A body of mass 5 xx 10^(-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 the time of descent .

A body of mass 5 xx 10^(-3) kg is launched upon a rough inclined plane

1.	A body of mass 5 xx 10^(-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 the time of descent .
Answer» Solution :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 ` .