Chapter 5. Answers to Even Numbered Problems m kj. 6. (a) 900 J (b) (a) 31.9 J (b) 0 (c) 0 (d) 31.9 J. 10.

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1 Answers to Even Numbered Problems Chapter m 4..6 J 6. (a) 9 J (b) (a) 3.9 J (b) (c) (d) 3.9 J. 6 m s. (a) 68 J (b) 84 J (c) 5 J (d) 48 J (e) 5.64 m s J 6. (a). J (b) 5. m s (c) 6.3 J 8.. m..5 m. (a).768 m (b) 5.68 J m s m 8. (a) 9.9 m s (b) 7.67 m s 3. (a) v 5.94 m s, v 7.67 m s (b) 47 J B m s m 36. (a) 9.9 m s (b).8 J 38. (a) No, µ ( m sn θ) (b) cos θ, µ ( sn θ) C x m x (c) normal orce, ravtatonal orce, and vertcal component o appled orce (d) 4.3 N, 47.9 J, 6.9 J 43

2 44 CHAPTER m s 4. (a).9 m s (b) 5.6 J h ( 4sn θ + ) m (meased alon the nclne) 48. (a) J (b).9 hp 5..9 m s m 54. (a) 7.9 hp (b) 4.9 hp 56. (a) 7.5 J (b) 5. J (c) 7.5 J (d) 3. J m s 6..6 m 6..4 m s J 66. (a) 58 trps (b) 9.5 (. hp) J 7. (a) 3. J (b).5 J (c) (d).5 J 7. (a) 3.8 J or javeln, (b).9 N on javeln, (c) Yes J (b).6 m s 7.3 J or dscus, 3.6 N on dscus, 8. J or shot 4. N on shot 8. (a) J (b).6 8. (a) 5.8 m (b).77 or 7. m s

3 Enery (a) Choose the sprn constant so the weht o a tray stretches all o sprns a dstance equal to the thcness o a tray. (b) 36 N m, The lenth and wdth o a tray are not needed. 86. (a). m s (b) 6. m s 88. (a) 4. m s (b) 7.9 J (c) 8 N (d) 77 N m m s 49. N and s drected 5.5 (a) The orce o ravty s ven by downwards. The anle between the orce o ravty and the drecton o moton s θ , and so the wor done by ravty s ven as ( θ) cos s 49. N cos6..5 m 6.3 J (b) The normal orce exerted on the bloc by the nclne s rcton orce s n mcos3., so the µ n N cos N Ths orce s drected opposte to the dsplacement (that s θ 8 ), and the wor t does s ( θ) cos s 8.5 N cos8.5 m 46.3 J (c) Snce the normal orce s perpendcular to the dsplacement; θ 9, cos θ, and the wor done by the normal orce s zero.

4 46 CHAPTER (a) Σ snθ + n m y n m sn θ Σ cos θ µ n x cos θ n µ cos θ m sn θ µ s. m n m n 8. q. m (.5 )( 8. )( 9.8 m s ) µ m 79.4 N µ snθ+ cosθ.5 sn. + cos. cos θ s 79.4 N cos.. m.49 J.49 J (b) 3 (c) cos θ 74.6 N 3 ( ) cos θ s 74.6 N cos8. m.49 J.49 J 5.5 (a) The nal netc enery o the bullet s KE (b) e now that. 3 mv 3 m s 9 J KE, and also ( cos θ av ) s. Thus, av KE 9 J scos θ.5 m cos.8 N net road cos s+ resst cos s N cos s+ 95 N cos8 s 5.7 ( θ ) ( θ ) net 3 N 95 N m. J Also, KE KE mv net, so 3 ( ). J v net m. m s

5 Enery hle the motorcycle s n the ar, only the conservatve ravtatonal orce acts on cycle and rder. Thus, mv + my mv + my, whch ves ( 35. m s) ( 33. m s) v v y y y 9.8 m s 6.94 m 5.4 Let m be the mass o the ball, R the radus o the crcle, and the 3. N orce. th KE+ PE KE+ PE yelds y at the bottom o the crcle, nc ( cos ) πr mv + mv + m( R) or Thus, ( πr) v + v + 4R m v π 3. N.6 m ( 5. m s) 4( 9.8 m s )(.6 m ) vn v 6.5 m s 5.7 Snce no non-conservatve orces do wor, we use conservaton o mechancal enery, wth the zero o potental enery selected at the level o the base o the hll. Then, mv+ my mv + my wth y yelds v v 3. m s y 9.8 m s.459 m Note that ths result s ndependent o the mass o the chld and sled.

