A 65.0 kg skier slides down a 37.20 slope with mu = 0.107.What is the friction force?

Physics

1 answer:

zmey [24]3 years ago

3 0

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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