v T R x m Version PREVIEW Practice 7 carroll (11108) 1

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1 Version PEVIEW Prctice 7 crroll (08) his print-out should he 5 questions. Multiple-choice questions y continue on the next colun or pge find ll choices before nswering. Atwood Mchine points A 6.3 kg ss is ttched to light cord tht psses oer ssless, frictionless pulley. he other end of the cord is ttched to 3.4 kg ss. he ccelertion of grity is 9.8 /s. ω Fro consertion of energy K i + U i K f + U f 0 + g l g l + + ( ) g l ( + ) herefore ( ) ( + ) g l [ 6.3 kg 3.4 kg 6.3 kg kg (9.8 /s )(5.6 ) ] / /s. 6.3 kg kg Use consertion of energy to deterine the finl speed of the first ss fter it hs fllen (strting fro rest) 5.6. Correct nswer: /s. Let : 6.3 kg, 3.4 kg, l 5.6. Consider the free body digrs nd Frictionless Verticl Circle 00 (prt of 3) 0.0 points A block of ss 0.5 kg is pushed ginst horizontl spring of negligible ss, copressing the spring distnce of x s shown in the figure. he spring constnt is 76 N/. When relesed, the block trels long frictionless, horizontl surfce to point B, the botto of erticl circulr trck of rdius 0.9, nd continues to oe up the trck. he speed of the block t the botto of the trck is 5 /s, nd the block experiences n erge frictionl force of 5 N while sliding up the trck. he ccelertion of grity is 9.8 /s. 6.3 kg 3.4 kg g g B x k Let the figure represent the initil configurtion of the pulley syste (before flls down). B Wht is x? Correct nswer:

2 Version PEVIEW Prctice 7 crroll (08) Fro consertion of energy, the initil potentil energy of the spring is equl to the kinetic energy of the block t B. herefore, we write k ( x) B B x k (0.5 kg) (5 /s) (76 N/) Since then E g h, (4.8 J) 0.5 kg (9.8 /s ) (.8 ) 33.7 /s, 33.7 /s.54 /s. 003 (prt of 3) 0.0 points Wht is the speed of the block t the top of the trck? Correct nswer:.54 /s. he chnge in the totl energy of the block s it oes fro B to is equl to the work done by the frictionl force E W f E E B W f. he totl energy t B is E B B (0.5 kg) (5 /s) 56.5 J. he work done by the frictionl force is W f f π (5 N) (π) (0.9 ) 4.37 J. herefore, the totl energy t is E E B + W f 56.5 J + ( 4.37 J) 4.8 J. We cn find now the speed of the block t fro E g h. 004 (prt 3 of 3) 0.0 points Wht is the centripetl ccelertion of the block t the top of the trck? Correct nswer: /s. he centripetl ccelertion t is c (.54 /s) /s. Bed on Loop the Loop points A bed slides without friction round loopthe-loop. he bed is relesed fro height of fro the botto of the loop-theloop which hs rdius he ccelertion of grity is 9.8 /s A 9.999

3 Version PEVIEW Prctice 7 crroll (08) 3 Wht is its speed t point A? Correct nswer:.996 /s. h h Let : nd h Fro consertion of energy, we he herefore K i + U i K f + U f 0 + g h + g ( ) g (h ). g (h ) [ ] (9.8 /s ) (9.999 ).996 /s. h 3.6 Choose the point where the block lees the trck s the origin of the coordinte syste. While on the rp, K b U t x g h x g h x g h ( 9.8 /s ) (.8 ) /s. Block Jup p (prt of 3) 0.0 points A block strts t rest nd slides down frictionless trck. It lees the trck horizontlly, flies through the ir, nd subsequently strikes the ground. 007 (prt of 3) 0.0 points Wht horizontl distnce does the block trel in the ir? Correct nswer: g x Wht is the speed of the bll when it lees the trck? he ccelertion of grity is 9.8 /s. Correct nswer: /s. Let : g 9.8 /s, 44 g, nd h.8. Let : h.8. With the point of lunch s the origin, hus h g t h t g. h x x t x g ( /s) 3.6. (.8 ) 9.8 /s

