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 - Page 257: 29


$v = 6.3~m/s$

Work Step by Step

From $x=2$ to $x = 4$, the particle's kinetic energy is transformed into potential energy. To find the maximum speed at $x = 2$, we can assume that the kinetic energy at $x = 4$ is zero. We can find the maximum speed the particle could have at $x = 2$ and not reach $x = 6$: $K_2+U_2 = K_4+U_4$ $\frac{1}{2}mv^2= 0+U_4 - U_2$ $v^2= \frac{(2)(U_4 - U_2)}{m}$ $v= \sqrt{\frac{(2)(U_4 - U_2)}{m}}$ $v= \sqrt{\frac{(2)(8~J -4~J)}{0.200~kg}}$ $v = 6.3~m/s$
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