Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 7 - Exercises and Problems - Page 128: 61

Answer

$v=\sqrt{\frac{4ax^{1.5}}{3m}}$

Work Step by Step

We know that the potential energy is the integral of the force with respect to distance, so we find: $ U = \int (a\sqrt x)dx=\frac{2}{3}ax^{1.5}$ We know that the kinetic energy is equal to the change in potential energy. Thus: $\frac{1}{2}mv^2 = \frac{2}{3}ax^{1.5}$ $v=\sqrt{\frac{4ax^{1.5}}{3m}}$
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