University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 6 - Work and Kinetic Energy - Problems - Exercises - Page 195: 6.11

Answer

(a) W = 374 J (b) W = -333 J (c) W = 0 (d) The net work is 41 J (e) W = 352 J

Work Step by Step

(a) $W = F~d = (72.0~N)(5.20~m)$ $W = 374~J$ (b) $W = mg~d~cos(\theta)$ $W = (128.0~N)(5.20~m)~cos(120.0^{\circ})$ $W = -333~J$ (c) The normal force does zero work because it makes an angle of $90^{\circ}$ with the direction of motion. (d) The net work done on the carton is $374~J-333~J$, which is $41~J$. (e) $W = F~d~cos(\theta)$ $W = (72.0~N)(5.20~m)~cos(20.0^{\circ})$ $W = 352~J$
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