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qz2sol_3706Hs11 - nd law applied to the vertical direction...

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TA: Tomoyuki Nakayama Tuesday, February 8, 2011 PHY 2048: Physic 1, Discussion Section 3706H Quiz 2 (Homework Set #4) Name: UFID: Formula sheets are not allowed. Do not store equations in your calculator. You have to solve problems on your own; memorizing final algebraic expressions from homework assignments and just plugging numbers into them will not give you full credit. Leave all your work. ________________________________________________________________________________ A 20.0-kg block is initially at rest on a horizontal surface. A horizontal force F of magnitude 50.0 N and a vertical force P are then applied to the block. The coefficient of friction for the block and surface are μ s = 0.400 and μ k = 0.200. a) Determine the magnitude of the frictional force acting on the block if the magnitude of P is 60.0 N. Newton’s 2
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Unformatted text preview: nd law applied to the vertical direction yields 0 = P + N – mg ⇒ N = mg – P = 136 N The maximum static friction is f s,max = μ s N = 54.4 N The magnitude of horizontal force F is smaller than the maximum static friction. Therefore, the static friction keeps the block from sliding. Newton’s 2 nd law applied to the horizontal direction yields the friction: 0 = F – f f = 50.0 N b) Determine the magnitude of the frictional force acting on the block if the magnitude of P is 80.0 N. The maximum static friction is f s,max = μ s N = μ s (mg – P)= 46.4 N This is smaller than the applied force. Hence the static friction cannot hold the block and the block starts to slide. The friction exerted on the block is kinetic friction. The magnitude of the friction is f = μ k N = μ k (mg – P)= 23.2 N...
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