Question

a block of mass m takes time t to slide down on a smooth inclined plane of angle of inclination θ and height h. if same block slide down on a rough inclined plane of same angle of inclination and same height and takes time n times of initial value, then coefficient friction between block and inclined plane is

Answer 1

Height of the inclined plane is "h"

so the length of the inclined plane that the block will slide down is given as

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

now the acceleration of the block along the inclined plane is given as

$a = gsin\theta$

now the time taken by the block to slide down the inclined plane is given by

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

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

now when object will slide on rough inclined plane then the acceleration of the block is given by

$a = g(sin\theta - \mu cos\theta)$

now to slide down same length of inclined plane the time taken by the object is given by

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

now given that the time taken on rough inclined plane is n times more than the time on smooth incline

$t' = nt$

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

now by solving above equation

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

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

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

so above is the friction coefficient for the inclined plane