College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 6 - Problems - Page 230: 60


(a) $Work = 1.5~J$ (b) $Work = 1.125~J$

Work Step by Step

(a) The area under the force versus displacement graph is equal to the work done by the force: $Work = \frac{1}{2}F_{max}~x$ $Work = \frac{1}{2}(15~N)(0.20~m)$ $Work = 1.5~J$ (b) On the graph, we can see that the area under the graph between $x= 0.10~m$ and $x = 0.20~m$ is 75% of the total area under the graph between $x = 0$ and $x = 0.20~m$: $Work = (0.75)(1.5~J) = 1.125~J$
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