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

Consider the diagram where a body slides down from along an inclined plane of inclination `theta ( = 45^(@))`
On smooth inclined plane Acceleration of a body sliding down a smooth inclined plane
`a = g sin theta `
Here , `" " theta = 45^(@)`
`therefore " " a = g sin 45^(@) = (g)/(sqrt2)`
Let the travelled distance be s .
Using equation of motion ,s = `ut + (1)/(2) at^(2)` , we get
s = `0.t + (1)/(2) (g)/(sqrt2) T^(2)`
or `" " s = (gT^(2))/(2sqrt2)`
On rough inclined plane Acceleration of the body a = g (sin `theta - mu cos theta`)
= g (sin `45^(@) - mu cos 45^(@))`
`= (g(1-mu))/(sqrt2) " " ("As sin" 45^(@) = cos 45^(@) = (1)/(sqrt2))`
Again using equation of motion , s = ut + `(1)/(2) at^(2)` , we get
s = `0.t + (1)/(2) (g)/(sqrt2) T^(2)`
or `" " s = (gT^(2))/(2sqrt2)" " ....(i)`
On rough inclined plane Acceleration of the body a = g (sin `theta - mu cos theta`)
= g (sin`45^(@) - mu cos 45^(@)`)
= `(g(1-mu))/(sqrt2) " " ("As sin" 45^(@) = "cos" 45^(@) = (1)/(sqrt2))`
Again using equation of motion , s = ut + `(1)/(2) at^(2)` , we get
` s = 0 (pT) + (1)/(2) (g (1-mu))/(sqrt 2) (pT^(2)) `
or `" " s = (g(1-mu)p^(2)T^(2))/(2sqrt2) " " ......(ii)`
From Eqs. (i) and (ii) , we get
`(gT^(2))/(2sqrt2) = (g(1-mu)p^(2)T^(2))/(2sqrt2)`
or `" " (1- mu) p^(2) = 1 `
or `" " 1- mu = (1)/(p^(2))`
or `" " mu = (1 - (1)/(p^(2)))`