A skier traveling 12.0 m/s reaches the foot of a steady upward 18° incline...?

[Dudetter03]

A skier traveling 12.0 m/s reaches the foot of a steady upward 18° incline and glides 15.2 m up along this slope before coming to rest. What was the average coefficient of friction?

Physics problem. Please help! If possible, please plug numbers into equations! Also, explanations are appreciated!

 

[Charles]

Ok, so the component of acceleration of gravity that is parallel to the incline is

a(gravity) = mgsin(18)

friction,
a(friction) = kmgcos(18) where k = coefficient of friction

v(intial)^2 = 2ax v(final) = 0

v(initial)^2 = 2g(15.2m)(sin(18) + kcos(18))


----> k = [v(initial)^2 / (2(15.2^2)g)] - tan(18)

?