Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 6 - Dynamics I: Motion Along a Line - Exercises and Problems: 49

Answer

$\mu_k = 0.12$

Work Step by Step

Since the sled is moving at a constant speed, the net force on the sled is zero. Therefore, the force of kinetic friction is equal in magnitude to the horizontal component of the applied force. $F_f = (75~N)~cos(30^{\circ})$ $F_N~\mu_k = (75~N)~cos(30^{\circ})$ $[mg-(75~N)~sin(30^{\circ})]~\mu_k = (75~N)~cos(30^{\circ})$ $\mu_k = \frac{(75~N)~cos(30^{\circ})}{mg-(75~N)~sin(30^{\circ})}$ $\mu_k = \frac{(75~N)~cos(30^{\circ})}{(60~kg)(9.80~m/s^2)-(75~N)~sin(30^{\circ})}$ $\mu_k = 0.12$
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