Answer
The correct answer is:
B. $1500~W$
Work Step by Step
The total work done is equal to the apparent increase in gravitational potential energy:
$U_g = mgh$
$U_g = mgd~sin~\theta$
$U_g = mgvt~sin~\theta$
$U_g = (100~kg)(10~m/s^2)(3~m/s)(300~s)~sin~30^{\circ}$
$U_g = 4.5\times 10^5~J$
This energy is expended in a time of 300 seconds. We can find the power output:
$power = \frac{energy}{time} = \frac{4.5\times 10^5~J}{300~s} = 1500~W$
The correct answer is:
B. $1500~W$
