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## On a planet where g = 10.0 m/s2 and air resistance is negligible, a sled is at rest on a rough inclined hill rising at 30°. The initial posi

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On a planet where g = 10.0 m/s2 and air resistance is negligible, a sled is at rest on a rough inclined hill rising at 30°. The initial position is 8.0 m up the hill along the slope from the bottom. The object is allowed to move and it stops on a rough horizontal surface, at a distance of 4.0 m from the bottom of the hill. The coefficient of kinetic friction on the hill is 0.40. What is the coefficient of kinetic friction between the horizontal surface and the sled?

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Physics
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2021-08-30T18:59:59+00:00
2021-08-30T18:59:59+00:00 2 Answers
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## Answers ( )

Answer:Coefficient of kinetic friction = 0.31

Explanation:g = 10 m/s²

distance covered up the hill, S = 8 m

Horizontal distance covered, d = 4 m

From the free body diagram attached:

The velocity at the bottom can be calculated by:

Work done through to frictional force = change in kinetic energy:

Answer:Explanation:a is the acceleration

μ is the coefficient of friction

Acceleration of the object is given by

Velocity at the bottom

after travelling 4m , its velocity becomes 0

Coefficient of kinetic friction

μ = F/N

Therefore, the Coefficient of kinetic friction is 0.31