# Coefficents of Friction

1. Dec 18, 2008

### sonoftunk

1. The problem statement, all variables and given/known data
A young skier (25kg) pushes off with ski poles to give herself an initial velocity of 3.5 m/s down a hill with 5o slope with a coefficient of friction of 0.20.
Find the time until skier comes to stop and her displacement.
Therefore
m = 25kg
vi = 3.5m/s
$$\theta$$ = 5o
$$\mu$$ = 0.20

2. Relevant equations
All kinematic-based equations
Newton's Laws (specifically $$\stackrel{\rightarrow}{F}$$ = m($$\Delta$$$$\stackrel{\rightarrow}{v}/$$$$\Delta$$t)
FF = $$\mu$$FN

3. The attempt at a solution
My idea for this was to find the FNET and sub that into Newton's Second Law, solving for $$\Delta$$t (assuming that $$\stackrel{\rightarrow}{v}$$f is zero)
My solution for $$\Delta$$t was 1.3 seconds, correct to two significant digits (to me, that seems quite off)

For the second half of the question, finding the distance, I used $$\Delta$$$$\stackrel{\rightarrow}{d}$$ = $$\stackrel{\rightarrow}{v}$$i + $$\frac{1}{2}$$$$\stackrel{\rightarrow}{a}$$($$\Delta$$t)2 using $$\stackrel{\rightarrow}{a}$$ as the component force of gravity acting parallel to the hill and I got 23m, correct to 2 significant digits.

I was then told I completed the question completely wrong. Now I'm lost. What was I supposed to do?

2. Dec 18, 2008

### LowlyPion

Welcome to PF.

Maybe consider it is a Kinetic Energy to work from friction problem?

Work would be the (μ*mg*cos5° - mg*sin5°) times the distance and that would equal the KE.

3. Dec 18, 2008

### sonoftunk

I haven't learned kinetic energy yet...

BTW, this is a correspondence coarse for Grade 12 Physics in Ontario..

4. Dec 19, 2008

### buffordboy23

You know the frictional force, so find the net acceleration on the skier due to friction and gravity.