2006MechSolutions

# 2006MechSolutions - 2006 AP Physics Mechanics Exam FN1 1 a...

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2006 AP Physics Mechanics Exam 1) a) b) block: gt v at v v μ - = - = 0 0 slab: t M M gM at v v S B B + - = - = μ 0 0 t M M gM gt v v S B B + = - = μ μ 0 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 s t t t t t 79 . 1 28 . 0 96 . 1 4 5 . 0 0 . 3 8 . 9 5 . 0 2 . 0 8 . 9 2 . 0 4 = = - + = - ( 29 ( 29 s m gt v at v v 50 . 0 79 . 1 96 . 1 4 0 0 = - = - = - = μ c) ( 29 ( 29 m at t v x 45 . 0 79 . 1 28 . 0 2 1 0 2 1 2 2 0 = + = + = d) ( 29 ( 29 J mv mv K W 375 . 0 0 5 . 0 3 2 1 2 1 2 1 2 2 0 2 = - = - = = 2) a) Graph force F vs. square of compression distance x 2 . b) x (m) F (N) x 2 ( m 2 ) 0.05 4 0.25 x 10 -2 0.10 17 1.00 x 10 -2 0.15 28 2.25 x 10 -2 0.20 68 4.00 x 10 -2 0.25 106 6.25 x 10 -2 d) Slope of graph ≈ A ≈ 1700 N/m 2 e) J dx x dx F W 57 . 0 1700 1 . 0 0 2 = = = f) 2 0 2 0 2 0 2 2 1 2 1 0 2 1 2 1 mv mv mv mv K W = - = - = = ( 29 s m v m W v 5 . 1 27 . 2 5 . 0 57 . 0 2 2 2 = = = = F N1 μ M B g M B g F N2 μ M B g M S g

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3) a) ma F mra mr I r F ma F mg F f f f = = = = = - α α τ θ 2 : sin : θ θ θ θ sin 2 1 2 sin sin sin g a ma mg ma ma mg F mg f = = = - = - b) ( 29 θ θ θ sin sin sin 2 1 2 2 2 2 0 2 gL v gL L g v ax v v = = = = - c) drop time: g H t gt H gt t v y y 2 2 1 2 1 2 2 0 = = = + = Horizontal displacement: ( 29 θ θ sin 2 sin 2 1 0 2 0 gL g H gL t t v at t v x x x = = = + = θ sin 2 HL x = c) ma F mra mr I r F ma F mg F f f f 2 1 2 1 : sin : 2 = = = = = - α α τ θ θ θ θ θ sin 3 2 2 3 sin
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