Question

A block of mass m takes time t to slide down on a smooth inclined plane of angle of inclination theta and height h. If same block slids down on a rough inclined plane of same angle of inclination and some height and takes time n times of initial value then coefficient of friction b/w block and inclined plane is ????

Answer 1

Well, there is a hard way and an easy way.

Easy way
height = h
angle = A

with no friction
KE gained in slide = m g h

(1/2) m v^2 = m g h
v = sqrt (2 g h)
average speed = v/2 = .5 sqrt (2 g h)
time = average speed * h/sin A
t = .5 h sqrt (2 g h) /sin A

now with work done against friction

Ke = m g h - mu m g cos A (h/sinA)
(1/2) m v^2 = m g h (1 - mu cot A)
v = sqrt [2 g h (1 - mu cot A)]
average speed = v/2
time = .5 h sqrt [2 g h(1 - mu cot A)]/sin A
so
time ratio = n
= sqrt (1 - mu cot A)/sqrt(1)

n^2 = 1 - mu cot A
mu cot A = 1- n^2
mu = (1 - n^2)/cotA
CHECK MY ALGEBRA !!!

Answer 2

Your explanation is good but [1+1/n^2]tanA is the answer . Can you explain how ? I am confused .