Question

a block of mass m takes time t to slide down on a smooth inclined plane of inclination theta and height h. If the same block slids down on a rough inclined plane of same angle of inclination and same height and takes time n times of its initial value, the coefficient of friction between block and inclined plane is?

Answer 1

Length of the inclined plane is given by

$L = \frac{h}{sin\theta}$

acceleration along the inclined plane due to gravity

$a = gsin\theta$

now the time taken to slide down will be given as

$t =\sqrt{ \frac{2L}{a}}$

$t = \sqrt{\frac{2h}{gsin^2\theta}}$

now when it slides down the rough plane we will have

acceleration is as

$a = gsin\theta - \mu gcos\theta$

now here the total time taken to slide down is

$t' = \sqrt{\frac{2h}{(gsin\theta - \mu gcos\theta)sin\theta}}$

given that

$t' = n t$

$\sqrt{\frac{2h}{(gsin\theta - \mu gcos\theta)sin\theta}} = n \sqrt{\frac{2h}{gsin^2\theta}}$

now we have

$\frac{1}{sin^2/theta - \mu sin\theta cos\theta} = n^2\frac{1}{sin^2\theta}$

$n^2 sin^2\theta - n^2\mu sin\theta cos\theta = sin^2\theta$

$(n^2 - 1)sin^2\theta = n^2\mu sin\theta cos\theta$

so coefficient of friction is given as

$\mu = \frac{n^2 - 1}{n^2} tan\theta$

so above is the friction coefficient on the rough plane