The minute hand of the clock on the wall is about 12.0cm long. The tip of the minute hand has a mass $\displaystyle m$. 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 $\displaystyle W = mgh_{2} - mgh_{1}$, thought I'm confused about what the h value should be. Does the 12cm (0.12m) even have anything to do with any $\displaystyle h$ value? Or will I have to somehow calculate some angle of the clock to find the displacement using the equation $\displaystyle W = Fcos\Theta\Delta{d}$? I think the fact that the tips displacement from 10:00 to 10:30 being $\displaystyle \frac{1}{24}$ of the clock's circumference having something to do with it, but I'm not sure.