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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Options

A. `(1-(1)/(P^(2)))`
B. `(1+(1)/(P^(2)))`
C. `(1)/(P^(2))`
D. `-(1)/(P^(2))`

Answer

A

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^(@)`
`:. 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)`
One rough inclined plane Acceleration of the body
`a = g("sin" theta - mu "cos" theta)`
`=(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))/(sqrt2)(pT)^(2)`
or `s = (g(1-mu)p^(2)T^(2))/(2sqrt2)`
From eqs (1) 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 -(2)/(p^(2)))` .