Question

A block of mass m placed on an inclined plane of angle of inclination theta slides down the plane with constant speed

Answer 1

When book is placed on inclined plane and its angle of inclination is some angle theta. It is given that velocity of book is uniform and its going down the incline with constant speed.

This shows that the net force on the book must be balanced and hence the component of weight along the incline must be counterbalanced by friction force which is opposite to the motion of the book

Now here we can say

$F_f = F_g$

$F_f = mgsin\theta$

so the kinetic friction must be equal to the component of weight along the incline plane

Now if we have to find the minimum friction coefficient so that above condition may hold good then we can use the formula of friction force

$F_f = \mu N$

here on inclined plane we can say

$N = mgcos\theta$

now by above equation

$F_f = \mu mgcos/theta$

now by above equation we can equate it with the component of weight

$mg sin\theta = \mu mg cos\theta$

$tan\theta = \mu$

so above is the relation between friction coefficient and angle of inclination