o the spacecrat ro its deinition and copare its energ to the annual consuption in the United States b eaining the ratio K/E. Finall, we can use the deinition o power to ind the average power required o the engines.

9 ork and Energ 537 (a) The work required b the propulsion sste equals the kinetic energ o the spacecrat: Substitute nuerical values and evaluate : (. c) ΔK K 4 8 ( kg) [(.)(.998 /s) ] J (b) Epress this aount o energ as a percentage o the annual consuption in the United States: (c) The average power is the rate at which work is done (or energ is delivered): K E P Δ Δt av J J 8 ΔK Δt % Substitute nuerical values and evaluate P : av, in P av, in J s The ass o the Space Shuttle orbiter is about 8 4 kg and the period o its orbit is 9 in. Estiate the kinetic energ o the orbiter and the work done on it b gravit between launch and orbit. (Although the orce o gravit decreases with altitude, this aect is sall in low-earth orbit. Use this act to ake the necessar approiation; ou do not need to do an integral.) The orbits are about 5 iles above the surace o Earth. Picture the Proble e can ind the orbital speed o the Shuttle ro the radius o its orbit and its period and its kinetic energ ro K v. e ll ignore the variation in the acceleration due to gravit to estiate the change in the potential energ o the orbiter between its value at the surace o the earth and its orbital value. Appl the work-kinetic energ theore to the orbiter to obtain: The kinetic energ o the orbiter in orbit is given b: Relate the orbital speed o the orbiter to its radius r and period T: b gravit ΔK K orbital K launch or, because K launch, K () b gravit orbital K orbital v () π r v T

10 538 Chapter 6 Substitute or v in equation () and π r π r sipli to obtain: Korbital T T Substitute nuerical values and evaluate K : orbital K orbital π 4 ( 8 kg)( [ i i)(.69 k/i) ] [( 9in)( 6s / in) ] TJ Substitute or K in equation () to obtain: orbital b gravit TJ Ten inches o snow have allen during the night, and ou ust shovel out our 5-t-long drivewa (Figure 6-9). Estiate how uch work ou do on the snow b copleting this task. Make a plausible guess o an value(s) needed (the width o the drivewa, or eaple), and state the basis or each guess. Picture the Proble Let s assue that the width o the drivewa is 8 t. e ll also assue that ou lit each shovel ull o snow to a height o, carr it to the edge o the drivewa, and drop it. e ll ignore the act that ou ust slightl accelerate each shovel ull as ou pick it up and as ou carr it to the edge o the drivewa. hile the densit o snow depends on the etent to which it has been copacted, one liter o reshl allen snow is approiatel equivalent to L o water. Epress the work ou do in liting the snow a distance h: ΔU gh ρ snow V snow gh where ρ is the densit o the snow. Using its deinition, epress the densities o water and snow: snow ρ snow and Vsnow ρ water V water water Divide the irst o these equations b the second to obtain: ρ ρ snow water V V water snow ρ ρ snow water V V water snow Substitute nuerical values and L kg/ kg/ L snow evaluate ρ snow ( ) : Calculate the volue o snow covering the drivewa: Vsnow ( 5t)( 8t) ρ t L 75t 3 t L 3. 3

11 ork and Energ 539 Substitute nuerical values in the epression or to obtain an estiate (a lower bound) or the work ou would do on the snow in reoving it: 3 3 ( kg/ )(. )( 9.8/s )( ) kj Rearks: Note that the work done on the snow is zero when it is carried to the end o the drivewa because F applied is perpendicular to the displaceent. ork, Kinetic Energ and Applications A 5-g piece o space junk has a speed o. k/s. (a) hat is its kinetic energ? (b) hat is its kinetic energ i its speed is halved? (c) hat is its kinetic energ i its speed is doubled? Picture the Proble e can use space junk. v to ind the kinetic energ o the piece o (a) The kinetic energ o the piece o space junk is given b: Substitute nuerical values and evaluate K: K v K (.5kg)(. k/s) kj (b) Because K v : K' 4 K.7 kj (c) Because K v : K' 4 K 43 kj 3 Find the kinetic energ o (a) a.45-kg baseball oving with a speed o 45. /s, and (b) a 6.-kg jogger running at a stead pace o 9. in/i. Picture the Proble e can use and the jogger. v to ind the kinetic energ o the baseball (a) Use the deinition o K to ind the kinetic energ o the baseball: K baseball (.45kg)( 45. /s) 47J (b) Convert the jogger s pace o 9. in/i into a speed: i in 69 v 9. in 6s i.98/s

12 54 Chapter 6 Use the deinition o K to ind the jogger s kinetic energ: K jogger ( 6. kg)(.98/s) 66J 4 A 6.-kg bo is raised a distance o 3. ro rest b a vertical applied orce o 8 N. Find (a) the work done on the bo b the applied orce, (b) the work done on the bo b gravit, and (c) the inal kinetic energ o the bo. Picture the Proble The work done b the orce acting on the bo is the scalar product o the orce producing the displaceent and the displaceent o the bo. Because the weight o an object is the gravitational orce acting on it and this orce acts downward, the work done b gravit is negative. e can use the workkinetic energ theore to ind the inal kinetic energ o the bo. (a) Use the deinition o work to obtain: on the bo F Δ () F and Δ are given b: F ( 8 N)j ˆ and Δ ( 3. )j ˆ Substitute or F and Δ in equation () and evaluate on the bo : (b) Appl the deinition o work again to obtain: on the bo b gravit ( 8 N) ˆj ( 3. ) ˆj.4 kj F Δ () g F and Δ are given b: F gj ˆ ( 6. kg)( 9.8 /s ) ˆ g j ( 58.9 N)j ˆ and Δ ( 3. )j ˆ Substituting or F and sipliing ields: Δ and b gravit ( 58.9 N) ˆj ( 3. ) ˆj.8 kj (c) According to the work-kinetic energ theore: ΔK + on the bo or, because K i, K + on the bo b gravit b gravit

13 ork and Energ 54 Substitute nuerical values and evaluate K : K.4 kj.8 kj.6 kj 5 A constant 8-N orce acts on a 5. kg bo. The bo initiall is oving at /s in the direction o the orce, and 3. s later the bo is oving at 68 /s. Deterine both the work done b this orce and the average power delivered b the orce during the 3.-s interval. Picture the Proble The constant orce o 8 N is the net orce acting on the bo and the work it does is equal to the change in the kinetic energ o the bo. The power o the orce is the rate at which it does work on the bo. Using the work-kinetic energ theore, relate the work done b the constant orce to the change in the kinetic energ o the bo: Substitute nuerical values and evaluate : K K i ( v ) v ( 5.kg) ( 68/s) ( /s).6 kj kj i [ ] The average power delivered b the orce is given b: Substitute nuerical values and evaluate P: Δ P Δt.6 kj P 3. s 3.5 k 6 You run a race with a riend. At irst ou each have the sae kinetic energ, but she is running aster than ou are. hen ou increase our speed b 5 percent, ou are running at the sae speed she is. I our ass is 85 kg, what is her ass? Picture the Proble e can use the deinition o kinetic energ to ind the ass o our riend. Using the deinition o kinetic energ and letting denote our ass and speed and our girlriend s, epress the equalit o our kinetic energies: v v v v

14 54 Chapter 6 Epress the condition on our speed that enables ou to run at the sae speed as our girlriend: v.5v Substitute or v and sipli to obtain: v.5 v.5 Substitute the nuerical value o.5 and evaluate : ( 85kg) 54kg 7 [SSM] A 3.-kg particle oving along the ais has a velocit o +. /s as it passes through the origin. It is subjected to a single orce, F, that varies with position, as shown in Figure 6-3. (a) hat is the kinetic energ o the particle as it passes through the origin? (b) How uch work is done b the orce as the particle oves ro. to 4.? (c) hat is the speed o the particle when it is at 4.? Picture the Proble The pictorial representation shows the particle as it oves along the positive ais. The particle s kinetic energ increases because work is done on it. e can calculate the work done on it ro the graph o F vs. and relate its kinetic energ when it is at 4. to its kinetic energ when it was at the origin and the work done on it b using the work-kinetic energ theore. (a) Calculate the kinetic energ o the particle when it is at : K v 6.J ( 3.kg) (./s) (b) Because the orce and displaceent are parallel, the work done is the area under the curve. Use the orula or the area o a triangle to calculate the area under the F() graph: 4 ( base)( altitude) ( 4.)( 6. N) J

