A small body was launched up an inclined plane set at an angle `alpha=15^@` against the horizontal. Find the coefficient of friction, if the time of the ascent of the body is `eta=2.0` times less than the time of its descent.

Explanation

Case 1. When the body is launched up:
Let k be the coefficient of friction, `u` the velocity of projection and l the distance traversed along the incline. Retarding force on the block `=mg sin alpha+k mg cos alpha` and hence the retardation =`g sin alpha+kg cos alpha`.
Using the equation of particle kinematics along the incline,
`0=u^2-2(g sin alpha+kg cos alpha)l`
or, `l=(u^2)/(2(gsin alpha+kg cos alpha))` (1)
and `0=u-(g sin alpha+kg cos alpha)t`
or, `u=(g sin alpha+kg cos alpha)t`
Using (2) in (1) `l=1/2(g sin alpha+kg cos alpha)t^2` (3)
Case (2). When the block comes downward, the net force on the body `=mg sin alpha-km g cos alpha` and hence its acceleration `=gsin alpha-kg cos alpha`
Let, `t` be the time required then,
`l=1/2(g sin alpha-k g cos alpha)t^('^2)` (4)
From Eqs. (3) and (4)
`(t^2)/(t^('^2))=(sin alpha-k cos alpha)/(sin alpha+k cos alpha)`
But `t/t^'=1/eta` (according to the question),
Hence on solving we get
`k=((eta^2-1))/((eta^2+1)) tan alpha=0*16`