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.

Options

A. 0.56
B. `0.60`
C. 0.75
D. 0.364

Answer

A

Explanation

For ascending Motion
Retardation ltBrgt `a_(A)=gsintheta{1+mucottheta}`
Let after distance S it stop
`V^(2)=0=u^(2)-2aS" "...(1)`
`V=0=u-at_(A)" "...(2)`
So from (1) & (2) `rArrt_(A)= sqrt((2S)/(a_(A)))" "...(3)`
For descending `a_(D)=gsintheta{1-mucostheta}` ltBrgt S = 0 + `(1)/(2) a_(D)t_(D)^(2)rArrt_(D)=sqrt((2S)/(a_(D)))" "...(4)`
because `t_(A)=(1)/(2) t_(D)`
So `mu=((3)/(4))tantheta=0.56`