A given object takes n times as much time to slide down a `45^(@)` rough incline as it takes to slide down a perfectly smooth `45^(@)` incline. The coefficeint of kinetic friction between the object and the incline is given by:

Options

A. `(1 - (1)/(n^(2)))`
B. `(1)/(1 - n^(2))`
C. `sqrt((1 - (1)/(n^2)))`
D. `sqrt((1)/(1 - n^2))`

Answer

A

Explanation

With friction - `a_(1) = g sin theta - mu_(k) g cos theta`
`therefore S = (1)/(2) a_(1) t_(1)^(2) …… (i)`
Without friction - `a_(2) = g sin theta`
`S = (1)/(2) a_(2) t_(2)^(2) ….. (ii) `
By eq. (i) and (ii)
`(1)/(2) a_(1) t_(1)^(2) = (1)/(2) a_(2) t_(2)^(2) " "` [ Since `t_(1) = n t_(2)`]
`implies (1)/(2) a_(1) n^(2) t_(2)^(2) = (1)/(2) a_(2) t_(2)^(2)`
`implies a _(2) = n^(2) a , implies (a_(2))/(a_(1)) = n^(2)`
`implies (g sin theta)/(g sin theta - mu_(k) g cos theta) = n^(2)`
`implies 1 -0 mu_(k) cot theta = (1)/(n^2)`
`mu_(k) cot theta = 1 - (1)/(n^2)`
as `theta = 45^(@) , mu_(k) = (1 - (1)/(n^2))`
Aliter :
`mu_(k) = tan theta (1 - (1)/(n^2)) = (1 - (1)/(n^2))`. (Use direct formula )