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: 42

Answer

$k = 2600~N/m$

Work Step by Step

$v = (95~km/h)(\frac{1000 ~m}{1~km})(\frac{1~h}{3600~s}) = 26.4~m/s$ The force exerted by a spring increases as the spring is compressed. Since the acceleration is proportional to the force, the car will experience the greatest acceleration at the point of maximum compression $x$. We can find an expression for the maximum compression $x$. Let $a = 4.0~g$. Then, $\sum F = ma$ $kx = 4.0~mg$ $x = \frac{4.0~mg}{k}$ We can use this expression for $x$ to find the spring constant $k$. At the point of maximum compression $x$, the elastic potential energy stored in the spring must equal the original kinetic energy of the car: $KE = EPE$ $\frac{1}{2}mv^2 = \frac{1}{2}kx^2$ $mv^2 = (k)(\frac{4.0~mg}{k})^2$ $k = \frac{16.0~mg^2}{v^2} = \frac{(16.0)(1200~kg)(9.80~m/s^2)^2}{(26.4~m/s)^2}$ $k = 2600~N/m$
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