SOL9 - Thevalingam Donald Homework 9 Due Dec 2 2006...

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Thevalingam, Donald – Homework 9 – Due: Dec 2 2006, midnight – Inst: Eslami 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. You want to master physics? Solve more problems. You want to solve problems but you cannot? Ask For help aFter you think about the problem. You do not want to learn physics? Tough luck! 001 (part 1 oF 1) 10 points The magnitude oF the vectors ~ F is 28 N, the Force on the right is applied at an angle 55 and the the mass oF the block is 24 kg. 24 kg F 55 F IF the surFace is Frictionless, what is the magnitude oF the resulting acceleration? Correct answer: 1 . 83584 m / s 2 . Explanation: Given : m = 24 kg , F = 28 N , and θ = 55 . Solution Consider the Free body diagram For the situation. 24 kg N mg F Two Forces are accelerating the block to the right: the Force ~ F on the leFt and the cosine component oF the Force ~ F on the right. Thus the net Force is F net = ma = F + F cos θ a = F + F cos θ m = (28 N)(1 + cos55 ) 24 kg = 1 . 83584 m / s 2 . keywords: 002 (part 1 oF 1) 10 points A 626 kg block is pushed on the slope oF a 35 Frictionless inclined plane to give it an initial speed oF 50 cm / s along the slope when the block is 1 . 2 m From the bottom oF the incline. The acceleration oF gravity is 9 . 8 m / s 2 . 626 kg 1 . 2 m 35 What is the speed oF the block at the bot- tom oF the plane? Correct answer: 3 . 70682 m / s. Explanation: Let : m = 626 kg , θ = 35 , and v 0 = 50 cm / s = 0 . 5 m / s . Note: It doesn’t make any di±erence whether the block is initially pushed up or down the slope. Consider the Free body diagram For the block M g sin θ N = cos M g We thereFore have a x = g sin θ . The fnal speed is v 2 = v 2 0 + 2 a x ( x - x 0 )
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Thevalingam, Donald – Homework 9 – Due: Dec 2 2006, midnight – Inst: Eslami 2 = v 2 0 + 2 g sin θ ( x - x 0 ) = (0 . 5 m / s) 2 + 2(9 . 8 m / s 2 )sin35 (1 . 2 m) = 13 . 7405 m 2 / s 2 , so v = q 13 . 7405 m 2 / s 2 = 3 . 70682 m / s . keywords: 003 (part 1 of 1) 10 points A woman pushes a 28 kg lawnmower at a steady speed. She exerts a 88 N force in a direction 20 below the horizontal. The acceleration of gravity is 9 . 8 m / 2 . 28 kg 88 N 20 Find the magnitude of the vertical force this lawnmower exerts on the lawn. Correct answer: 304 . 498 N. Explanation: 28 kg F α mg N The lawnmower moves at a steady speed and hence is in dynamic equilibrium. This implies, among other things, that all the ver- tical forces acting on it must balance: 0 = X F y = N - F sin α - M g where N is the normal (vertical) force exerted by the lawn upon the lawnmower; according to Newton’s Third Law, this is also the verti- cal force exerted by the lawnmower upon the lawn. Solving the above equation for N , we ±nd N = M g + F w sin α = (28 kg)(9 . 8 m / 2 ) + (88 N)sin(20 ) = 304 . 498 N . keywords:
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This note was uploaded on 03/04/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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SOL9 - Thevalingam Donald Homework 9 Due Dec 2 2006...

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