Question

A block has been placed on an inclined plane
with the slope angle θ, block slides down the
plane at constant speed. The coefficient of kinetic
friction is equal to
(a) sin θ (b) cos θ
(c) g (d) tan θ

Answer 1

Answer:

let the angle be x

answer =tanx

Explanation:

let the coefficient of friction be u, N - normal reaction force

=> u X N = mgsinx

=> u X mgcosx= mgsinx

=> u = sinx/cosx

=> u = tanx

let me know if the answer is wrong

Answer 2

$\Large\boxed{\sf{\mu=\tan\theta}}$

Since the angle of inclination is $\sf{\theta,}$ the block moves along the surface with the acceleration $\sf{g\sin\theta}$ if there is no friction.

If there is a frictional force $\sf{f=\mu mg\cos\theta}$ where $\sf{\mu}$ is the coefficient of friction of the surface and $\sf{mg\cos\theta}$ is the normal reaction acting on the block, then the acceleration acting on the block is,

$\longrightarrow\sf{a=g\sin\theta-\dfrac{f}{m}}$

$\longrightarrow\sf{a=g\sin\theta-\mu g\cos\theta}$

$\longrightarrow\sf{a=g(\sin\theta-\mu\cos\theta)}$

Here the acceleration is zero as the block slides down with constant speed.

$\longrightarrow\sf{a=0}$

$\longrightarrow\sf{g(\sin\theta-\mu\cos\theta)=0}$

$\longrightarrow\sf{\sin\theta-\mu\cos\theta=0}$

$\longrightarrow\sf{\underline{\underline{\mu=\tan\theta}}}$