15 ork and Energ 543 (c) Epress the kinetic energ o the particle at 4. in ters o its speed and ass: K v 4 s v4 s 4 s K 4 s () Using the work-kinetic energ theore, relate the work done on the particle to its change in kinetic energ: Solve or the particle s kinetic energ at 4. : K 4. K 4 s K K + 4 s 4 Substitute nuerical values and evaluate K 4 s : K 6.J + J 4 s 8J Substitute nuerical values in equation () and evaluate v 4 : v ( ) 8J 3.kg 4 3.5/s 8 A 3.-kg object oving along the ais has a velocit o.4 /s as it passes through the origin. It is acted on b a single orce, F, that varies with, as shown in Figure 6-3. (a) Find the work done b the orce ro. to.. (b) hat is the kinetic energ o the object at.? (c) hat is the speed o the object at.? (d) hat is the work done on the object ro. to 4.? (e) hat is the speed o the object at 4.? Picture the Proble The work done on an object can be deterined b inding the area bounded b its graph o F as a unction o and the ais. e can ind the kinetic energ and the speed o the particle at an point b using the workkinetic energ theore. (a) Epress, the area under the curve, in ters o the area o one square, A square, and the nuber o squares n: Deterine the work equivalent o one square: Estiate the nuber o squares under the curve between and. : na A square () (.5 N)(.5 ).5 J square n

16 544 Chapter 6 Substitute or n and A square in equation () and evaluate : ( )(.5 J).8 J.75 J (b) Relate the kinetic energ o the object at., K, to its initial kinetic energ, K, and the work that was done on it between and. : K K + Substitute nuerical values and evaluate K : K ( 3.kg)(.4 /s).4 J J +.75J (c) The speed o the object at. is given b: v K Substitute nuerical values and evaluate v : v ( ).4J 3.kg.8/s (d) Estiate the nuber o squares under the curve between and 4. : Substitute nuerical values and evaluate 4 : (e) Relate the kinetic energ o the object at 4., K 4, to its initial kinetic energ, K, and the work that was done on it between and 4. : n 6 (.5J) 3.5J 3.3J 4 6 K K Substitute nuerical values and evaluate K 4 : K 4 ( 3.kg)(.4 /s).9j + 3.5J The speed o the object at 4. is given b: K 4 v4

17 ork and Energ 545 Substitute nuerical values and evaluate v 4 : v ( ).9J 3.kg 4.8/s 9 One end o a light spring (orce constant k) is attached to the ceiling, the other end is attached to an object o ass. The spring initiall is vertical and unstressed. You then ease the object down to an equilibriu position a distance h below its initial position. Net, ou repeat this eperient, but instead o easing the object down, ou release it, with the result that it alls a distance H below the initial position beore oentaril stopping. (a) Show that h g/k. (b) Use the work kinetic-energ theore to show that H h. Tr this eperient on our own. Picture the Proble The ree-bod diagra shows the object at its equilibriu point. e can use Newton s nd law under equilibriu conditions to show that h g/k and the work-kinetic energ theore to prove that H h. (a) Appl F a to the object in its equilibriu position: (b) Appl the work-kinetic energ theore to the spring-object sste to obtain: The work done b the net eternal orce is the su o the work done b gravit and b the spring: Because the orce eerted on the object b the spring is upward and the displaceent o the object is downward, b spring is negative. Substituting or and b gravit ields: b spring F s Fg or, because kh g F r s F r g F s kh and F g g, h g k net, et ΔK or, because the object begins and ends at rest, net, et b gravit +b spring gh kh H h

18 546 Chapter 6 3 A orce F acts on a particle that has a ass o.5 kg. The orce is related to the position o the particle b the orula F C 3, where C.5 i is in eters and F is in newtons. (a) hat are the SI units o C? (b) Find the work done b this orce as the particle oves ro 3. to.5. (c) At 3., the orce points opposite the direction o the particle s velocit (speed is. /s). hat is its speed at.5? Can ou tell its direction o otion at.5 using onl the work-kinetic-energ theore? Eplain. Picture the Proble The work done b this orce as it displaces the particle is the area under the curve o F as a unction o. e can integrate F to ind the work done on the particle and then use the work-kinetic energ theore to ind the particle s speed at 3.. (a) Solve the orce equation or C to F N C where C has units o obtain: 3 3. (b) Because F varies with position non-linearl, we need to epress the work it does as an integral: net, et F d F d where the inus sign is a consequence o the act that the direction o the orce is opposite the displaceent o the particle. Substituting or F ields: net, et.5 3. C 3 d C d Evaluate the integral to obtain: net, et N J 4 [ ].5 4 N 4 [ ].5 (.5) ( 3.) J (c) Appl the work-kinetic-energ theore to obtain: net, et ΔK K.5 K3. Substituting or ields: K.5 and K 3. net, et v.5 v 3. Solve or v.5 to obtain: v.5 et v net, 3. + ()

19 ork and Energ 547 Substitute nuerical values and evaluate v 3. : v ( /s) ( 9.49 J) +.5 kg.5 3 /s No. All we can conclude is that the particle gained kinetic energ. 3 You have a vacation cabin that has a nearb solar (black) water container used to provide a war outdoor shower. For a ew das last suer, our pup went out and ou had to personall haul the water up the 4. ro the pond to the tank. Suppose our bucket has a ass o 5. kg and holds 5. kg o water when it is ull. However, the bucket has a hole in it, and as ou oved it verticall at a constant speed v, water leaked out at a constant rate. B the tie ou reached the top, onl 5. kg o water reained. (a) rite an epression or the ass o the bucket plus water as a unction o the height clibed above the pond surace. (b) Find the work done b ou on the bucket or each 5. kg o water delivered to the tank. Picture the Proble e can epress the ass o the water in our bucket as the dierence between its initial ass and the product o the rate at which it loses water and our position during our clib. Because ou ust do work against gravit in liting and carring the bucket, the work ou do is the integral o the product o the gravitational ield and the ass o the bucket as a unction o its position. (a) Epress the ass o the bucket and the water in it as a unction o its initial ass, the rate at which it is losing water, and our position,, during our clib: Δ Find the rate, r, at which our Δ bucket loses water: Substitute or r to obtain: ( ).kg r Δ.kg r.5kg/ Δ 4. kg ( ).kg.5

20 548 Chapter 6 (b) Integrate the orce ou eert on the bucket, ()g, between the liits o and 4. : kj [ ] ( 9.8/s )(. kg) (.5kg/) kg g. kg.5 d 4. Rearks: e could also ind the work ou did on the bucket, at least approiatel, b plotting a graph o ()g and inding the area under this curve between and A 6.-kg block slides.5 down a rictionless incline that akes an angle o 6º with the horizontal. (a) Draw the ree-bod diagra o the block, and ind the work done b each orce when the block slides.5 (easured along the incline). (b) hat is the work done on the block? (c) hat is the speed o the block ater it has slid.5, i it starts ro rest? (d) hat is its speed ater.5, i it starts with an initial speed o. /s? Picture the Proble e can use the deinition o work as the scalar product o orce and displaceent to ind the work done b each orce acting on the block. hen we have deterined the work done on the block, we can use the workkinetic energ theore the ind the speed o the block at an given location and or an initial kinetic energ it a have. (a) The orces acting on the block are a gravitational orce, eerted b the earth, that acts downward, and a noral orce, eerted b the incline, that is perpendicular to the incline. F r n 6 r r F g g The work done b Substituting or F g ields: F g is given b: F F s F s cosθ g g g where s is the displaceent o the block and θ is the angle between F g and s. F gs cosθ g

21 Substitute nuerical values and evaluate : Fg F g ork and Energ 549 ( 6.kg)( 9.8/s )(.). kj cos3 The work done b b: F n is given Substitute nuerical values and evaluate : Fn (b) The work done on the block is the su o the work done b F and the work done b F : n g F F s F s cosθ n n n where s is the displaceent o the block and θ is the angle between F and s. (.) cos9 F F n n F + tot g Fn n Substitute nuerical values and evaluate tot : tot +. kj. kj (c) Appl the work-kinetic energ theore to the block to obtain: Substituting or and K ields: Fg (ro Part (b)) F g ΔK K K () or, because v i, F g K gs cosθ v v gscosθ i Substitute nuerical values and evaluate v or s.5 : v ( )(.5) 9.8/s 5. /s cos3 (d) I K i, equation () becoes: F g K K i v v i Solving or v and sipliing ields: v gs cosθ + v Substitute nuerical values and evaluate v : i ( )(.5) cos3 + (. /s) 5.4/s v 9.8/s

22 55 Chapter 6 33 [SSM] You are designing a jungle-vine swinging sequence or the latest Tarzan ovie. To deterine his speed at the low point o the swing and to ake sure it does not eceed andator saet liits, ou decide to odel the sste o Tarzan + vine as a pendulu. Assue our odel consists o a particle (Tarzan, ass kg) hanging ro a light string (the vine) o length attached to a support. The angle between the vertical and the string is written as φ. (a) Draw a ree-bod diagra or the object on the end o the string (Tarzan on the vine). (b) An ininitesial distance along the arc (along which the object travels) is dφ. rite an epression or the work d done on the particle as it traverses that distance or an arbitrar angle φ. (c) I the 7., and i the particle starts ro rest at an angle 5, deterine the particle s kinetic energ and speed at the low point o the swing using the work kinetic-energ theore. Picture the Proble Because Tarzan s displaceent is alwas perpendicular to the tension orce eerted b the rope, this orce can do no work on hi. The net work done on Tarzan is the work done b the gravitational orce acting on hi. e can use the deinition o work and the work-kinetic energ theore to ind his kinetic energ and speed at the low point o his swing. (a) The orces acting on Tarzan are a gravitational orce F g (his weight) eerted b the earth and the tension orce T eerted b the vine on which he is swinging. T r r dφ φ (b) The work d done b the gravitational orce F g as Tarzan swings through an arc o length ds is: d φ drφ F r g Fg ds Fg ds cos F dssinφ g ( 9 φ) Note that, because dφ is decreasing as Tarzan swings toward his equilibriu position: ds dφ Substituting or ds and F g ields: d gsinφ dφ

