A piece of ics slides down a `45^(@)` incline in twice the time it takes to slide down a frictionless `45^(@)` incline . What is the coefficient of friction between the ice and the incline ? .

Explanation

Here , `theta = 45^(@) , s_(1) = s_(2) , u = theta `
On the rough incline, ` a_(1) = g (sin theta - mu cos theta )`
`t_(1)` = time taken
On the frictionless incline , ` a_(2) = g sin theta`
`t_(2)` = time taken, and `t_(1) = 2 t_(2)`
From `s = ut + (1)/(2)` at 2
`s_(1) = 0 + (1)/(2) g sin (sin theta - mu cos theta) t_(2)^(1)`
and `s_(2) = 0 + (1)/(2) g sin 0.t_(2)^(2)`
As `s_(1) = s_(2)`
`:. (1)/(2) g (sin theta - mu cos theta ) t_(1)^(2)= (1)/(2) g sin 0 . t_(2)^(2) `
`(sin theta - mu cos theta )/(sin theta ) = t_(2)^(2)/(t_(2)^(1)) = (t_(2)^(2))/((2t_(2))^(2)) = (1)/(4) `
` 1 - mu cot theta = (1)/(4)` or `mu cot theta= 1 - (1)/(4)= (3)/(4)`
`mu = (3)/(4 cot theta )`.