Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 6 - Work and Energy - Problems - Page 166: 44

Answer

(a) $v_{max} = \sqrt{\frac{kx_0^2 + mv_0^2}{m}}$ (b)$x_{max} = \sqrt{\frac{kx_0^2 + mv_0^2}{k}}$

Work Step by Step

$Energy = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ (a) The maximum speed will occur when the potential energy in the spring is 0 and all the energy in the system is kinetic energy: $\frac{1}{2}mv_{max}^2 = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ $v_{max} = \sqrt{\frac{kx_0^2 + mv_0^2}{m}}$ (a) The maximum stretch will occur when the kinetic energy in the spring is 0 and all the energy in the system is potential energy: $\frac{1}{2}kx_{max}^2 = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ $x_{max} = \sqrt{\frac{kx_0^2 + mv_0^2}{k}}$
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