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# HW8 - schneiter(tcs639 HW08 Ross(89152 This print-out...

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schneiter (tcs639) – HW08 – Ross – (89152) 1 This print-out should have 6 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A block accelerates 3 . 5 m / s 2 down a plane inclined at angle 26.0 . The acceleration of gravity is 9 . 81 m / s 2 . m μ k 3 . 5 m / s 2 26 Find μ k between the block and the inclined plane. Correct answer: 0 . 0907798. Explanation: Given : a = 3 . 5 m / s 2 , θ = 26 , and g = 9 . 81 m / s 2 . Consider the free body diagram for the block m g sin θ N = m g cos θ μ N a m g Basic Concepts: vector F net = mvectora Parallel to the ramp: F x,net = ma x = F g,x - F k F g,x = mg sin θ F k = μ k F n Perpendicular to the ramp: F y,net = F n - F g,y = 0 F g,y = mg cos θ Solution: Consider the forces parallel to the ramp: F k = F g,x - ma x = mg sin θ - ma x Consider the forces perpendicular to the ramp: F n = F g,y = mg cos θ Thus the coefficient of friction is μ k = F k F n = m g sin θ - m a m g cos θ = g sin θ - a g cos θ = 9 . 81 m / s 2 sin 26 - 3 . 5 m / s 2 9 . 81 m / s 2 cos 26 = 0 . 0907798 . 002 10.0 points The magnitude of each force is 283 N, the force on the right is applied at an angle 35 and the mass of the block is 23 kg. The coefficient of friction is 0 . 148. The acceleration of gravity is 9 . 8 m / s 2 . 23 kg μ = 0 . 148 283 N 35 283 N What is the magnitude of the resulting ac- celeration?

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HW8 - schneiter(tcs639 HW08 Ross(89152 This print-out...

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