Answer
The correct answer is (c) 350 W.
Work Step by Step
We can find the angle $\theta$ when a person walks on a 5.0% grade.
$tan(\theta) = \frac{5.0}{100}$
$\theta = arctan(\frac{5.0}{100})$
$\theta = 2.86^{\circ}$
When a person walks a distance of 1.4 meters on a 5.0% grade, the vertical increase in height is $(1.4~m)~sin(2.86^{\circ})$ which is 0.070 meters.
We can find the work required to climb vertically a distance of 0.070 meters.
$W = mgh = (70~kg)(9.80~m/s^2)(0.070~m)$
$W = 50~J$
Since the person does this amount of work to climb 0.070 meters vertically each second, the power is 50 watts.
When a person walks a distance of 1.4 meters on a 5.0% grade, the horizontal distance is $(1.4~m)~cos(2.86^{\circ})$, which is 1.398 meters.
We can find the power required to walk a distance of 1.398 meters on a flat surface.
$P = \frac{1.398~m}{1.4~m}~300~W$
$P = 299.6~watts$
The total power output is 299.6 watts + 50 watts, which is 350 watts. The correct answer is (c) 350 watts
