Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 6 - Work and Energy - Problems: 9

Answer

The net work done on the box is 390 J

Work Step by Step

We first find the final speed of the box: $v = a~t = (2.0~m/s^2)(7.0~s)$ $v = 14~m/s$ Then, we find the net work done on the box: $KE_1+Work = KE_2$ $0+Work = \frac{1}{2}mv^2$ $Work = \frac{1}{2}(4.0~kg)(14~m/s)^2$ $Work = 390~J$ The net work done on the box is 390 J
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