# Homework Help: This problem has got me stumped. I need the answer to make an A on a test

1. Nov 11, 2009

### tdcman92

1. The problem statement, all variables and given/known data
this straight from a textbook.
A skier starts from rest at the top of a hill that is inclined at 10.5degrees with the horizontal. The hillside is 200.0 m long, and the coefficient of friction between the snow and the skies is 0.075. At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. How far does the skier move along the horizontal portion of thesnow before coming to rest?

2. Relevant equations

3. The attempt at a solution
i tried different thing but i cant enderstand how the problem can be solved without the mass of the skier.

Last edited: Nov 11, 2009
2. Nov 12, 2009

### ehild

How would you solve the problem if you knew the mass of the skier?

ehild

3. Nov 12, 2009

### tdcman92

idk how

4. Nov 12, 2009

### matt_crouch

try resolving the forces up and down the slope

5. Nov 12, 2009

### Master7731

use the conservation of mechanical energy, i think the masses should cancel out

6. Nov 12, 2009

### Master7731

i think this is the way to do it:

mgh = 1/2mv^2 + Wf
mgh = 1/2mv^2 + mu(Fn)cos theta (r)
mgh = 1/2mv^2 + mu(mg)cos theta (r) *h can be found by using sin of 10.5
Solve for Vf = ?

plug in Vf as Vi
1/2mv^2 = mg(d) , in this case d is the distance you want...

and i think you should get the answer

7. Nov 12, 2009

### tdcman92

i dont know what mu(Fn)cos theta (r) or mu(mg)cos theta (r) means.
i suck a physics.

8. Nov 12, 2009

### indr0008

u can use the law of energy conversion here, what do u think the initial energy that the man has?

9. Nov 12, 2009

### tdcman92

i redid it and i figured that the vf is 26.72m/s at the bottom of the hill

now im on the horizontal part of the equation.
the skier starts from 26.72m/s and the coefficient of friction is 0.075
i still need help i feel so dumb

Last edited: Nov 12, 2009
10. Nov 13, 2009

### Master7731

the coeffiecient of friction is mu and theta is basically the angle, bottom part of the triangle, and to find use your SohCahToa

11. Nov 14, 2009

### indr0008

can u show us ur calculation? because if u have found the speed at the botton of the hill, then u are supposed to be able to find the rest of the question