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SOL21 - Thevalingam Donald Homework 21 Due Mar 4 2007...

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Thevalingam, Donald – Homework 21 – Due: Mar 4 2007, midnight – Inst: Eslami 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A 36 kg child on a 2 . 3 m long swing is released from rest when the swing supports make an angle of 25 with the vertical. The acceleration of gravity is 9 . 8 m / s 2 . If the speed of the child at the lowest po- sition is 1 . 83818 m / s, what is the mechani- cal energy dissipated by the various resistive forces ( e.g. friction, etc.)? Correct answer: 15 . 2051 J. Explanation: The total energy at the top of the swing is potential, and the total energy at the bottom is kinetic. At the top of the swing, the swing is a cos θ below the support, so it is (1 - cos θ ) above its lowest point. The mechanical energy lost due to friction would thus be Δ E = E top - E bottom = mgh - 1 2 mv 2 = (36 kg)(9 . 8 m / s 2 )(0 . 215492 m) - 1 2 (36 kg)(1 . 83818 m / s) 2 = 15 . 2051 J keywords: 002 (part 1 of 3) 10 points The coefficient of friction between the 6 . 1 kg mass and the table is 0 . 55, and the coefficient of friction between the 5 kg mass and the table is 0 . 3. Consider the motion of the 39 kg mass which descends by the amount of Δ y = 0 . 3 m after releasing the system from rest. The acceleration of gravity is 9 . 8 m / s 2 . 6 . 1 kg μ = 0 . 55 5 kg μ = 0 . 3 39 kg T 12 T 13 a g Find the work done against friction.
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