A skier traveling 12.0 m/s reaches the foot of a steady upward 18 degree incline and glides 12.2 m up along this slope before coming to rest. What was the average coefficient of friction?

Answer 1

ke at bottom = (1/2) m v^2
= (1/2) m (144) = 72 m Joules

pe at top = m g h= m (9.81)(12.2 sin 18
= 37 m Joules
so
Lost 72-37 = 35 m Joules to friction

Friction force
= mu m g cos 18= 9.33 mu m
work done by friction
= 9.33 mu m * 12.2 = 114 mu m
so
35 m = 114 mu m

mu = 35/114 = .307
sticky snow :(

Answer 2

The process is correct.