A body of mass `5xx10^(-3)` kg is launched upon a rough inclined plane making an angle of `30^(@)` with the horizontal. Obtain the coefficient of friction between the body and the plane if the time of ascent is half of the time of descent.

Explanation

Here, ` m = 5 xx 10^(-3) kg , theta = 30^(@) , mu = ? T_(1) = (1)/(2) t 2 `
If f is the force of friction between the body and inclined plane , then
` f = mu R = mu mg cos theta`
Let `a_(1)` be the acceleration while ascending the plane as shown in Here friction is acting down the incline .
`mg sin theta + f = m a_(2)`
let `a_(2)` be acceleration while descending down the plane as shown in Here , friction is up the plane
`mg sin theta - f = m a_(2) `
Divide (ii) by (iii) `(mg sin theta + f )/(mg sin theta - f ) = a_(1)/(a_(2)) `
` (mg sin theta + mu mg cos theta)/(mg sin theta - mu mg cos theta )= a_(1)/(a_(2)) `
` (mg cos theta (tan theta + mu ))/(mg cos theta(tan theta - mu ) )=(a_1)/(a_(2)) `
If `t_(1)` is time of ascent and `t_(2)` is time of descent for length of plane `(l)` then
`l = (1)/(2) a_(1) t_(1)^(2) = (1)/(2) a_(2) t_(2)^(2)`
`a_(1)/(a_(2))= t_(2)^(2)/(t_(1)^(2)) = ((1)/(2))^(2) = (4)/(1)`
From (iv) , `(tan theta + mu )/(tan theta - mu )= (4)/(1)`
` tan theta + mu = 4 tan theta - 4 mu `
` 5 mu = 3 tan theta `
` mu = (3)/(5) tan 30^(@) = (3)/(sqrt3) `
` mu = (1.732)/(5) = 0.346 `
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