physics

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 ????

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asked by urgent !!!!!!!!!
  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 !!!

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

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    posted by Keerthana

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