4 Version PEVIEW Prctice 7 crroll (08) (prt 3 of 3) 0.0 points Wht is the speed of the block when it hits the ground? Correct nswer: /s. ω Let : h.8. Now choose ground leel s the origin. nergy consertion gies us K f U i f g h (7) f g h ( 9.8 /s ) (.8 ) /s. 4.5 kg kg Find the speed of ech object just s they pss ech other. Correct nswer: /s. Let : 4.5 kg,.7 kg, h 4.5. Consider the free body digrs nd Alternte Solution: y g h (9) ( 9.8 /s ) (.8 ) /s, so f x + y (0) ( /s) + ( /s) /s. Serwy CP (prt of 3) 0.0 points wo objects re connected by light string pssing oer light frictionless pulley s shown in the figure. he 4.5 kg object is relesed fro rest t point 4.5 boe the floor. he ccelertion of grity is 9.8 /s..7 kg 4.5 kg g g When they eet both re h fro the floor, nd they he the se speed since they re connected by the string. Applying consertion of energy, (K + U g ) i (K + U g ) f g h + ( ) + g h ( ) + g h g h ( + )

5 Version PEVIEW Prctice 7 crroll (08) 5 + ( + ) g h g h g h ( + ) ( ) g h +. hus ( ) g h + (4.5 kg.7 kg) (9.8 /s ) (4.5 ).7 kg kg /s 00 (prt of 3) 0.0 points b) Wht is the speed of the ech object t the instnt before the 4.5 kg hits the floor? Correct nswer: /s. Gien : y f y f h y f 0 Consertion of echnicl energy gies g h ( + ) f + g h f ( )gh + ( ) g h f + (4.5 kg.7 kg).7 kg kg (9.8 /s ) (4.5 ) /s. 0 (prt 3 of 3) 0.0 points c) How uch higher does the.7 kg object trel fter the 4.5 kg object hits the floor? Correct nswer: he.7 kg ss becoes projectile lunched stright upwrd with iy /s. fy iy g y 0 t the xiu height, so y x iy g ( /s) (9.8 /s ) Sliding Down Doe 0 (prt of 3) 0.0 points A sll box of ss is t the top of sphericl doe with rdius. Strting fro rest fter slight push, the box slides down long the frictionless sphericl surfce (see figure). Ignore the initil kinetic energy (obtined fro the slight push). Apply the principle of consertion of energy to select the correct eqution tht reltes the speed of the box to the polr ngle θ nd the rdius while the box is sliding down on the surfce. θ. g ( cos θ) correct. g 3. g sin θ 4. g 5. g cotθ 6. g ( sin θ) 7. g sin θ

6 Version PEVIEW Prctice 7 crroll (08) 6 8. g ( + tnθ) 9. g cos θ 0. g tn θ Bsic Concepts: Potentil Energy, Kinetic Energy, Centripetl Force θ θ g Since the surfce is frictionless, the echnicl energy of the box is consered, so the initil echnicl energy is the se s the finl echnicl energy g + i g cos θ +. It strts fro rest (the push is slight ) so i 0, nd we find g ( cos θ) g ( cos θ). 03 (prt of 3) 0.0 points Select the inequlity reltion which corresponds to the condition tht the sll box will sty on the surfce.. < g ( + cos θ). > g sin θ 3. < g cos θ correct 4. > g ( + cot θ) 5. < g ( sin θ) 6. < g tnθ 7. > g cotθ 8. < g sin θ 9. < g sin θ 0. sin θ > g ( + cos θ) he rdil ccelertion of the box is while it is on the surfce. So the net rdil force on the box is. he norl force the surfce exerts on the box cn only be outwrd so grity ust supply the totl force inwrd, with ny extr countercted by the norl force, which ust be positie. hus g cos θ N g cos θ. herefore, g cos θ. 04 (prt 3 of 3) 0.0 points Find the criticl ngle t which the box lees the surfce of the doe.. θ θ θ 48. correct 4. θ θ θ θ θ θ θ 38.6

7 Version PEVIEW Prctice 7 crroll (08) 7 he box lees the surfce when the norl force is zero. At this instnt the coponent of the grity force long the norl to the surfce will be equl to the ss ties the Fro the di- centripetl ccelertion. gr, we see tht Fro Prt : g cos θ. () g ( cos θ) At height of 0 eters, 50 J g h g (0 ) g 5 N nd the totl energy is 00 Joules. 0 t the xiu height, so the kinetic energy is 0. he totl energy will not chnge, so t the top, 00 J g h (5 N) h h 0. g ( cos θ). () Putting Eqs. () nd () together, we he g cos θ g ( cos θ) 3 g cos θ g cos θ 3 θ 48.. AP M 993 MC points A bll is thrown upwrd. At height of 0 eters boe the ground, the bll hs potentil energy of 50 Joules (with the potentil energy equl to zero t ground leel) nd is oing upwrd with kinetic energy of 50 Joules. Wht is the xiu height h reched by the bll? Consider ir friction to be negligible.. h 30. h 0 correct 3. h h 0 5. h 40

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