# Energy - Ski sliding down a ramp

1. Nov 18, 2014

### Ritzycat

1. The problem statement, all variables and given/known data
A ski starts from rest and slides down a 28 incline 65m long. If the coefficient of friction is 0.090, what is the ski's speed at the base of the incline? If the snow is level at the foot of the incline and has the same coefficient of friction, how far will the ski travel along the level?

2. Relevant equations
KE = (1/2)mv2
PEg = mgh

Note that I want to solve this problem using energy relations, not kinematics.

3. The attempt at a solution
Taking Point A to be the top of the ramp, point B to be right before the bottom.

PEg = KE + Ediss
mgh = (1/2)mv2 + Ff
mgh = (1/2)mv2 + μFN
mgh = (1/2)mv2 + μmg(sin 30)
gh = (1/2)v2 + μg(sin 30)
(9.8m/s2)(30.5157m) = (1/2)(v2) + (0.090)(9.8m/s2)(sin 30)

v = 24.4 m/s.

That answer was incorrect. I have not yet attempted the second question since I don't have a correct answer for the first one.

2. Nov 18, 2014

### Tanya Sharma

First thing, The angle of incline is 28 ,not 30. Secondly work done by friction is not μN.

Last edited: Nov 18, 2014
3. Nov 18, 2014

### ehild

The dissipated energy is not equal to the force of friction.

4. Nov 18, 2014

### Ritzycat

Thanks for the reminder - although when I plugged in the new value, the answer changed very minimally...

What should I set E diss equal to then??

5. Nov 18, 2014

### Tanya Sharma

Energy dissipated is equal to the work done by friction . How do you calculate the work done by a force ?

6. Nov 18, 2014

### ehild

Do you think that energy is the same as force?????????

7. Nov 18, 2014

### Ritzycat

Ff * d = W

I think that will yield me the right answer. I would use the length of the incline, since that is the direction Friction was acting in. Then I'll plug it in. I'm right about to go to bed, so I'll do that in the morning.

No I forgot to multiply it by Distance................!!! i get it now Silly Ritzycat

8. Nov 18, 2014

### ehild

Such things do happen :)