# Work Done by Gravity

1. Feb 28, 2012

### PeachBanana

1. The problem statement, all variables and given/known data

A 265-kg load is lifted 24.0 m vertically with an acceleration =0.190 g by a single cable.

Determine the work done by gravity on the load.

2. Relevant equations

Force of Gravity = mg
W = F * d * cos (theta)

3. The attempt at a solution
I think I know why I did this incorrectly but I want to make sure.

First attempt: Force gravity = (265 kg)(9.8 m/s^2) = 2.6 * 10^3 N.
W = (2.6 x 10^3 N) (24.0 m) * cos (0°)
W = 6.24 x 10^4 J.

The correct answer is the same magnitude but the opposite sign. Is that because the angle should have been 180° because gravity always points down? The displacement is up so therefore the angle between the net force and the displacement would have been 180°, not 0°?

2. Feb 28, 2012

### SammyS

Staff Emeritus
That looks fine. What's your question?

3. Feb 28, 2012

### PeachBanana

I did it incorrectly the first time but I wanted to make sure I knew why it was wrong.

4. Feb 28, 2012

### Ericv_91

Yes, if your displacement is in the positive direction, then in this case gravity must be in the opposite direction (180 degrees from positive direction).