Question

Find the work done on a 48,500 g object to move it on a 22.55 m up a 25.50 degrees incline.

Answer 1

The net work done against gravity only acounts change in the height above the ground. If we take the initial height of the object $h_i = 0$ then its final height is (see the figure):

$$h_f = 22.55m\times \sin25.5\degree \approx 8.6295m$$

Then, the work done is:

$$A = mg(h_f - h_i) = mgh_f$$

where $m = 48.5g = 0.0485\space kg$ is the mass of the object and $g = 9.81m/s^2$ is the gravitational acceleration. Thus, obtain:

$$A = 0.0485\times 9.81\times 8.6295 \approx 4.106\space J$$

Answer. 4.106 J.