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: 30

Answer

(a) $W = 206~J$ (b) $P = 68.7~watts$

Work Step by Step

(a) Since the block moves at a steady speed, the force applied by the person must be equal in magnitude to the force of kinetic friction. We can find the applied force $F$. We can assume that the coefficient of kinetic friction is 0.70. $F = F_f$ $F = mg~\mu_k$ $F = (10~kg)(9.80~m/s^2)(0.70)$ $F = 68.6~N$ We can find the work done by the person's force. Note that the block moves a distance of 3.0 meters in 3.0 seconds. $W = F~d$ $W = (68.6~N)(3.0~m)$ $W = 206~J$ (b) We can find the person's power output. $P = \frac{Work}{time}$ $P = \frac{206~J}{3.0~s}$ $P = 68.7~watts$
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