College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 6 - Problems - Page 232: 82


The engine's average power output is $9.30\times 10^5~Watts$

Work Step by Step

We can find the drag racer's change in kinetic energy: $KE = \frac{1}{2}mv^2 = \frac{1}{2}(500.0~kg)(125~m/s)^2 = 3.906\times 10^6~J$ The engine does this much work in a time of 4.2 seconds. We can find the power output: $Power = \frac{Energy}{Time} = \frac{3.906\times 10^6~J}{4.2~s} = 9.30\times 10^5~Watts$ The engine's average power output is $9.30\times 10^5~Watts$.
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.