6 48 CHAPTER Realze that all three masses have dentcal speeds at each pont n the moton and that v. Then, conservaton o mechancal enery ves KE PE PE, or m m m v m y y m y y m y y ( + + 3) ( ) + ( ) + 3( 3 3) m m 9.8 m s Thus, v ( ) + + ( + ) yeldn v 5. m s 5.33 (a) Use conservaton o mechancal enery rom when the projectle s at rest wthn the un untl t reaches maxmum heht. Then, ( KE PE PEs) ( KE PE PEs) becomes + mymax x or max 3 ( ) my. 9.8 m s. m x (. m ) 544 N m (b) Ths tme, we use conservaton o mechancal enery rom when the projectle s at rest wthn the un untl t reaches the equlbrum poston o the sprn. Ths ves ( ) KE PE + PE PE + PE m x + x + s s v x x m 544 N m 3. (. m ) ( 9.8 m s )(. m ) yeldn v 9.7 m s

7 Enery e shall tae PE at the lowest level reached by the dver under the water. The dver alls a total o 5 m, but the non-conservatve orce due to water resstance acts only dn the last 5. m o all. The wor-enery theorem then ves KE+ PE KE+ PE nc av cos8 5. m m s 5 m or ( ) ( + ) + Ths ves the averae resstance orce as 3 av. N. N 5.44 Choose PE at water level and use ( KE PE) ( KE PE) + + or the trp down the cved slde. Ths ves h mv m + + mh, so the 5 speed o the chld as she leaves v 4h 5 the end o the slde s h q h/5 y The vertcal component o ths launch velocty s 4h v y vsnθ snθ 5 At the top o the arc, y vy v y + ay y ves the maxmum heht the chld reaches dn the arborne trp as v. Thus, 4h h θ + ( ) 5 5 sn ymax Ths may be solved or max h 5 y to yeld ymax ( 4sn θ + )

8 5 CHAPTER Choose PE at the level o the base o the hll and let x represent the dstance the ser moves alon the horzontal porton beore comn to rest. The normal orce exerted on the ser by the snow whle on the hll s n mcos.5 and, whle on the horzontal porton, n m. Consder the entre trp, startn rom rest at the top o the hll untl the ser comes to rest on the horzontal porton. The wor done by rcton orces s nc cos8 m cos8 + x ( mcos.5 )( m ) µ µ Applyn nc ( KE PE ) ( KE PE ) m x + + to ths complete trp ves ( m ) ( m) x [ ] m µ cos.5 m µ + + m sn.5 sn.5 µ or x cos.5 ( m ). I µ.75, then x 89 m v v 8. m s 5.53 (a) The acceleraton o the car s.5 m s a. Thus, the t. s constant orward orce due to the enne s ound rom Σ enne ar ma as enne ar + ma N.5.5 m s.65 N The averae velocty o the car dn ths nterval s v averae power nput rom the enne dn ths tme s P av v+ v 3 hp ennevav (.65 N )( 9. m s) 3. hp 746 av 9. m s, so the (b) At t. s, the nstantaneous velocty o the car s v 8. m s and the nstantaneous power nput rom the enne s 3 hp P ennev (.65 N )( 8. m s) 63.9 hp 746

9 Enery The wor done on the partcle by the orce as the partcle moves rom x x to x x s the area under the cve rom x to x. (a) or x to x 8. m, area otranle ABC AC alttude x (N) B 6 4 A C E D x (m) (b) or 8 ( 8. m )( 6. N ) 4. J x 8. m to x. m, 8 area otranle CDE CE alttude (. m )( 3. N ) 3. J (c) J+ 3. J. J 5.6 (a) rom v v a y ( y) +, we nd the speed just beore touchn the round as v m s. m 4.4 m s (b) Choose PE at the level where the eet come to rest. Then KE+ PE KE+ PE becomes nc s + mv + ms ( av cos8 ) ( ) or ( 75 )( 4.4 m s) ( ) mv + m + s 5. m av m s.5 N

