First slide

Friction on inclined plane of angle more than angle of repose

Question

A block of mass 2 kg is sliding down a rough inclined plane having inclination $30^{0}$ with constant velocity of 0.5 m/s. Then coefficient of kinetic friction between the block and the same is

Easy

A

0.577
B

0.672
C

0.425
D

0.867

Solution

(1)
Since velocity of the block is constant,
$\begin{array}{l} m g \sin θ - f_{k} = 0 \Rightarrow m g \sin θ - μ_{k} . m g \cos θ = 0 \Rightarrow μ_{k} = \tan θ \\ ∴ μ_{k} = \tan 30^{0} = \frac{1}{\sqrt{3}} = 0.577. \end{array}$

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