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Q. A body of mass M starts sliding down on the inclined plane, where the critical angle is $\angle ACB =30^{\circ}$ as shown in figure. The coefficient of kinetic friction will bePhysics Question Image

Laws of Motion

Solution:

We now that,
$\theta=$ angle of repose
$\mu=\tan \theta=\tan 30^{\circ}=\frac{1}{\sqrt{3}}$
Therefore, coefficient of kinetic friction
$= Mg \times \frac{1}{\sqrt{3}}=\frac{ Mg }{\sqrt{3}}$