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

Published by Pearson

# Chapter 10 - Interactions and Potential Energy - Exercises and Problems - Page 256: 18

#### Answer

The spring will be compressed a distance of 4.0 cm

#### Work Step by Step

The energy $U_s$ stored in the spring after the block compresses the spring will be equal to the block's initial kinetic energy. We can find an expression for the distance that the spring is compressed. Therefore; $U_s = KE$ $\frac{1}{2}kx_0^2=\frac{1}{2}mv^2$ $x_0^2=\frac{mv^2}{k}$ $x_0=\sqrt{\frac{mv^2}{k}}$ We can find the distance $x_2$ that the spring is compressed when the initial speed of the block is $2v$ as; $U_s = KE$ $\frac{1}{2}kx_2^2=\frac{1}{2}m(2v)^2$ $x_2^2=\frac{m(2v)^2}{k}$ $x_2=\sqrt{\frac{m(2v)^2}{k}}$ $x_2=2~\sqrt{\frac{mv^2}{k}}$ $x_2 = 2~x_0$ $x_2 = (2)(2.0~cm)$ $x_2 = 4.0~cm$ The spring will be compressed a distance of 4.0 cm.

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