#### Answer

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.
$$P=\frac{W}{t}=\frac{mgh}{t}=\frac{(62 kg)(9.80\frac{m}{s^2})(5.0 m)}{9.0s}=340 W$$

Published by
Pearson

ISBN 10:
0-32162-592-7

ISBN 13:
978-0-32162-592-2

340 W.

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