Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 10 - Energy - Exercises and Problems - Page 272: 4


(a) $KE = 6.75\times 10^5~J$ (b) h = 45.9 m (c) The answer to part (b) does not depend on the mass.

Work Step by Step

(a) We can find the kinetic energy of the car. $KE = \frac{1}{2}mv^2$ $KE = \frac{1}{2}(1500~kg)(30~m/s)^2$ $KE = 6.75\times 10^5~J$ (b) The potential energy at the height $h$ would be equal to the kinetic energy just before impact. $PE = KE$ $mgh = \frac{1}{2}mv^2$ $h = \frac{v^2}{2g}$ $h = \frac{(30~m/s)^2}{(2)(9.80~m/s^2)}$ $h = 45.9~m$ (c) Since the required height $h$ is equal to $\frac{v^2}{2g}$, the answer to part (b) does not depend on the mass.
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