23 ork and Energ 55

24 55 Chapter 6 (c) Appl the work-kinetic energ theore to Tarzan to obtain: ΔK K Ki or, because K i, K Substituting or ields: K gsinφ dφ 5 g [ cosφ] 5 Substitute nuerical values and evaluate the integral to ind K : K ( kg)( 9.8 /s )( 7. )[ cosφ] ( kg)( 9.8 /s )( 7. )( cos5 ).453 kj.5 kj 5 Epress K as a unction o v : K v v K Substitute nuerical values and evaluate v : v (.453 kj) kg 7. /s 34 Siple achines are requentl used or reducing the aount o orce that ust be supplied to peror a task such as liting a heav weight. Such achines include the screw, block-and-tackle sstes, and levers, but the siplest o the siple achines is the inclined plane. In Figure 6-3, ou are raising a heav bo to the height o the truck bed b pushing it up an inclined plane (a rap). (a) The echanical advantage MA o the inclined plane is deined as the ratio o the agnitude o the orce it would take to lit the block straight up (at constant speed) to the agnitude o the orce it would take to push it up the rap (at constant speed). I the plane is rictionless, show that MA /sin θ L/H, where H is the height o the truck bed and L is the length o the rap. (b) Show that the work ou do b oving the block into the truck is the sae whether ou lit it straight up or push it up the rictionless rap.

25 ork and Energ 553 Picture the Proble The ree-bod diagra, with F representing the orce required to ove the block at constant speed, shows the orces acting on the block. e can appl Newton s nd law to the block to relate F to its weight w and then use the deinition o the echanical advantage o an inclined plane. In the second part o the proble we ll use the deinition o work. F r θ F r g F r n (a) Epress the echanical advantage MA o the inclined plane: Fg MA () F Appl F a to the block: F Fg sinθ a or, because a, F F sinθ F sinθ g F g Substitute or F in equation () and sipli to obtain: Reer to the igure to obtain: Fg MA () F sinθ sinθ g H sin θ L Substitute or sinθ in equation () and sipli to obtain: (b) The work ou do pushing the block up the rap is given b: MA H L FL rap L H Because F sinθ and F g g: ( F sinθ ) L glsinθ F g rap g The work ou do liting the block into the truck is given b: liting F H g Because F g g and H Lsinθ : θ rap liting liting glsin 35 Particle a has ass, is initiall located on the positive ais at and is subject to a repulsive orce F ro particle b. The location o particle b is ied at the origin. The orce F is inversel proportional to the square o the distance between the particles. That is, F A/, where A is a positive constant. Particle a is released ro rest and allowed to ove under the inluence o the

26 554 Chapter 6 orce. Find an epression or the work done b the orce on a as a unction o. Find both the kinetic energ and speed o a as approaches ininit. Picture the Proble Because the orce varies inversel with distance, we need to use the integral or o the epression or the work done on the particle. e can then appl the work-kinetic energ theore to the particle to ind its kinetic energ and speed as unctions o. The work done b the orce on the cos particle is given b: F diˆ F ( θ ) Substituting or F and noting that d where θ is the angle between F and d î. θ : ( cos ) d A A d Evaluate this integral to obtain: A A A Appling the work-kinetic energ theore to the particle ields: ΔK K K or, because the particle is released ro rest at, K Substitute or to obtain: K A A As : K and v A A K A 36 You eert a orce o agnitude F on the ree end o the rope. I the load oves up a distance h, (a) through what distance does the orce ove? (b) How uch work is done b the rope on the load? (c) How uch work do ou do on the rope? (d) The echanical advantage (deined in Proble 34) o this sste is the ratio F/F g, where F g is the weight o the load. hat is this echanical advantage?

27 ork and Energ 555 Picture the Proble The object whose weight is w is supported b two portions o the rope resulting in what is known as a echanical advantage o. The work that is done in each instance is the product o the orce doing the work and the displaceent o the object on which it does the work. (a) I w oves through a distance h: F oves a distance h (b) Assuing that the kinetic energ o the weight does not change, relate the work done on the object to the change in its potential energ to obtain: ΔU wh cos θ wh (c) Because the orce ou eert on the rope and its displaceent are in the sae direction: Deterine the tension in the ropes supporting the object: ( h) cos F( h) F θ Fvertical F w F w Substitute or F: F( h) w( h) wh (d) The echanical advantage o the inclined plane is the ratio o the weight that is lited to the orce required to lit it, that is: MA w F w w Rearks: Note that the echanical advantage is also equal to the nuber o ropes supporting the load. Scalar (Dot) Products 37 hat is the angle between the vectors A and B i A B AB? Picture the Proble Because A B AB cosθ, we can solve or cosθ and use the act that A B AB to ind θ. Solve A B AB cosθ, or θ to obtain: θ cos A B AB

28 556 Chapter 6 Substitute or and evaluate θ : AB cos cos ( ) 8 A B θ 38 Two vectors A and B each have agnitudes o 6. and the angle between their directions is 6º. Find A B. Picture the Proble e can use the deinition o the scalar product to evaluate A B. Epress the deinition o the scalar product A B : Substitute nuerical values and evaluate A B : A B A B AB AB cosθ ( 6. )( 6.) cos Find A B or the ollowing vectors: (a) A 3iˆ 6 ˆj, B 4 iˆ + ˆj ; (b) A 5 iˆ + 5 ˆj, B iˆ 4 ˆj ; and (c) A 6 iˆ + 4 ˆj, B 4iˆ 6 ˆj. Picture the Proble The scalar product o two-diensional vectors A and B is A B + A B. (a) For A 3iˆ 6 ˆj and B 4 iˆ + ˆj : A B ( 3)( 4) + ( 6)( ) 4 (b) For A 5 iˆ + 5 ˆj B iˆ 4 ˆj : and A B ( 5)( ) + ( 5)( 4) (c) For A 6 iˆ + 4 ˆj and B 4iˆ 6 ˆj : A B ( 6 )( 4) + ( 4)( 6) 4 Find the angles between the vectors A and B given: (a) A 3iˆ 6 ˆj, B 4 iˆ + ˆj ; (b) A 5 iˆ + 5 ˆj, B iˆ 4 ˆj ; and (c) A 6 iˆ + 4 ˆj, B 4iˆ 6 ˆj. Picture the Proble The scalar product o two-diensional vectors A and B is AB cos θ A B + A B. Hence the angle between vectors A and B is given b A + cos B AB θ AB.

29 ork and Energ 557 (a) For A 3iˆ 6 ˆj and B 4 iˆ + ˆj : θ cos 3 ()( 3 4) + ( 6)() + ( 6) ( 4) + ( ) 43 (b) For A 5 iˆ + 5 ˆj and B iˆ 4 ˆj : θ cos 5 ()() 5 + ()( 5 4) ( 4) 8 (c) For A 6 iˆ + 4 ˆj and B 4iˆ 6 ˆj : θ cos 6 ( 6)( 4) + ( 4)( 6) ( 6) 9 4 A.-kg particle is given a displaceent o Δs ( 3. ) iˆ + ( 3. ) ˆj + (. ) kˆ. During the displaceent, a constant orce F (. N) iˆ (. N) ˆj + (. N)kˆ acts on the particle. (a) Find the work done b F or this displaceent. (b) Find the coponent o F in the direction o this displaceent. Picture the Proble The work done b a orce F during a displaceent Δ s or which it is responsible is given b F Δ s. (a) Using the deinitions o work and the scalar product, calculate the work done b the given orce during the speciied displaceent: F Δs [(. N) iˆ (. N) ˆj + (. N) kˆ ] ( 3.) iˆ + ( 3.) ˆj (.) [(.)( 3.) + (.)( 3.) + (.)(.) ] N.J [ ˆ] k (b) Use the deinition o work that includes the angle between the orce and displaceent vectors to epress the work done b F : ( F cos ) Δs FΔs cos θ θ Solve or the coponent o F in the direction o Δ s : F cos θ Δs

