Online Homework Answers 6 (Oct. 19, 2006)

Online Homework Answers 6 (Oct. 19, 2006) - Hood Charles...

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Hood, Charles – Homework 6 – Due: Oct 19 2006, 11:00 pm – Inst: Jan Opyrchal 1 This print-out should have 9 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 block is pushed upward along a frictionless inclined plane. m F θ Determine the horizontal force F on the block which causes it to move upward with a constant velocity. 1. F = m g sin θ 2. F = m g cos θ 3. F = m g cos θ 4. F = m g tan θ correct 5. F = m g cot θ 6. F = m g sin θ Explanation: The block is subject to three forces: its weight ~ W , the normal force ~ N of the surface, and the external force ~ F which pushes it to the right. The external force acts in the hor- izontal direction, the weight acts vertically, and the normal force acts in the direction nor- mal (perpendicular) to the surface — leftward from vertically up by angle θ . Let the x axis point horizontally to the left while the y axix points vertically up. In these coordinates, the net force on the block has components F net x = F - N sin θ, F net y = N cos θ - W. Since the block does not accelerate, both com- ponents must vanish, hence N = W cos θ = mg cos θ and F = N sin θ = mg sin θ cos θ = mg tan θ. keywords: 002 (part 1 of 1) 10 points The system shown below is released from rest and moves 18 . 3 cm in 1 . 80351 s. The acceleration of gravity is 9 . 8 m / s 2 . 3 kg M What is the value of the mass M ? Assume all surfaces are frictionless. Correct answer: 0 . 0348461 kg. Explanation: Given : m = 3 kg , y 0 = 0 m , y = 18 . 3 cm = 0 . 183 m , t 0 = 0 s , and t = 1 . 80351 s . m M a T N m g a T M g From kinematics, we have y - y 0 = v 0 + 1 2 a t 2 = 1 2 a t 2 a = 2 y t 2
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Hood, Charles – Homework 6 – Due: Oct 19 2006, 11:00 pm – Inst: Jan Opyrchal 2 The only external force acting on the sys- tem is the weight M g suspended from the
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