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

$P = 463~watts$
The work $W$ that the firefighter does is equal to the change in potential energy. $W = PE$ $W = mgh$ $W = (90~kg+40~kg)(9.80~m/s^2)(20~m)$ $W = 25,480~J$ We can find the required power as; $P = \frac{Work}{time}$ $P = \frac{25,480~J}{55~s}$ $P = 463~watts$