Q5: An ice skater moving at 12 m/s coasts to a halt in 95m on an ice surface. What is the coefficient
of (kinetic) friction between ice and skates?​

u = 0.077

Explanation:

Work done by friction is

Wf = ∆KE + ∆PE

-umgx = ∆KE,. ∆PE =0 (level ice surface)

-umgx = KEf - KEi = -(1/2)mv^2

Solving for u,

u = v^2/2gx

= (12 m/s)^2/2(9.8 m/s^2)(95 m)

= 0.077