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

For upward motion: upward retardation `a_(1)=(muN+mgsintheta)/(m)`
`a_(1)=mugcos30^(@)+gsin30^(@)=(sqrt3mu+1)g/2`

`becauses=1/2a_(1)t_(1)^(2)thereforet_(1)=sqrt((2s)/(a_(1)))=sqrt((4s)/((sqrt3mu+1)g))`
For downward motion: downward acceleration `a_(2)=(mgsintheta-muN)/(m)`
`s_(2)=sin30^(@)-gcos30^(@)=(1-sqrt3mu)g/2`

`impliest_(2)=sqrt((2s)/(a_(2)))=sqrt((4s)/((1-sqrt3mu)g))`
Now according to question `2t(1)=t_(2)`
`implies2sqrt((4s)/((sqrt3mu+1)g))=sqrt((4s)/((1-sqrt3mu)g))`
`implies(1-sqrt3mu)/(1+sqrt3mu)=1/4impliesmu=(sqrt3)/(5)`