## Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson

# Chapter 13 - Exercises and Problems - Page 238: 36

.1 percent

#### Work Step by Step

Plugging in $v=\omega A$ and $\omega=2\pi f$ into the equation $E=\frac{1}{2}mv^2$, we find: $E = 2\pi^2mf^2A^2 \\ E= 2\pi^2(1400)(.67)^2(.18)^2 =401.93 \ J$ We divide this by the kinetic energy of the car to find: $= \frac{401.6\ J}{\frac{1}{2}mv^2}=\frac{401.6\ J}{\frac{1}{2}(1400)(20)^2}=.001 \approx\fbox{.1 percent}$

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