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 - General Problems: 81


340 W.

Work Step by Step

The work done on the athlete is the change in PE, mgh. The power is the work done divided by time. $$P=\frac{W}{t}=\frac{mgh}{t}=\frac{(62 kg)(9.80\frac{m}{s^2})(5.0 m)}{9.0s}=340 W$$
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.