Question

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

• A. $\sin \theta$
• B. $\cos \theta$
• C. $g$
• D. $\tan \theta$

Answer 1

Answer: (D)

$\tan \theta$

Answer 2

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
$\therefore \mu_k=tan\ \theta$.

So, the correct option is (D): $\tan \theta$