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

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 10 - Interactions and Potential Energy - Exercises and Problems: 4

Answer

(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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.