10 5 CHAPTER The normal orce the nclne exerts on bloc A s n ( m )cos37, and the rcton orce s µ n µ m cos37. The vertcal dstance bloc A rses s A A m sn37 m, whle the vertcal dsplacement o bloc B s m. y A e nd the common nal speed o the two blocs by use o KE+ PE KE+ PE KE+ PE nc Ths ves ( µ ) m Acos37 s m A + m B v + m A( ya) + m B( yb) or A A y B v ( µ ) m B yb m A ya m A cos37 s m + m A B 9.8 m s m 5 m.5 5 m cos37 5 whch yelds v 57 m s The chane n the netc enery o bloc A s then 3 KEA m Av ( 5 )( 57 m s ) 3.9 J 3.9 J 5.7 rst, determne the mantude o the appled orce by consdern a ree-body daram o the bloc. Snce the bloc moves wth constant velocty, Σ Σ rom Σ x, we see that n cos3 x y 3 n 5 m Thus, µ n µ cos3 and Σ y becomes sn 3 m+ µ cos3, or ( 5. )( 9.8 m s ) m sn 3 µ cos3 sn3.3 cos3. N

11 Enery 53 (a) The appled orce maes a 6 anle wth the dsplacement up the wall. Thereore, cos6. N cos6 3. m 3. J s mcos8 s 49 N. 3. m.5 J (b) ncos9 s (c) n PE m y 49 N 3. m.5 J (d) 5.73 The potental enery assocated wth the wnd s PEw x, where x s meased horzontally rom drectly below the pvot o the swn and postve when movn nto the wnd, neatve when movn wth the wnd. e choose PE at the level o the pvot as shown n the e. Also, note that D Lsnφ + Lsn θ so D φ sn sn θ L, or L cos wnd drecton q L x L sn L snq v L cosq 5. m φ sn sn m D (a) Use conservaton o mechancal enery, ncludn the potental enery assocated wth the wnd. The nal netc enery s zero Jane barely maes t to the other sde, and ( KE PE PEw ) ( KE PE PEw ) becomes + m Lcos + + Lsn mv + m Lcos + Lsn ( φ) ( φ) ( θ) ( θ) L m or v L( cosθ cosφ) + ( snθ+ sn φ) where m s the mass o Jane alone. Ths yelds v 6.5 m s

12 54 CHAPTER 5 (b) Aan, usn conservaton o mechancal enery wth KE ( KE PE PEw ) ( KE PE PEw ) ves + M Lcos + Lsn Mv + M Lcos + + Lsn where M 3 s the combned mass o Tarzan and Jane. Thus, ( θ) ( θ) ( φ) ( φ) L v L( cosφ cosθ) ( snθ+ sn φ) whch ves v 9.87 m s M 5.76 hen the bloc moves dstance x down the nclne, the wor done by the rcton orce s cos8 x µ nx µ mcos θ x. rom the wor-enery theorem, KE+ PE + PE KE+ PE + PE, we nd nc s s ( cos ) nc µ m θ x KE+ PE + PEs. Snce the bloc s at rest at both the start and the end, ths ves µ ( 9.6 N cos37. )(. m ) + ( 9.6 N )(. m sn37. ) + ( N m )(. m ) or µ Careul examnaton o e P5.83 reveals that, dn some tme nterval, bloc B moves upward. cm, bloc A wll move downward. cm and the dstance separatn the two blocs ncreases by 3. cm. Generalzn, we conclude that when the vertcal separaton between the blocs ncreases by h, bloc B moves upward dstance h 3 and bloc A moves downward dstance h 3. Also, at any nstant dn the moton bloc A has speed v, the speed o bloc B wll be v. Choosn y at the level where both blocs start rom rest and man use o the above observatons, conservaton o mechancal enery ves ( A) + ( B) + ( A) + ( B) ( A) + ( B) + ( A) + ( B) m v m v m y m y m v m v m y m y or mv m( v ) m( h ) m( h ) Ths reduces to 5 h 8 3 v, or h 8 8h v 3 5 5