When body slides down from rest along smooth inclined plane making angle of `45^(@)` with the horizontal, it takes time `T` When the same body slides down from rest along a rough inclined plane making the same angle and through the same distance it is seen to take time `pT`, where p is some number greater that 1. Calculate late the coefficient of friction beween the body and the rough plane.
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Explanation

When a body slides down a smooth inclined plane `a = g sin 45^(@) = (g)/(sqrt(2))" "( :' sin 45^@ = cos45^(@)= (1)/(sqrt(2)))`
From `s = ut + (1)/(2)at^(2)`
`s = 0 + (1)/(2) (g)/sqrt2 T^(2)`
When the same body slides down the rought inclined plane
`a = g (sin theta - mu cos theta) = g (1-mu)/sqrt(2)`
Again, from `s = ut + (1)/(2) g (1-mu)/sqrt(2) (pT)^(2)`
From (i) and (ii) `g (1-mu)/sqrt(2) p^(2) T^(2) = (gT^(2))/(2 sqrt(2))` .