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
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
