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A 6.0-kg block slides from point A down a frictionless curve to point B. After the block passes point B, a friction force opposes the motion of the block so that it comes to a stop 2.5 m from B. Calculate the coefficient of kinetic friction between the block and the surface after position B.

A) 0.10

B) 0.20

C) 0.30

D) 0.40

User PrimRock
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1 Answer

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The coefficient of kinetic friction between the block and the surface after point B is approximately 0.30. Hence the correct option is c.

As the block moves down the frictionless curve from point A to point B, its kinetic energy is converted into potential energy. After passing point B, a friction force acts against the motion, causing the block to decelerate and eventually come to a stop. The work done by friction is equal to the initial kinetic energy of the block.

Using the work-energy principle, we can equate the work done by friction to the change in kinetic energy. The formula for work done by friction is W friction​=μk​⋅m⋅g⋅d​ is the coefficient of kinetic friction, m is the mass of the block, g is the acceleration due to gravity, and d is the distance traveled after point B.

The change in kinetic energy is given by ΔKE= 1/2 mv^2, where v is the final velocity of the block after coming to a stop.

Setting these two expressions equal and solving for the coefficient of kinetic friction (μk​), we find μk ≈0.30. Therefore, the correct answer is option C, indicating that the coefficient of kinetic friction between the block and the surface after point B is approximately 0.30. Hence the correct option is c.

User Kirill Gamazkov
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