Posted by **urgent !!!!!!!!!** on Thursday, August 7, 2014 at 1:56pm.

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 ????

- physics -
**Damon**, Thursday, August 7, 2014 at 2:30pm
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 !!!

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