act160.key - L 2 = 1 12(3.75(4 2 = 5.0 kg m 2 I ball = M...

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Physics I – Exam 2 Answer Key Part A – 1: A, 2: A, 3: D, 4: C, 5: C, 6: H, 7: F, 8: F, 9: H, 10: F Part B B-1 Potential energy at the top = potential energy at the bottom: m g h = ½ k (-h) 2 k = 2 m g / h = 98. N/m B-2 Straight line for force as shown. Parabolas for energy as shown. Max/min energy at 5 cm Features Force: straight line slope = -98 N/m F = 0 at y = 5 cm Total PE: parabola 0.49 J at y = 0 or y = 10. min value at 5 cm = 0.3675 J same value at y = 10 and y = 0. KE: inverted shape of PE curve 0 J at y = 0 or y = 10. max value at 5 cm = PE max – PE min same value at y = 10 and y = 0. 10 5 Total PE (J) 0.1225 y (cm) +4.9 10 5 0 10 y (cm) y (cm) KE (J) Net Force (N) 0 5 -4.9 0 0 0.49 0.3675
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Part C C-1 : No There is an external torque from gravity on the system, or can just say angular momentum starts at zero, ends at not zero, and so it is not conserved. C-2: Yes Only the force of gravity does work on the system. Gravity is a conservative force. C-3: I stick = 1 / 12 M stick
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Unformatted text preview: L 2 = ( 1 / 12 ) (3.75) (4) 2 = 5.0 kg m 2 . I ball = M ball r 2 = (5.0) (2) 2 = 20.0 kg m 2 . I system = I stick + I ball = 25.0 kg m 2 . The stick is balanced and so the only force that counts is gravity on the ball. C-4: Torque and angular acceleration are into page (–Z) by r.h.r. or cross product. τ = r × F = (2) (M ball g) sin 90° = (2) (9.8) (5) = 98 N m. τ = I α → α = τ / I = (98) / (25) = 3.92 rad/s. C-5: Torque is zero because r and F are in the same direction. α = 0 because τ = 0. (no units needed for 0) C-6: Ball goes down 2 m while stick center of mass stays at the same height. ∆ U = M ball g ∆ h = (5) (9.8) (–2) = –98 J. C-7: ∆Κ = – ∆ U = +98 J. C-8: K final = 1 / 2 I system ϖ 2 = 98 → ϖ 2 = (2) (98) / 25 = 7.84 → ϖ = 2.8 rad/s. L final = I ϖ = (25) (2.8) = 70 kg m 2 /s. direction is into the page (–Z) by r.h.r....
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act160.key - L 2 = 1 12(3.75(4 2 = 5.0 kg m 2 I ball = M...

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