30 558 Chapter 6 Substitute nuerical values and evaluate Fcosθ : F cosθ.j ( 3.) + ( 3.) + (.).N 4 (a) Find the unit vector that is in the sae direction as the A.ˆ i. ˆj.kˆ. (b) Find the coponent o the vector A.ˆ i. ˆj.kˆ in the direction o the vector B 3.ˆ i + 4. ˆj. Picture the Proble A unit vector that is in the sae direction as a given vector and is the given vector divided b its agnitude. The coponent o a vector that is in the direction o another vector is the scalar product o the orer vector and a unit vector that is in the direction o the latter vector. (a) B deinition, the unit vector that is in the sae direction as A is: Substitute or A and A and evaluate ˆ u A A A.ˆ i. ˆj.kˆ u û A : A (.) + (.) + (.) ˆ.ˆ i. ˆj.kˆ 6.8ˆ i.4ˆj.4ˆ k (b) The coponent o A in the direction o B is given b: A A ˆ () u direction o B B The unit vector in the direction B 3.ˆ i + 4. ˆj o B is: B ( 3.) + ( 4.) ˆ u B 3 iˆ ˆj Substitute or B and ûb in equation () and evaluate A : direction o B A direction o B (. ˆ. ˆ ) ˆ ˆ.ˆ i j k i + (.) + (.) + (.)( ) j [SSM] (a) Given two nonzero vectors A and B, show that i A + B A B, then A B. (b) Given a vector A 4ˆ i 3 j ˆ, ind a vector in the

31 ork and Energ 559 plane that is perpendicular to A and has a agnitude o. Is this the onl vector that satisies the speciied requireents? Eplain. Picture the Proble e can use the deinitions o the agnitude o a vector and the dot product to show that i A + B A B, then A B. (a) Epress + : ( ) A B A + B A + B Epress : ( ) A B A B A B A + B Equate these epressions to obtain: ( ) ( ) Epand both sides o the equation to obtain: A B A + A B + B A A B + B Sipli to obtain: 4 A B or A B Fro the deinition o the dot product we have: Because neither A nor B is the zero vector: (b) Let the required vector be B. The condition that A and B be perpendicular is that: Epress B as: A B AB cosθ where θ is the angle between A and B. cos θ θ 9 and A B. A B () B B iˆ + B ˆj () Substituting or A and B in equation () ields: ( 4 iˆ 3 ˆj ) ( B iˆ + B ˆj ) or 4 B 3B (3) Because the agnitude o B is : B B (4) + Solving equation (3) or B and substituting in equation (4) ields: 3 ( B ) B 4 + Solve or B to obtain: B ± 8

32 56 Chapter 6 Substituting or B in equation (3) and solving or B ields: B ±6 Substitute or B and B in equation () to obtain: B ± 6 iˆ ± 8 ˆj No. Because o the plus-and-inus signs in our epression or B, the required vector is not the onl vector that satisies the speciied requireents. 44 Unit vectors A ˆ and B ˆ are in the plane. The ake angles o θ and θ, respectivel, with the + ais, respectivel. (a) Find the and coponents o the two vectors directl. (Your answer should be in ters o the angles.) (b) B considering the scalar product o A ˆ and B ˆ, show that cos (θ θ ) cos θ cos θ + sin θ sin θ. Picture the Proble The diagra shows the unit vectors Aˆ and Bˆ arbitraril located in the st quadrant. e can epress these vectors in ters o the unit vectors î and ĵ and their and coponents. e can then or the scalar product o Âand Bˆ to show that cos(θ θ ) cosθ cosθ + sinθ sinθ. (a) Epress  in ters o the unit vectors î and ĵ : Aˆ A iˆ + A ˆj where A cosθ and A sinθ Proceed as above to obtain: (b) Evaluate Bˆ B iˆ + B ˆj where B cosθ and B sinθ A ˆ B ˆ : Aˆ Bˆ ( cosθ ˆ ˆ i + sinθ j) cosθ iˆ + sinθ ˆj ( ) cosθ cosθ + sinθ sinθ Fro the diagra we note that: A ˆ B ˆ cos( θ θ )

33 ork and Energ 56 Substitute or A ˆ B ˆ to obtain: cos ( θ θ ) cosθ cosθ + sinθ sinθ 45 In Chapter 8, we shall introduce a new vector or a particle, called its linear oentu, sbolized b p. Matheaticall, it is related to the ass and velocit v o the particle b p v. (a) Show that the particle s kinetic p p energ K can be epressed as K. (b) Copute the linear oentu o a particle o ass.5 kg that is oving at a speed o 5 /s at an angle o 5 clockwise ro the + ais in the plane. (c) Copute its kinetic energ using p p both K v and K and veri that the give the sae result. Picture the Proble The scalar product o two vectors is the product o their agnitudes ultiplied b the cosine o the angle between the. (a) e re to show that: p p K () The scalar product o p with itsel is: p p p Because p v: p p v Substitute in equation () and sipli to obtain: (b) The linear oentu o the particle is given b: v K p v v Substitute or and v and sipli to obtain: p [ sin 335 ˆj ] (.5 kg)( 5 /s) cos335 iˆ + ( 5 /s) ( 34 kg /s) iˆ + ( 6 kg /s)j ˆ (c) Evaluating K v ields: K (.5 kg)( 5 /s).8 kj

34 56 Chapter 6 p p Evaluating K ields: K [( 34 /s) iˆ + ( 6 /s) ˆj ] ( 34 /s) iˆ + ( 6 /s) (.5 kg).8 kj [ ˆj ] ( 34 /s) + ( 6 /s) 5. kg 46 (a) Let A be a constant vector in the plane with its tail at the origin. Let r i ˆ + j ˆ be a vector in the plane that satisies the relation A r. Show that the points with coordinates (,) lie on a straight line. (b) I A ˆ i 3ˆ j, ind the slope and intercept o the line. (c) I we now let A and r be vectors in three-diensional space, show that the relation A r speciies a plane. Picture the Proble e can or the dot product o A and r and require that A r to show that the points at the head o all such vectors r lie on a straight line. e can use the equation o this line and the coponents o A to ind the slope and intercept o the line. (a) Let A a iˆ ˆ + a j. Then: A r ( a ˆ + ˆ) ( ˆ + ˆ i a j i j) a + a Solve or to obtain: (b) Given that A iˆ 3 ˆj : (c) The equation A r speciies all vectors r whose coponent r A (the coponent o r parallel to A ) has the constant agnitude A. a + a a which is o the or + b and hence is the equation o a straight line. a a and b 3 a ( Aˆ ) A ˆ A r AA r A r Ar so ra A

35 Let B represent all vectors r such that r A Aˆ + B. Multipling both sides b  shows that B is perpendicular to A : r Aˆ Because ˆ Because B A ˆ : B Aˆ The relation r A Aˆ + B is shown graphicall to the right: ( A Aˆ + B) A A ork and Energ 563 Aˆ Aˆ Aˆ + B Aˆ + B Aˆ r A ra A : A A + B A r r ˆ B A ˆ A (,, z) A ˆ B r Fro the igure we can see that the tip o r is restricted to a plane that is perpendicular to  and contains the point a distance A ro the origin in the direction o Â. The equation A r speciies this plane. 47 [SSM] hen a particle oves in a circle that is centered at the origin and the agnitude o its position vector r is constant. (a) Dierentiate r r r constant with respect to tie to show that v r, and thereore v r. (b) Dierentiate v r with respect to tie and show that a r + v, and thereore ar v / r. (c) Dierentiate v v v with respect to tie to show that a v dv dt, and thereore a dv dt. t Picture the Proble The rules or the dierentiation o vectors are the sae as those or the dierentiation o scalars and scalar ultiplication is coutative. (a) Dierentiate r r r constant: Because v r : d dt ( r r ) v r dr dr r + r v r dt dt d ( constant) dt

36 564 Chapter 6 (b) Dierentiate v v v constant with respect to tie: Because a v : d dt ( v v ) a v dv dv v + v a v dt dt d ( constant) dt The results o (a) and (b) tell us that a is perpendicular to v and parallel (or antiparallel to r. (c) Dierentiate v r with respect to tie: d dt ( v r ) dr dv v + r dt dt d v + r a dt ( ) Because v + r a : r a v () Epress a r in ters o θ, where θ is the angle between r and a : a r a cosθ Epress r a : r a ra cosθ ra r Substitute in equation () to obtain: ra v r Solving or a r ields: ork and Power a r v r 48 Force A does 5. J o work in s. Force B does 3. J o work in 5. s. hich orce delivers greater power, A or B? Eplain. Picture the Proble The power delivered b a orce is deined as the rate at d which the orce does work; i.e., P. dt Calculate the rate at which orce A does work: 5. J P A s.5

37 ork and Energ 565 Calculate the rate at which orce B does work: P B 3. J 5.s.6 Force B delivers greater power because, although it does onl 6% o the work orce A does, it does it in 5% o the tie required b orce A. 49 A single orce o 5. N in the + direction acts on an 8.-kg object. (a) I the object starts ro rest at at tie t, write an epression or the power delivered b this orce as a unction o tie. (b) hat is the power delivered b this orce at tie t 3. s? Picture the Proble e can use Newton s nd law and the deinition o acceleration to epress the velocit o this object as a unction o tie. The power input o the orce accelerating the object is deined to be the rate at which it does work; that is, F v P d dt. (a) The power o the orce as a unction o tie is given b: Epress the velocit o the object as a unction o its acceleration and tie: d P () t F v Fv() t cosθ dt or, because the orce acts in the sae direction as the velocit o the object, P t Fv t () ( ) ( ) v ( t) at Appling ields: F a to the object a F Substitute or a in the epression or v to obtain: Substituting in equation () ields: Substitute nuerical values and evaluate P(t): F v () t t P P F () t t () t ( 5. N) t 8. kg ( 3. /s)t ( 3.5 /s) t

