# A little assistance with Friction mu k

## Homework Statement

A skier with a mass of 62 kg starts from rest and skis down an icy (frictionless) slope that has a length of 69 m at an angle of 32° with respect to the horizontal. At the bottom of the slope, the path levels out and becomes horizontal, the snow becomes less icy, and the skier begins to slow down, coming to rest in a distance of 160 m along the horizontal path.

a) What is the speed of the skier at the bottom of the slope?
m/s
i solved a) and got that it was 26.7 m/s for the speed

B is what i am having trouble with
b) What is the coefficient of kinetic friction between the skier and the horizontal surface?
µk =

## Homework Equations

fmax= mu k N i think

## The Attempt at a Solution

I tried many different ways...
26.7/607.6 = uk .04
uk mgcos(theta) = ma
uk 62(9.8cos32) = (5.19)(62) = .6 ... something

I am still confused about how to solve this any hints or techniques would greatly be appreciated

P.S. this is my first post please be kind :)

Doc Al
Mentor
fmax= mu k N i think
Close enough: f_k = mu_k N (No need for max--you were thinking of static friction.)

Hint: The frictioned section of the path is horizontal.

What's the skier's acceleration?
What's the only horizontal force acting on the skier?
Apply Newton's 2nd law and solve for mu_k. (What's N?)