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 ????
Well, there is a hard way and an 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
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 !!!
posted by Damon