38 566 Chapter 6 (b) Evaluate P(3. s): P ( 3. s) ( 3.5 /s)( 3.s) Find the power delivered b a orce F acting on a particle that oves with a velocit v, where (a) F ( 4. N)ˆ i + ( 3. N) k ˆ and v ( 6. / s)ˆ i ; (b) F ( 6. N)ˆ i ( 5. N)ˆ j and v 5. ( / s)ˆ i + ( 4. / s)ˆ j ; and (c) F ( 3. N)ˆ i + ( 6. N)ˆ j and v (. / s)ˆ i + ( 3. / s)ˆ j. Picture the Proble The power delivered b a orce is deined as the rate at d which the orce does work; i.e., P F v. dt (a) For F (4. N) î + (3. N) kˆ and v (6. /s) î : [( N) iˆ + ( 3. N) kˆ ] ( 6. /s) [ ˆ] 4 P F v 4. i (b) For F (6. N) î (5. N) ĵ and v (5. /s) î +( 4. /s) ĵ : [(. N) iˆ ( 5. N) ˆj ] ( 5. /s) iˆ + ( 4. /s) [ ˆ] 5 P F v 6 j (c) For F (3. N) î + (6. N) ĵ and v (. /s) î + (3. /s) ĵ : [(. N) iˆ + ( 6. N ) ˆj ] (. /s) iˆ + ( 3. /s) [ ˆ] 4 P F v 3 j 5 [SSM] You are in charge o installing a sall ood-service elevator (called a dubwaiter in the ood industr) in a capus caeteria. The elevator is connected b a pulle sste to a otor, as shown in Figure The otor raises and lowers the dubwaiter. The ass o the dubwaiter is 35 kg. In operation, it oves at a speed o.35 /s upward, without accelerating (ecept or a brie initial period, which we can neglect, just ater the otor is turned on). Electric otors tpicall have an eicienc o 78%. I ou purchase a otor with an eicienc o 78%, what iniu power rating should the otor have? Assue that the pulles are rictionless. Picture the Proble Choose a coordinate sste in which upward is the positive direction. e can ind P in ro the given inoration that P out.78pin. e can epress P out as the product o the tension in the cable T and the constant speed v o the dubwaiter. e can appl Newton s nd law to the dubwaiter to epress T in ters o its ass and the gravitational ield g. Epress the relationship between the otor s input and output power: P out. 78Pin in. 8Pout P ()

39 ork and Energ 567 Epress the power required to ove the dubwaiter at a constant speed v: Appl F a to the dubwaiter: Substitute or T in equation () to obtain: P Tv out T g a or, because a, T g P.8Tv. 8gv in Substitute nuerical values and evaluate P in : P in (.8)( 35kg)( 9.8/s ) (.35/s).5k 5 A cannon placed at the edge o a cli o height H ires a cannonball directl upward with an initial speed v. The cannonball rises, alls back down (issing the cannon b a sall argin) and lands at the oot o the cli. Neglecting air resistance, calculate the velocit v as a unction o tie, and show eplicitl that the integral o Fnet v over the tie that the cannonball spends in light is equal to the change in the kinetic energ o the cannonball over the sae tie. Picture the Proble Because, in the absence o air resistance, the acceleration o the cannonball is constant, we can use a constant-acceleration equation to relate its velocit to the tie it has been in light. e can appl Newton s nd law to the cannonball to ind the net orce acting on it and then or the dot product o F and v to epress the rate at which the gravitational ield does work on the cannonball. Integrating this epression over the tie-o-light T o the ball will ield the desired result. Epress the velocit o the cannonball as a unction o tie while it is in the air: Appl F a to the cannonball to epress the orce acting on it while it is in the air: Evaluate F v v t) ˆ i + F ( v gt) ˆj ( v gt)j ˆ ( net g ˆj net : F v g ˆj ( v gt) net gv + g t ˆj

40 568 Chapter 6 Relate Fnet v to the rate at which work is being done on the cannonball: d F v gv g net + t dt Separate the variables and integrate over the tie T that the cannonball is in the air: T ( gv + g t) dt () g T gv T Using a constant-acceleration equation, relate the tie-o-light T to the initial and ipact speeds o the cannonball: v v gt T v v g Substitute or T in equation () and sipli to evaluate : g v vv + v g v v gv g v v ΔK 53 A particle o ass oves ro rest at t under the inluence o a single constant orce F. Show that the power delivered b the orce at an tie t is P F t/. Picture the Proble I the particle is acted on b a single orce, that orce is the net orce acting on the particle and is responsible or its acceleration. The rate at which energ is delivered b the orce is F v P. e can appl the workkinetic energ theore to the particle to show that the work done b the constant orce ater a tie t is equal to the gain in kinetic energ o the particle. Epress the rate at which this orce F v P () does work in ters o F and v : The velocit o the particle, in ters o its acceleration and the tie that the orce has acted is: v at Using Newton s nd law, substitute or a : v F t Substitute or v in equation () and sipli to obtain: F P F t F F t F t

41 ork and Energ A 7.5-kg bo is being lited b eans o a light rope that is threaded through a single, light, rictionless pulle that is attached to the ceiling. (a) I the bo is being lited at a constant speed o. /s, what is the power delivered b the person pulling on the rope? (b) I the bo is lited, at constant acceleration, ro rest on the loor to a height o.5 above the loor in.4 s, what average power is delivered b the person pulling on the rope? Picture the Proble Because, in (a), the bo is being lited at constant speed; the orce acting on it is equal to the orce due to gravit. e re given the speed and the orce and thus calculation o the power is straightorward. In (b) we have a constant acceleration and are given the height to which the bo is lited and the tie during which it is lited. The work done in this case is the orce ties the displaceent and the average power is the ratio o the work done to the tie required. e can ind the orce acting on the bo b appling Newton s nd law and the acceleration o the bo b using a constant-acceleration equation. (a) The power eerted b the person pulling on the rope is given b: P T v Tv cosθ or, because T and v are in the sae direction, P Tv () Appl F a obtain: to the bo to Substituting or T in equation () ields: Substitute nuerical values and evaluate P: (b) The average power eerted b the person pulling the rope is given b: Use a constant-acceleration equation to relate Δ to the interval o acceleration Δt: T F g a or, because F g g and a, T g and T g P gv P P ( 7.5 kg)( 9.8 /s )(. /s).5 k Δ FΔ () Δt Δt av ( t) Δ v Δt + a Δ or because the bo starts ro rest, ( ) Δ Δ a Δt a Δt ( ) Appl F a to the bo to Δ F a obtain: ( Δt)

42 57 Chapter 6 Substitute or F in equation () to Δ obtain: ( Δt) ( Δ) P av Δt ( Δt) 3 Δ Substitute nuerical values and evaluate P av : P ( 7.5 kg)(.5 ) 3 (.4 s) av.46 k Center o Mass ork and Center o Mass Translational Kinetic Energ 55 You have been asked to test drive a car and stud its actual perorance relative to its speciications. This particular car s engine is rated at 64 hp. This value is the peak rating, which eans that it is capable, at ost, o providing energ at the rate o 64 hp to the drive wheels. You deterine that the car s ass (including test equipent and driver on board) is kg. (a) hen cruising at a constant 55. i/h, our onboard engine-onitoring coputer deterines that the engine is producing 3.5 hp. Fro previous coasting eperients, it has been deterined that the coeicient o rolling riction on the car is.5. Assue that the drag orce on the car varies as the square o the car s speed. That is, F d Cv. (a) hat is the value o the constant, C? (b) Considering the peak power, what is the aiu speed (to the nearest i/h) that ou would epect the car could attain? (This proble can be done b hand analticall, but it can be done ore easil and quickl using a graphing calculator or spreadsheet.) Picture the Proble hen the car is oving at a constant speed, the static riction orce eerted b the road on the tires ust be balancing the orce o rolling riction eerted b the roadwa and the drag orce eerted b the air. The power the engine produces is the product o this orce and the velocit o the car. F r d r r F r n F r g F r static riction (a) Appl F a to the car as it cruises at constant velocit: Solve or F static riction and to obtain: and substitute or F d F F static Fd riction static riction Cv r + μ F or, because F n F g g, F Cv + μ g static riction r g

43 ork and Energ 57 Multipling both sides o this equation b the speed v o the car ields an epression or the power P developed b the engine: F static riction v P ( Cv + μ g)v () r Solve or C to obtain: P C 3 v μrg v Epress P 3.5 hp in watts: Epress v 55 i/h in /s: 746 P 3.5 hp.7 k hp i 69 v 55 h i 4.58 /s h 36 s Substitute nuerical values and evaluate C: C ( 4.58 /s) (.5)( kg)( 9.8 /s ).7 k.388 kg/ 3 ( 4.58 /s).38 kg/ (b) Rewriting equation () with v as the variable ields the cubic equation: Epress 64 hp in watts: v 3 μrg + v C P C 746 P 64 hp.3 k hp Substitute nuerical values in the cubic equation and sipli to obtain: or v 3 (.5)( kg)( 9.8 /s ).3 k + v.388 kg/.388 kg/ v ( 47.4 / s ) 3.3 / s + v Use our graphing calculator to solve this cubic equation and then use conversion actors ro our tet to convert our answer to i/h: i.688 v 66. h s.348 s 48 i/h 56 As ou drive our car along a countr road at night, a deer jups out o the woods and stands in the iddle o the road ahead o ou. This occurs just as ou are passing ro a 55-i/h zone to a 5-i/h zone. At the 5-i/h speed-

