Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 10 - Interactions and Potential Energy - Exercises and Problems: 26

Answer

The maximum speed of the particle is 63.2 m/s

Work Step by Step

We know the total energy $E$ is equal to the sum of the potential energy $U$ and the kinetic energy $K$. If the particle oscillates between x = 2.0 mm and x = 8.0 mm, then the kinetic energy at those two points is zero. We can consider the point x = 2.0 mm to find the total energy; $E = U+K$ $E = 5.0~J+0$ $E = 5.0~J$ The total energy is 5.0 J. The maximum speed occurs when the kinetic energy is at a maximum. This occurs when the potential energy is at a minimum. From the graph, we can see that the potential energy is at a minimum at the point x = 4.0 mm. We can find the kinetic energy at the point x = 4.0 mm; $K+U = E$ $K = E-U$ $K = 5.0~J - 1.0~J$ $K = 4.0~J$ We can find the speed of the particle at point x = 4.0 mm; $K = 4.0~J$ $\frac{1}{2}mv^2 = 4.0~J$ $v^2 = \frac{(2)(4.0~J)}{m}$ $v = \sqrt{\frac{(2)(4.0~J)}{m}}$ $v = \sqrt{\frac{(2)(4.0~J)}{0.0020~kg}}$ $v = 63.2~m/s$ The maximum speed of the particle is 63.2 m/s.
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