Kinetic friction.

1. Sep 27, 2007

Bob Loblaw

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

An 60-kg skier is sliding down a ski slope at a constant velocity. The slope makes an angle of 11° above the horizontal direction.
(a) Neglecting any air resistance, what is the force of kinetic friction acting on the skier?

(b) What is the coefficient of kinetic friction between the skis and the snow?

2. Relevant equations

Fr = μ*Nf

3. The attempt at a solution

I was able to calculate Nf from the information given as mg*cos(24) or 537.16N. Without knowing the force I was unable to answer the problem. Any ideas where to begin? Thanks!

Last edited: Sep 27, 2007
2. Sep 27, 2007

bel

If he is sliding down with constant velocity, he is not accelerating as he would be if there were no friction, hence the force of friction is equal in magnitude and opposite in direction to the component of gravitational force parallel to the incline.

3. Sep 27, 2007

Bob Loblaw

So I took mg*sin(11) which gave me 112.31 which is correct.

Now as I need to find the coefficient of kinetic friction between the skis and the snow. Using my equation, I tried Fr = μ*Nf where Fr is 112.31 and Nf was 537.16. I arrived at .021 which was not correct. Where did I go wrong?

4. Sep 27, 2007

bel

Calculate the magnitude of your normal force again, I got a different answer.

5. Sep 27, 2007

Bob Loblaw

Nomal force should be mg*cos(11) in this case. which would be 577.8. I tried using that in my equation and came up with .19 as the kinetic coefficient for part B which is still not the correct answer. Am I using the right equation?

6. Sep 27, 2007

bel

I got 60*9.81*cos(11)= 577.79 N, keep a few more digits in calculation, by the way, and round at the end, especially since you are looking at small numerical answers.

7. Sep 27, 2007

Bob Loblaw

0.1943 was accepted as an answer in web assign. Thanks for your help!