A skier slides horizontally along the snow for a distance of 12 m before coming to rest. the coefficient of kinetic friction between the skier and the snow is μk = 0.060. initially, how fast was the skier going?
We can use work-energy to solve this question. The work done by friction on the skier will be equal in magnitude to the skier's initial kinetic energy.
KE = Work
(1/2)mv^2 = mg ÎĽk d
v^2 = 2g ÎĽk d
v = sqrt{ 2g ÎĽk d }
v = sqrt{ (2) (9.80 m/s^2) (0.060) (12 m) }
v = 3.76 m/s
Initially the skier was moving at a speed of 3.76 m/s
