# Thread: Calculate change in gravitational potential energy.

1. ## Calculate change in gravitational potential energy.

The minute hand of the clock on the wall is about 12.0cm long. The tip of the minute hand has a mass $m$. The change in gravitational potential energy of the tip of the minute hand as the time changes from 10:00a.m. to 10:30a.m. is:

a. 0.24 mg
b. -0.24 mg (Correct Answer)
c. 9.12 mg
d. 0.12 m

My question is, how was the answer acquired? I was thinking along the lines of using the equation $W = mgh_{2} - mgh_{1}$, thought I'm confused about what the h value should be. Does the 12cm (0.12m) even have anything to do with any $h$ value? Or will I have to somehow calculate some angle of the clock to find the displacement using the equation $W = Fcos\Theta\Delta{d}$? I think the fact that the tips displacement from 10:00 to 10:30 being $\frac{1}{24}$ of the clock's circumference having something to do with it, but I'm not sure.

2. In this case, the “angle” is rather easy to calculate, since we are talking about the minute hand. At 10:00, the minute hand is pointing straight up, and at 10:30 it is pointing straight down. So the change in height of the tip is just one diameter of the clock (i.e., twice the radius).

Remember the absolute height doesn’t matter (indeed the potential is defined so it can be shifted by any amount), but the height difference: $W = mgh_2 - mgh_1=mg(h_2-h_1)$. You know $h_2-h_1=-2r$ where $r=12cm$.