44 57 Chapter 6 liit sign, ou sla on the car s brakes, causing the to lock up, and skid to a stop inches ro the startled deer. As ou breathe a sigh o relie, ou hear the sound o a police siren. The policean proceeds to write ou a ticket or driving 56 i/h in 5-i/h zone. Because o our preparation in phsics, ou are able to use the 5--long skid arks that our car let behind as evidence that ou were not speeding. hat evidence do ou present? In orulating our answer, ou will need to know the coeicient o kinetic riction between autoobile tires and dr concrete (see Table 5-). Picture the Proble e can appl the work-kinetic energ theore to the car as it slides to a halt to relate the work done b kinetic riction to its initial speed v i. Because the su o the orces acting on the car in the direction is zero, we can use the deinition o the coeicient o kinetic riction to epress the kinetic riction orce k in ters o the gravitational orce F g g. v i? r k F r n r r F g g Appl the work-kinetic energ theore to our car to obtain: The work done on our car is done b the kinetic riction orce that brings our car to a stop: Substituting or K i and in equation () ields: Appl F a obtain: to our car to Substituting or F n in equation () ields: ΔK K Ki or, because K, () K i d b riction k d cosθ or, because k and d are in opposite directions (θ 8 ), d k kd vi or, because μ F, k n k i μ F d v () F F F F g n g k i k n n g μ gd v v μ gd i k k

45 ork and Energ 573 Reerring to Table 5- or the coeicient o kinetic riction or rubber on dr concrete, substitute nuerical values and evaluate v i : v i (.8)( 9.8 /s )( 5 ) 9.8 /s Convert v i to i/h to obtain: i.688 v 9.8 h 44 i/h s.348 s Thus, at the tie ou applied our brakes, which occurred as ou passed the 5 i/h sign, ou were doing 6 i/h under the liit. Your eplanation should convince the judge to void the ticket. General Probles 57 In Februar, a o 6.7 billion k h o electrical energ was generated b nuclear power plants in the United States. At that tie, the population o the United States was about 87 illion people. I the average Aerican has a ass o 6 kg, and i 5% o the entire energ output o all nuclear power plants was diverted to suppling energ or a single giant elevator, estiate the height h to which the entire population o the countr could be lited b the elevator. In our calculations, assue that g is constant over the entire height h. Picture the Proble 5 % o the electrical energ generated is to be diverted to do the work required to change the potential energ o the Aerican people. e can calculate the height to which the can be lited b equating the change in potential energ to the available energ. Epress the change in potential energ o the population o the United States in this process: Letting E represent the energ generated in Februar, relate the change in potential to the energ available to operate the elevator: ΔU Ngh.5E Ngh. 5E h Ng

46 574 Chapter 6 Substitute nuerical values and 9 evaluate h: (.5)( 6.7 k h) h h 6 ( 87 )( 6kg)( 9.8/s ) s 58 One o the ost powerul cranes in the world operates in Switzerland. It can slowl raise a 6-t load to a height o.. (Note that t one tonne is soeties called a etric ton. It is a unit o ass, not orce, and is equal to kg.) (a) How uch work is done b the crane during this lit? (b) I it takes. in to lit this load to this height at constant velocit, and the crane is percent eicient, ind the (gross) power rating o the crane. Picture the Proble e can use the deinition o the work to ind the work done b the crane in liting this load and the deinition o power to ind the power rating o the crane. (a) The work done b the crane in liting the load ro position to position is given b: Because F and ds are in the sae direction: b crane F ds where F is the orce eerted b the crane. ( s s ) Fh b crane Fds F or, because F F g g, gh b crane Substitute nuerical values and evaluate b crane : 6 ( 6. kg)( 9.8/s )(.) 76.3 MJ 76MJ b crane (b) I the eicienc o the crane is η, the power rating o the crane is given b: P Δ η ( Δt) where Δ is the work we calculated in part (a). Substitute nuerical values and evaluate P : (.)( 6.s) P 76.3MJ 59M

47 ork and Energ In Austria, there once was a 5.6-k-long ski lit. It took about 6 in or a gondola to travel up its length. I there were gondolas going up, each with a cargo o ass 55 kg, and i there were ept gondolas going down, and the angle o ascent was 3º, estiate the power P the engine needed to deliver in order to operate the ski lit. Picture the Proble The power P o the engine needed to operate this ski lit is equal to the rate at which it does work on the gondolas. Because as an ept gondolas are descending as are ascending, we do not need to know their ass. Epress the rate at which work is done as the cars are lited: Δ FΔs Pav Δt Δt where F is the orce eerted b the engine on the gondola cars and Δs is the vertical displaceent o the cars. For N gondola cars, each o ass M, NMgΔs Pav F F g NMg : Δt Relate Δs to the angle o ascent θ and the length L o the ski lit: Δs Lsinθ Substituting or Δs ields: P av NMgLsinθ Δt Substitute nuerical values and evaluate P: P ( 55kg)( 9.8/s )( ) ( 6in)( 6s/in) 5.6k sin3 5k 6 To coplete our aster s degree in phsics, our advisor has ou design a sall linear accelerator capable o eitting protons, each with a kinetic energ o. kev. (The ass o a single proton is.67 7 kg.) In addition,. 9 protons per second ust reach the target at the end o the.5--long accelerator. (a) hat the average power ust be delivered to the strea o protons? (b) hat orce (assued constant) ust be applied to each proton? (c) hat speed does each proton attain just beore it strikes the target, assuing the protons start ro rest? Picture the Proble e can use the deinition o average power and the workkinetic energ theore to ind the average power delivered to the target. A second application o the work-kinetic energ theore and the use o the deinition o

48 576 Chapter 6 work will ield the orce eerted on each proton. Finall, we can use the deinition o kinetic energ to ind the speed o each proton just beore it hits the target. (a) The average power delivered b the strea o protons is given b: Appl the work-kinetic energ theore to the proton bea to obtain: Δ Pav () Δt ΔK Substituting ΔK or Δ in equation ΔK Pav () ields: Δt Substitute nuerical values and evaluate P av : P av 9. kev/particle.6 J s/particle ev.6 μ (b) Appl the work-kinetic energ theore to a proton to obtain: The work done b the accelerating orce F on each proton is: Substituting or in equation () ields: ΔK () F d Fd cosθ or, because F and d are in the sae direction, Fd Fd ΔK F ΔK d Substitute nuerical values and evaluate F: F.6. kev ev N.7 N 9 J (c) The kinetic energ o each proton is related to its speed: K v v K

49 Substitute nuerical values and evaluate v : v.39 ork and Energ kev ev 7.67 kg 6 /s 9 J 6 The our strings pass over the bridge o a violin, as shown in Figure The strings ake and angle o 7. with the noral to the plane o the instruent on either side o the bridge. The resulting noral orce pressing the bridge into the violin is. 3 N. The length o the strings ro the bridge to the peg to which each is attached is 3.6 c. (a) Deterine the tension in the strings, assuing the tension is the sae or each string. (b) One o the strings is plucked out a distance o 4., as shown. Make a ree-bod diagra showing all o the orces acting on the segent o the string in contact with the inger (not shown), and deterine the orce pulling the segent back to its equilibriu position. Assue the tension in the string reains constant during the pluck. (c) Deterine the work done on the string in plucking it out that distance. Reeber that the net orce pulling the string back to its equilibriu position is changing as the string is being pulled out, but assue that the agnitudes o the tension orces reain constant. Picture the Proble The ree-bod diagra shows the orces acting on one o the strings at the bridge. The orce whose agnitude is F is one-ourth o the orce (. 3 N) the bridge eerts on the strings. e can appl the condition or equilibriu in the direction to ind the tension in each string. Repeating this procedure at the site o the plucking will ield the restoring orce acting on the string. e can ind the work done on the string as it returns to equilibriu ro the product o the average orce acting on it and its displaceent. (a) Noting that, due to setr T T, appl F to the string at the point o contact with the bridge: F T sin8. T F sin8.

