# HW7 solutions - HomeWork 7 solutions Physics 203 1 Chapter 6 problem 3 In part(b the question is asking about the work done by the weight of the

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HomeWork 7 solutions Physics 203 1. Chapter 6 problem 3. In part (b) the question is asking about the work done by the weight of the elevator. A better way to say this is “the work done by the gravitational force exerted by the earth”. Also, the problem is not saying it specifically, but the system of interest is the elevator (the Earth is not in the system). REASONING The work done by the tension in the cable is given by Equation 6.1 as ( ) cos WT s θ = . Since the elevator is moving upward at a constant velocity, it is in equilibrium, and the magnitude T of the tension must be equal to the magnitude mg of the elevator’s weight; T = mg . SOLUTION a. The tension and the displacement vectors point in the same direction (upward), so the angle between them is = 0 ° . The work done by the tension is ( ) ( ) ( )( ) ( ) 25 cos cos 1200 kg 9.80 m/s cos 0° 35 m 4.1 10 J W T s mg s θθ = = = = × (6.1) b. The weight and the displacement vectors point in opposite directions, so the angle between them is = 180 ° . The work done by the weight is ( ) ( )( ) ( ) cos 1200 kg 9.80 m/s cos180° 35 m 4.1 10 J W mg s = = = −× (6.1) 2. Problem 24. Before you do this problem, draw an energy bar chart for the situation. I’ll choose the system to be just the skier. The initial moment will be when she is at the beginning of the 21 m, and the final moment will be when she is at the end of the 21 m. Here’s the energy bar chart. The skier starts with kinetic energy, but the friction force exerted by the snow on her

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## This note was uploaded on 06/22/2009 for the course PHYS 1101 taught by Professor Richardson, b during the Fall '08 term at Cornell University (Engineering School).

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HW7 solutions - HomeWork 7 solutions Physics 203 1 Chapter 6 problem 3 In part(b the question is asking about the work done by the weight of the

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