Question

A skier slides horizontally along the snow for a distance of 11.9 m before coming to rest. The coefficient of kinetic friction between the skier and the snow is 0.0458. Initially, how fast was the skier going?

I do not know how to start this. Thank you.

Answer 1

ΔKE =KE2 –KE1 = 0 -m•v²/2.
W(fr) =μ•m•g•s•cosα,
where α is the angle between the friction force and displacement.
α =180º, cos α = -1.
-m•v²/2 = - μ•m•g•s,
v = sqrt(2•μ• g•s).

Answer 2

So for v would I do sqrt(2*11.9*9.8*0.0458) ?