50 578 Chapter 6 Substitute nuerical values and evaluate T: T 4 3 (. N) sin8. 45N 44.5 N (b) A ree-bod diagra showing the orces restoring the string to its equilibriu position just ater it has been plucked is shown to the right: Epress the net orce acting on the string iediatel ater it is released: Use trigonoetr to ind θ: F T cosθ () net θ tan 6.3c c Substitute nuerical values or T and θ in equation () and evaluate F net : F net ( 44.5N) 9.85N 9.9 N cos (c) Epress the work done on the string in displacing it a distance d : I we pull the string out a distance, the agnitude o the orce pulling it down is approiatel: d Fd' ' 4T F L L ( T ) ' Substitute to obtain: 4T d L ' d' Integrate d to obtain: 4T T d ' ' L L where is the inal displaceent o the string. Substitute nuerical values and evaluate : ( 44.5N ) 3 ( 4. ) J

51 ork and Energ The agnitude o the single orce acting on a particle o ass is given b F b, where b is a constant. The particle starts ro rest. Ater it travels a distance L, deterine its (a) kinetic energ and (b) speed. Picture the Proble e can appl the work-kinetic energ theore and the deinition o work to the particle to ind its kinetic energ when it has traveled a distance L. e can then use the deinition o kinetic energ to ind its speed at this location. (a) Appl the work-kinetic energ theore to the particle to obtain: The work done on the particle is also given b: ΔK K Ki or, because the particle is initiall at rest, K v () d F d Fd cosθ or, because F and d are in the sae direction, d K Fd Substituting or F and going ro dierential to integral or ields: K L L b d b d 3 bl 3 (b) Substitute or K in equation () to obtain: 3 3 bl v v bl [SSM] A single horizontal orce in the + direction acts on a cart o ass. The cart starts ro rest at, and the speed o the cart increases with as v C, where C is a constant. (a) Find the orce acting on the cart as a unction o. (b) Find the work done b the orce in oving the cart ro to. Picture the Proble e can use the deinition o work to obtain an epression or the position-dependent orce acting on the cart. The work done on the cart can be calculated ro its change in kinetic energ. (a) Epress the orce acting on the cart in ters o the work done on it: ( ) d F () d Appl the work-kinetic energ theore to the cart to obtain: ΔK K K i or, because the cart starts ro rest, K

52 58 Chapter 6 Substitute or in equation () to dk d F( ) ( v ) obtain: d d d [ ( C) ] C d (b) The work done b this orce changes the kinetic energ o the cart: ΔK v C v v ( ) C 64 A orce F (. N/ ) iˆ is applied to a particle initiall at rest in the plane. Assue it starts ro rest. Find the work done b this orce on the particle, and its inal speed, as it oves along a path that is (a) in a straight line ro point (.,. ) to point (., 7. ), and (b) in a straight line ro point (.,. ) to point (5., 6. ). (The given orce is the onl orce doing work on the particle.) Picture the Proble The work done b F on the particle depends on whether F causes a displaceent o the particle in the direction it acts. e can appl the work-kinetic energ theore to ind the inal speed o the particle in each case. (a) The work done on the particle is given b: Substitute or F and d to obtain: F d ( ) iˆ d ˆ 7.. N/ j. Appl the work-kinetic energ theore to the particle to obtain: ΔK K K i or, because the particle starts ro rest and, K v

53 (b) Calculate the work done b F during the displaceent o the particle ro. to 5. : (. N/ ) ) 5.. (. N/ ) (. N/ ) 78J ork and Energ iˆ d iˆ d 5.. Appl the work-kinetic energ theore to the particle to obtain: ΔK K K i or, because the particle starts ro rest, ( 78 J) 58 J K 78 J v where is the ass o the particle. 65 A particle o ass oves along the ais. Its position varies with tie according to t 3 4t, where is in eters and t is in seconds. Find (a) the velocit and acceleration o the particle as unctions o t, (b) the power delivered to the particle as a unction o t, and (c) the work done b the net orce ro t to t t. Picture the Proble The velocit and acceleration o the particle can be ound b dierentiation. The power delivered to the particle can be epressed as the product o its velocit and the net orce acting on it, and the work done b the orce and can be ound ro the change in kinetic energ this work causes. In the ollowing, i t is in seconds and is in kilogras, then v is in /s, a is in /s, P is in, and is in J. (a) The velocit o the particle is given b: The acceleration o the particle is given b: (b) Epress the rate at which energ is delivered to this particle as it accelerates: 3 ( t 4t ) ( 6t t) d d v 8 dt dt ( 6t 8t ) ( 8) dv d a t dt dt P Fv av

54 58 Chapter 6 Substitute or a and v and sipli to obtain: (c) Appl the work-kinetic energ theore to the particle to obtain: Substitute or K and K and sipli: P ( t 8)( 6t 8t) t( 9t 8t + 8) 8 ΔK K K [( v( t )) ( v( ) ) ] t [( 6t 8t )] ( 3t 4) Rearks: e could also ind b integrating P(t) with respect to tie. 66 A 3.-kg particle starts ro rest at.5 and oves along the ais under the inluence o a single orce F , where F is in newtons and is in eters. (a) Find the work done b the orce as the particle oves ro.5 to 3.. (b) Find the power delivered to the particle as it passes through the point 3.. Picture the Proble e can calculate the work done b the given orce ro its deinition. The power can be deterined ro F v P and v ro the change in kinetic energ o the particle produced b the work done on it. (a) Epress the work done on the particle: Substitute or F and ds and evaluate ˆ ˆ F ds F ds Fi ds i 3. ( ) the deinite integral: J 5.6J d 3..5 (b) Epress the power delivered to the particle as it passes through the point at 3. : Appl the work-kinetic energ theore to ind the work done on the particle as it oves ro the point at.5 to the point at 3. : P F v F 3. v 3. K K K Δ 3..5 K.5, K 3. v 3. or, because ()

55 ork and Energ 583 Solve or v3. to obtain: v 3. Evaluating ields: 3..5 ( ) J d 3..5 Now we can evaluate v 3. : ( 8.695J) v 3.kg.48/s Evaluate F3 to obtain: F ( 3.) 3.( 3.) 9. N Substitute or F3 and v 3. in equation () and evaluate P: P ( 9. N)(.48 /s) 67 The initial kinetic energ iparted to a.-kg bullet is J. (a) Assuing it accelerated down a.--long rile barrel, estiate the average power delivered to it during the iring. (b) Neglecting air resistance, ind the range o this projectile when it is ired at an angle such that the range equals the aiu height attained. Picture the Proble e ll assue that the iring height is negligible and that the bullet lands at the sae elevation ro which it was ired. e can use the equation R ( v g) sin θ to ind the range o the bullet and constantacceleration equations to ind its aiu height. The bullet s initial speed can be deterined ro its initial kinetic energ. (a) The average power delivered to the bullet during iring is given b: Appl the work-kinetic energ theore to the bullet to obtain: Δ Pav () Δt ΔK K Ki or, because the initial speed o the bullet is zero, K K v v

56 584 Chapter 6 Because the acceleration o the bullet is (assued to be) constant, its average speed in the barrel is: The tie-in-the-barrel is given b: v + v i v av Δt v v v av K Substituting or Δt in equation () and sipliing ields: P av K K Substitute nuerical values and J evaluate P av : av (. ) P 8 k ( ) J. kg (b) Using the equal-height range equation, epress the range o the bullet as a unction o its iring speed and angle o iring: Rewrite the range equation using the trigonoetric identit sinθ sinθ cosθ : Using constant-acceleration equations, epress the position coordinates o the projectile along its light path in ters o the paraeter t: Eliinate the paraeter t and ake use o the act that the aiu height occurs when the projectile is at hal the range to obtain: v R g sin θ v sin θ v sinθ cosθ R () g g ( v cosθ )t and ( ) v sin θ t gt h ( v sin ) θ g Equate R and h to obtain: ( v sinθ ) g v sinθ cosθ g Sipliing this equation ields: tan 4 θ θ tan ( 4) 76.

57 ork and Energ 585 Relate the bullet s kinetic energ to its ass and speed and solve or the square o its speed: K v v K Substitute or v in equation () to obtain: Substitute nuerical values and evaluate R: K sin θ R g R ( ) sin[ ( 76. )] J (. kg)( 9.8/s ) 5.74 k 68 The orce F acting on a.5-kg particle is shown as a unction o in Figure (a) Fro the graph, calculate the work done b the orce when the particle oves ro. to the ollowing values o : 4., 3.,.,., +., +., +3., and +4.. (b) I it starts with a velocit o. /s in the + direction, how ar will the particle go in that direction beore stopping? Picture the Proble The work done on the particle is the area under the orceversus-displaceent curve. Note that or negative displaceents, F is positive, so is negative or <. (a) Use either the orulas or the areas o siple geoetric igures or counting squares and ultipling b the work represented b one square (each square is. J) to coplete the ollowing table:, , J (b) The energ that the particle ust lose beore it stops is its kinetic energ at : Substitute nuerical values and evaluate K : K v K (.5 kg)(. /s). J

58 586 Chapter 6 The particle is accelerated (recall that F du d ) between and. and gains an additional. J o kinetic energ. hen it arrives at., however, it will have lost the kinetic energ it gained between and. and its kinetic energ will again be. J. Inspection o the graph leads us to conclude that it will have lost its reaining kinetic energ when it reaches (a) Repeat Proble 68(a) or the orce F shown in Figure (b) I the object starts at the origin oving to the right with a kinetic energ o 5. J, how uch kinetic energ does it have at 4.. Picture the Proble The work done on the particle is the area under the orceversus-displaceent curve. Note that or negative displaceents, F is negative, so is positive or <. In (b), we can appl the work-kinetic energ theore to ind the kinetic energ at 4.. (a) Use either the orulas or the areas o siple geoetric igures or counting squares and ultipling b the work represented b one square (each square is. J) to coplete the ollowing table:, , J (b) Appl the work-kinetic energ theore to the particle to obtain: K K K Δ 4 Solving or K 4 ields: K 4 + K + 5. J Fro to 4, the work done b the orce is:. J 3square square 3. J Substitute or or K to obtain: 4 in the epression K 4 3. J + 5. J 8. J 7 A bo o ass M is at rest at the botto o a rictionless inclined plane (Figure 6-38). The bo is attached to a string that pulls with a constant tension T. (a) Find the work done b the tension T as the bo oves through a distance along the plane. (b) Find the speed o the bo as a unction o. (c) Deterine the power delivered b the tension in the string as a unction o.

