Answer to Question #142300 in Physics for Sydney

Question

A 1.25 kg squirrel runs east across the ground for 7.00 meters. She climbs, 0.50 meters before running west on a limb for 3.25 meters. She jumps up .85 meters onto another branch, travels west 3.75 meters. She stops and eats a nut. Calculate net work. What is the squirrel’s gravitational potential energy?

Answer 1

We can consider the gravity field near the surface of Earth as a uniform (in x and y directions) conservative vector field. In such case the work done against gravity field depends only on the initial and final height of the squrrel above the ground level.

During her jorney, squrrel raised from the ground level $h_i = 0m$ to the level of $h_f = 0.5m + 85m = 85.5m$. She's done the work against the gravity force, which is equal to:

$$F = mg = 1.25kg\times 10m/s^2 = 12.5N$$

where $m = 1.25kg$ is the squrrel's mass and $g = 10m/s^2$ is the gravitational acceleration. Thus, the work is:

$$A = F(h_f -h_i) = 12.5N\times (85.5m-0m) = 1068.75\space J$$

The squirrel’s gravitational potential energy is:

$$W = mgh_f = 1068.75\space J$$

Answer. Work: 1068.75 J, potential energy: 1068.75 J.