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Sindrei [870]
3 years ago
11

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 be

tween the ice and the skates?
a.1.80

c.0.18

b.0.057

d.0.56
Physics
1 answer:
SVEN [57.7K]3 years ago
7 0
The horizontal force : f = k*N
k- coefficient of friction
k = f /N
N = m * g = 45 kg * 9.81 m/s² = 441.45 N
k = 25 N : 441.45 N = 0.057
Answer C) 0.057
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