59 ork and Energ 587 Picture the Proble The orces acting on the block are shown in the ree-bod diagra. e can use the deinition o work to ind the work done b T in displacing the block a distance up the incline and the work-kinetic energ theore to epress the speed o the bo as a unction o and θ. Finall, we can use the deinition o power to deterine the power produced b the tension in ters o and θ. F r n M F r g θ T r (a) Use the deinition o work to epress the work the tension T does oving the bo a distance up the incline: T d Td cosφ or, because the angle between T and d is zero, T d Evaluate this integral to obtain: T b T (b) Appl the work-kinetic energ theore to the bo to obtain: The work done on the bo is the work done b the net orce acting on it: Reerring to the orce diagra, we see that the net orce acting on the block is: ΔK K Ki or, because the bo is initiall at rest, K Mv v M Fnet d F net d cosφ or, because the angle between F net and d is zero, Fnet T Mg sinθ F net d

60 588 Chapter 6 Substitute or F net and evaluate the integral to obtain: ( T Mg sinθ ) ( T Mg sinθ ) d Substituting or in the epression or v and sipliing ields: v ( T Mg sinθ ) T M M g sinθ (c) The power produced b the tension in the string is given b: Substitute or v to obtain: P T v Tv cosφ or, because the angle between T and v is zero, P Tv T P T g sinθ M 7 [SSM] A orce acting on a particle in the plane at coordinates (, ) is given b F ( F r)( iˆ j ˆ ), where F is a positive constant and r is the distance o the particle ro the origin. (a) Show that the agnitude o this orce is F and that its direction is perpendicular to r iˆ + ˆj. (b) Find the work done b this orce on a particle that oves once around a circle o radius 5. that is centered at the origin. Picture the Proble (a) e can use the deinition o the agnitude o vector to show that the agnitude o F is F and the deinition o the scalar product to show that its direction is perpendicular to r. (b) The work done as the particle oves in a circular path can be ound ro its deinition. (a) Epress the agnitude o F : F F + F F F + r r F + r Because r + : F F r F r + r F

61 ork and Energ 589 For the scalar product o F and r : F F r ( iˆ ˆj ) ( iˆ + ˆj ) r F r ( ) Because F r : F r (b) The work done in an angular displaceent ro θ to θ is given b: rev θ θ θ θ F r F F ds F r θ θ ( iˆ ˆj ) ds ( r sinθ iˆ r cosθ ˆj ) θ ( sinθ iˆ cosθ ˆj ) θ ds ds Epress the radial vector r in ters o its agnitude r and the angle θ it akes with the positive ais: The dierential o r is tangent to the circle (ou can easil convince oursel o this b oring the dot product o r and dr ) and so we can use it as ds in our epression or the work done b F in one revolution: r r cosθ iˆ + r sinθ ˆj dr ds r sinθ dθ iˆ + r cosθ dθ ˆj ( r sinθ iˆ + r cosθ ˆj ) dθ Substitute or ds, sipliing, and integrate to obtain: rev,ccw F π rf ( sinθ iˆ cosθ ˆj ) ( r sinθ iˆ + r cosθ ˆj ) π π ( sin θ + cos θ ) dθ rf dθ rfθ ] πrf dθ π Substituting the nuerical value o r ields: rev, ccw π ( 5. ) F ( π ) F

62 59 Chapter 6 I the rotation is clockwise, the integral becoes: rev,cw rfθ ] π πrf Substituting the nuerical value o r ields: rev, cw ( 5. ) F ( π ) π F 7 A orce acting on a.-kg particle in the plane at coordinates (, ) 3 is given b F ( b r )( iˆ + ˆj ), where b is a positive constant and r is the distance ro the origin. (a) Show that the agnitude o the orce is inversel proportional to r, and that its direction is antiparallel (opposite) to the radius vector r ˆi+ ˆ j. (b) I b 3. N, ind the work done b this orce as the particle oves ro (.,. ), to (5.,. ) along a straight-line path. (c) Find the work done b this orce on a particle oving once around a circle o radius r 7. that is centered at the origin. Picture the Proble e can substitute or r and iˆ + ˆj in F to show that the agnitude o the orce varies as the reciprocal o the square o the distance to the origin, and that its direction is opposite to the radius vector. e can ind the work done b this orce b evaluating the integral o F with respect to ro an initial position., to a inal position 5.,. Finall, we can appl Newton s nd law to the particle to relate its speed to its radius, ass, and the constant b. (a) Substitute or r and obtain: Sipliing ields: iˆ + ˆj in F to b F + rˆ 3 ( ) + where rˆ is a unit vector pointing ro the origin toward the point o application o F. b F b rˆ rˆ + r That is, the agnitude o the orce varies inversel with the square o the distance to the origin and its direction (as indicated b the inus sign) is antiparallel (opposite) to the radius vector r iˆ + ˆj.

63 (b) Find the work done b this orce b evaluating the integral o F with respect to ro an initial position., to a inal position 5., : 5. b d b. 3. N.9 J ork and Energ (c)no work is done as the orce is perpendicular to the velocit. 73 A block o ass on a horizontal rictionless tabletop is attached b a swivel to a spring that is attached to the ceiling (Figure 6-39). The vertical distance between the top o the block and the ceiling is o, and the horizontal position is. hen the block is at, the spring, which has orce constant k, is copletel unstressed. (a) hat is F, the coponent o the orce on the block due to the spring, as a unction o? (b) Show that F is proportional to 3 or suicientl sall values o. (c) I the block is released ro rest at, where <<, what is its speed when it reaches? Picture the Proble e can use trigonoetr to epress the - coponent o F s and the Pthagorean theore to write it as a unction o. Epanding on o the ters in F binoiall reveals its cubic dependence on. In (c), we can appl the workkinetic energ theore and use the deinition o work to ind the speed o the block when it reaches. F r n ϕ F r g F r s (a) F, the -coponent o given b: F s, is Epress F s when the length o the spring is : F F s sinϕ () F k ( ) s Substituting or F s in equation () ields: F ( ) sinϕ k () Reerring to the igure, note that: sin ϕ

64 59 Chapter 6 Substitute or sinϕ to obtain: ( ) F k k (3) Reer to the igure again to see that: + + Substituting or in equation (3) ields: F k (4) + (b) Rewrite the second ter in parentheses in equation (4) b dividing nuerator and denoinator b to obtain: F k + k + k + Epanding + binoiall ields: higher order ters For <<, we can ignore the higher order ters and : + + Substitute or + in the epression or F and sipli to obtain: F k k + 3 (c) Appl the work-kinetic energ theore to the block to obtain: ΔK K Ki or, because the block is released ro rest, K v (5)

65 Integrate F to obtain: k 3 kl d 8 L ork and Energ Substitute or in equation (5): kl 4 L k v v 8 74 Two horses pull a large crate at constant speed across a barn loor b eans o two light steel cables. A large bo o ass 5 kg sits on the crate (Figure 6-4). As the horses pull, the cables are parallel to the horizontal loor. The coeicient o riction between the crate and the barn loor is.5. (a) hat is the work done b each horse i the bo is oved a distance o 5? (b) hat is the tension in each cable i the angle between each cable and the direction the crate oves is 5? Picture the Proble e can appl the work-kinetic energ theore to the crate to relate the work done b the horses and the work done b the riction orce. In part (b), we can use the deinition o work and our result ro Part (a) to ind the tension in the cables. (a) Appl the work-kinetic energ theore to the crate to obtain: The work done b riction is given b: horses + riction ΔK or, because the horses pull the crate at constant speed, ΔK, and + () horses riction riction k d kd cosφ where φ is the angle between k and d. Because k and d are in opposite directions: Because the noral orce acting on the crate equals the weight o the crate: riction riction k d μ F d μ gd k n k Substitute or riction in equation () to obtain: μ gd and μkgd horses k horses Because each horse does hal the work: each horse horses μ gd k

66 594 Chapter 6 Substitute nuerical values and evaluate each horse : (.5)( 5 kg)( 9.8 /s )( 5 ) kj 7.7 kj each horse (b) Epress the work done b each horse as a unction o the tension in each cable: T d Td cosθ each horse Solving or T ields: each horse T d cosθ Substitute nuerical values and T evaluate T: ( 5 ) kj cos5.3 kn

67 ork and Energ 595