Question

A 45.0 kg ice skater needs a 25 N horizontal force to get moving on a smooth ice surface. What is the coefficient of friction between the ice and the skates?

a.1.80
c.0.18
b.0.057
d.0.56

Answer 1

As we know f=k/N
k = f /N
So
N = m * g = putting values 45 kg * 9.81 m/s² = 441.45 N
k = 25 N : 441.45 N =
so the coeffcient of friction between the ice and the skates is 0.057