# Skier's gravitational energy

1. Nov 10, 2007

### rcmango

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

A 70.0 kg skier rides a 2700 m long lift to the top of a mountain. The lift makes an angle of 14.3° with the horizontal. What is the change in the skier's gravitational potential energy?

2. Relevant equations

3. The attempt at a solution

mg cos(?) * (h0 - hf)

again, not sure why this formula doesnot work.

2. Nov 10, 2007

### catkin

The GPE near the Earth's surface is mgh where h is the height above some reference height.

Thus a change in GPE is mgΔh where Δh is the change in height.

In your equation (h0 - hf) looks like a change in height -- which is what you want, so why should it be multipled by cos(?)?

3. Nov 10, 2007

### rl.bhat

Using trig. you can find that (hf - ho ) = 2700*sin14.3. Using this value you can find gravitational potential energy.

4. Nov 10, 2007

### rcmango

okay so gpe = m * g * (chng in height)

so this is what i get 70 * 9.8 * (667.)

and get something like: 4.6E-5

okay, thats correct. thanks alot.

5. Nov 10, 2007

### rl.bhat

A 70.0 kg skier rides a 2700 m long lift to the top of a mountain. Can you explain this sentence? Which is his starting point? "2700m long lift" Is it the distasance between the starting point and the end point?