Question

a block of mass 5kg is slipping down a rough inclined wedge of mass 15kg kept rest on horizontal surface with constant velocity
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Answer 1

In my opinion, the question is incomplete

Answer 2

Correct Question:

A block of mass 5 kg is slipping down a rough inclined wedge of mass 15kg kept at rest on horizontal surface with constant velocity and the inclined angle is 30°. Find the value of coefficient of friction ?

Calculation:

The block is coming down with constant velocity, this means that all the forces on it are balanced. So, the component of gravitation force is being balanced by the frictional force.

$\sf\therefore \: mg \sin( \theta) = f$

$\sf\implies\: mg \sin( \theta) = \mu(normal \: reaction)$

$\sf\implies\: mg \sin( \theta) = \mu[mg \cos( \theta) ]$

$\sf\implies\: \mu = \dfrac{ \sin( \theta) }{ \cos( \theta) }$

$\sf\implies\: \mu = \tan( \theta)$

$\sf\implies\: \mu = \tan( {30}^{ \circ} )$

$\sf\implies\: \mu = \dfrac{1}{ \sqrt{3} }$

So, value of coefficient of friction is 1/√3.