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## Homework Statement

A skier with a mass of 53 kg starts from rest and skis down an icy (frictionless) slope that has a length of 59 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 155 m along the horizontal path.

(a) What is the speed of the skier at the bottom of the slope?

(b) What is the coefficient of kinetic friction between the skier and the horizontal surface?

I just need help with (b)

## Homework Equations

Vf

^{2}- Vi

^{2}= 2ay

fk = u

_{k}*N

## The Attempt at a Solution

Don't need help with (a) but I'll list what I did anyway...

(a) What is the speed of the skier at the bottom of the slope?

Got the height first.

-->sin32 = y/59

Vi = 0m/s

-->Vf

^{2}= 2ay

-->Vf

^{2}= 2*-9.8m/s

^{2}*-31.265m

-->Vf = 24.755 m/s

(b)

Vi = 24.755m/s

Vf = 0m/s

x = 155m

-->Vf

^{2}-Vi

^{2}= 2*a*x

-->0 - 24.755

^{2}= 2*a*155m

-->a = -1.9768m/s

^{2}

-->F - F

_{k}= ma

-->mg - u

_{k}N = ma

-->mg - u

_{k}mg = ma

-->g - u

_{k}g = a

-->u

_{k}= (g-a) / g

-->u

_{k}= (-9.8m/s

^{2}- (-1.9768m/s

^{2})) / (-9.8m/s

^{2})

-->u

_{k}= 0.7983

Somehow this is wrong??? Any help is much appreciated!