#### Answer

(a) $W = 8.9\times 10^5~J$
(b) $P_o = 54 ~W$
$P_o = 0.072 ~hp $
(c) $P_I = 360~W$
$P_I = 0.48~hp$

#### Work Step by Step

(a) The work done by the hiker will be equal to the change in gravitational potential energy.
$W = mgh$
$W = (65~kg)(9.80~m/s^2)(1400~m)$
$W = 8.9\times 10^5~J$
(b) The average power output $P_o$ is equal to the total work done by the hiker divided by the total time of the hike.
$P_o = \frac{W}{t} = \frac{8.9\times 10^5~J}{(4.6~h)(3600~s/h)}$
$P_o = 54~W$
We can convert the average power output to units of horsepower.
$P_o = (54~W)(\frac{1~hp}{746~W}) = 0.072~hp$
(c) We can calculate the required rate of energy input $P_I.$ Note that only 15% of the power input by the hiker is used as power output.
$0.15\times P_I = P_o$
$P_I = \frac{P_o}{0.15}$
$P_I = \frac{54~W}{0.15} = 360~W$
We can convert the power input $P_I$ to units of horsepower.
$P_I = (360~W)(\frac{1~hp}{746~W}) = 0.48~hp$