Question

a block of mass m takes t time to slide down on a smooth inclined plane of inclination theta and height h. if same block slides 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

when an object is sliding down the incline plane of height "h" and inclination $\theta$

then total length of the inclined plane is

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

also the acceleration on the smooth inclined plane will be

$a = g sin\theta$

now we have time to slide down

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

$t = \frac{1}{sin\theta}\sqrt{\frac{2*h}{g}}$

now when it slide down the rough plane the acceleration will be less due to friction force

now in that case acceleration is given by

$a = gsin\theta - \mu g cos\theta$

again by above formula

$t' = \frac{1}{sin\theta}\sqrt{\frac{2*h}{g(1-\mu tan\theta)}}$

given that the time on rough plane is n time more than the smooth plane

$t' = nt$

$\frac{1}{sin\theta}\sqrt{\frac{2*h}{g(1-\mu tan\theta)}} = n*\frac{1}{sin\theta}\sqrt{\frac{2*h}{g}}$

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

so friction coefficient is given by

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

Answer 2

There's actually a tiny mistake in that answer... While taking sin theta as common from acceleration on rough surface it'll be g(1-mu.cot theta) sin theta