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 11 - Work - Exercises and Problems - Page 305: 48


(a) $\mu_k = \frac{v_0^2}{2~g~d}$ (b) $\mu_k = 0.037$

Work Step by Step

(a) The work done by friction is equal in magnitude to the initial kinetic energy of the box. We can find an expression for the coefficient of kinetic friction; $W_f = KE$ $F_f~d = \frac{1}{2}mv_0^2$ $mg~\mu_k~d = \frac{1}{2}mv_0^2$ $\mu_k = \frac{v_0^2}{2~g~d}$ (b) We can use the expression from part (a) to solve this question. $\mu_k = \frac{v_0^2}{2~g~d}$ $\mu_k = \frac{(1.2~m/s)^2}{(2)(9.80~m/s^2)(2.0~m)}$ $\mu_k = 0.037$
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