Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 9 - Work and Kinetic Energy - Exercises and Problems: 44

Answer

(a) $v = \sqrt{\frac{2~F~h~cot(\theta) - 2mgh}{m}}$ (b) v = 4.0 m/s

Work Step by Step

(a) We can find an expression for the distance $d$ that the crate moves along the slope; $\frac{h}{d} = sin(\theta)$ $d = \frac{h}{sin(\theta)}$ The work done by the force will be equal to the sum of the potential energy and the kinetic energy. Therefore; $PE+KE = W$ $mgh + \frac{1}{2}mv^2 = F~d~cos(\theta)$ $\frac{1}{2}mv^2 = F~(\frac{h}{sin(\theta)})~cos(\theta) - mgh$ $v^2 = \frac{2~F~h~cot(\theta) - 2mgh}{m}$ $v = \sqrt{\frac{2~F~h~cot(\theta) - 2mgh}{m}}$ (b) We can use the expression in part (a) to find the speed at the top of the slope: $v = \sqrt{\frac{2~F~h~cot(\theta) - 2mgh}{m}}$ $v = \sqrt{\frac{(2)(25~N)(2.0~m)~cot(20^{\circ}) - (2)(5.0~kg)(9.80~m/s^2)(2.0~m)}{5.0~kg}}$ $v = 4.0~m/s$
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