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Q. A block has been placed on a inclined plane with the slope angle $\theta$, block slides down the plane at constant speed. The coefficient of kinetic friction is equal to

AIPMTAIPMT 1993Laws of Motion

Solution:

The acceleration is nullified by force of kinetic friction/mass
mg sin$\theta$ is force downwards.
$\mu_k$ is the coefficient of kinetic friction.
$\mu_k mg \cos\theta $ is force acting upwards.
$\therefore mg \sin\theta=\mu_k mg \cos \theta=$ mass $\times$ acceleration $= 0$ as $v$ is constant
$\therefore \mu_k=\tan\theta$.