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Q. A block of mass $5 \, kg$ is placed on a rough inclined plane. The inclination of the plane is gradually increased until the block just begins to slide down. The sin of angle made by the plane with horizontal is $\frac{3}{5}$ when the block start sliding. The coefficient of friction between the block and the plane is ( $g \, = \, 10 \, m \, s^{- 2}$ )

NTA AbhyasNTA Abhyas 2022

Solution:

The block will just start sliding when the angle $\theta $ made by the inclined plane with the horizontal will be equal to the angle of repose.
$\theta=\theta_{\text {repose }}=\tan ^{-1}(\mu)$
$\Rightarrow tan\theta =\mu $
$\mu =\frac{3}{4}$