A skier slides horizontally along the snow for a distance of 12.1 m before coming to rest. The coefficient of kinetic friction between the skier and the snow is 0.0343. Initially, how fast was the skier going?

Answer 1

a = u*g = 0.0343 * (-9.8) = -0.336 m/s^2.

V^2 = Vo^2 + 2a*d = 0.
Vo^2 = -2a*d = -2*(-0.336)*12.1 = 8.13.
Vo = 2.85 m/s.