potential energy?

Question

A skier starts from rest at the top of a hill that is inclined at 11.1 degrees with the horizontal. The hillside is 174.8 m long, and the coefficient of friction between the snow and the skis is 0.067. At the bottom of the hill, the snow is level and the coefficient of friction is unchanged.
The acceleration of gravity is 9.81 m/s2
How far does the skier move along the horizontal portion of the snow before coming to rest? Answer in units of m.

Thanks for any help! :)

Answer 1

The net acceleration down the incline is
g [ sin θ- μ cos θ]
g [ sin 11.1 - 0.067 cos 11.1]
= 1.244 m/s^2.
The final velocity^2 at the bottom is
v^2 = 2as = 2 x1.244 x 174.8
v = 20.85 m/s
------------------------------------
Acceleration along the horizontal is
μ g = 0.65727
s = v^2 / 2a = 20.85^2 / 2* 0.65727
s = 330.7 m
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Answer 2

Let L = 174.8m
Angle @ = 11.1 deg
Mass of the skier and skis = M
g = acceleration of gravity
c = coefficient of friction
D = distance the skier moves before coming to rest

The vertical drop is 174.8 { sin(11.1 deg)} = V = Lsin@
The potential energy at the top is m g V
Friction does work on the skier over the distance L in an amount = perpendicular component of gravitational forceXcXL = Mg cos@ L c


At the bottom, the skier has kinetic energy mgV-Mgcos@ L c
= Mg(V-cL cos@) = KE (bottom)
The skier will glide until the work done by friction = KE (bottom) = Mg D c
MgDc = Mg( L sin@ - cLcos@ )
D = (Lsin@/c - L cos@ )= L ( sin@/c - cos@ )