A 65.0 kg skier slides down a 37.20 slope with mu = 0.107.What is the friction force?
1 answer:
Answer:
54.3N
Explanation:
The normal force is perpendicular to the slope, so:
Normal Force = cos(37.2)(9.8*65).......507.39N
F(friction)=mu*F(normal)
F(friction)=(0.107)(507.39)
F(friction)=54.3N
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