## Physics: Principles with Applications (7th Edition)

Published by Pearson

# 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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