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 - Page 303: 17


$F_{max} = 8.0~N$

Work Step by Step

We can use the work-energy theorem to find the work done on the particle from x = 0 to x = 2 m. $KE_0+W = KE_f$ $W = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_0^2$ $W = \frac{1}{2}(0.50~kg)(6.0~m/s)^2 - \frac{1}{2}(0.50~kg)(2.0~m/s)^2$ $W = 8.0~J$ The work done on the particle is equal to the area under the force versus position graph. We can find the value of $F_{max}$. $\frac{1}{2}F_{max}~x = W$ $F_{max} = \frac{2W}{x}$ $F_{max} = \frac{(2)(8.0~J)}{2~m}$ $F_{max} = 8.0~N$
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