## 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

Question

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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2 weeks 2021-08-30T18:59:59+00:00 2 Answers 0 views 0

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: 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