Homework Help: Potential Energy Problem!

1. Oct 2, 2005

Ginny Mac

A 1.50 kg snowball is fired from a cliff 12.5 m high. The snowball's initial velocity is 14.0 m/s, directed at 41.0 degrees above the horizontal. (a) How much work is done by the gravitational force during its flight to the flat ground below the cliff? (b) What is the change in gravitational potential energy of the snowball-Earth system during flight? (c) If the grav. potential energy is taken to be zero at the height of the cliff, what is its value when the snowball reaches the ground?

For (a), I have used
Wg= mgd (cos theta) to get Wg = (1.50 kg)(9.8 m/s^2)(12.5 m)(cos41), and arrived at -.987 J/m.

For (b), wouldn't this be indeterminate? I thought we needed a reference point for y to calculate the change in gravitational potential energy.

For (c), I used U-Uinitial = mg(y-yinitial), and arrived at -183.75 J. My reasoning is that the change in potential energy does not depend on choice of reference point for the snowball on the ground (at y=0), so instead it depends on change in y. Therefore, we get change in U = (1.50 kg)(9.8 m/s^2)(0-12.5 m)=-183.75 J.

I think I am getting caught up and confused in physical properties, etc. Please help if I am wrong or please let me know if I am correct in my reasoning! Thank you so much.

Gin

2. Oct 2, 2005

Staff: Mentor

Realize that gravity acts straight down, not at an angle. And since the displacement and the force are in the same direction, the work done will be positive. And the units of work would be J, not J/m.

The change in gravitational potential energy does not depend on the reference point used. Pick any reference point you like. (If they ask, like in the next part, for the potential energy--not just the change--then they must give a reference point. And they did for part c.)

Your answer is correct, but your reasoning is a bit convoluted (no need to calculate changes in PE). The gravitational PE is given by mgy, where y is measured from the reference point. If the point in question is below the reference point, then y is negative.