First slide

Kinetic friction

Question

A block has been placed on an inclined plane. The slope angle $θ$ of the plane is such that the block slides down the plane at a constant speed. The coefficient of kinetic friction is equal to

Moderate

A

sin $θ$
B

cos $θ$
C

g
D

tan $θ$

Solution

Angle of plane is just equal to the angle of repose.
i.e. $μ_{k} . m g \cos θ = m g \sin θ \Rightarrow μ_{k} = \tan θ$
Hence, the coefficient of kinetic friction, $μ_{k}$ = tan $θ$

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