Essential University Physics: Volume 1 (4th Edition)

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 13 - Exercises and Problems - Page 246: 28

Answer

$0.2\%$

Work Step by Step

We know, Energy in Simple Harmonic Motion $=\frac{1}{2}KA^{2}$ Where, K - spring constant of the shock absorber, A - Amplitude of the shock absorber. $E_{SHM}=\frac{1}{2}KA^{2}-(1)$ We know, frequency $(f)=\frac{1}{2\pi}\sqrt {\frac{K}{m}}$ in S.H.M. where, K - Spring constant, m - mass on the shock absorber. Let's plug known values into this equation. $1\space s^{-1}=\frac{1}{2\pi}\sqrt {\frac{K}{500\space kg}}=\gt1\space s^{-2}=\frac{1}{4}\times\frac{K}{500\space kg}$ $2000\space \pi^{2}\space N/m=K$ $(1)=\gt$ $E_{SHM}=\frac{1}{2}\times2000\space \pi^{2}\space N/m\times(0.2\space m)^{2}=394.78\space J$ Total energy in all shock absorbers $(E_{T})=E_{SHM}\times4$ $E_{T}=394.78\space J\times 4=1580\space J$ Let's find the Kinetic energy in the car, $E_{K}=\frac{1}{2}mV^{2}=\frac{1}{2}\times2000\space kg\times[(\frac{100\space km}{h})(\frac{1000\space m}{km})(\frac{h}{3600\space s})]^{2}$ $E_{K}=771.6\space kJ$ Energy percentage $=\frac{E_{T}}{E_{K}}\times100\%=\frac{1580\space J}{771.6\times10^{3}\space J}\times100\%=0.2\%$
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