A body of mass `5xx10^(-3)` kg is projected upwards along a plane inclined at an angle of `30^(@)` with the horizontal. If the time required by the body to move up the incline is half the time required for it to slide down find the coefficient of friction between the surface and the body.

Explanation

Let the mass of the body be m, and its upward acceleration be `a_(1)` . Let the force of friction be f.
`:. " " ma_(1) = "mg sin"theta +f = "mg sin"theta +mu "mg cos" theta`
or, `a_(1) = g("sin"theta+mu cos theta)`

If `a_(2)` is the acceleration during the downward motion, then
`ma_(2) = "mg sin"theta -f = mgsin theta -mu mgcos theta)`
or, `" " a_(2) = g(sin theta -mu cos theta)`

`:. a_(1)/(a_(2))= (sintheta+mucostheta)/(sintheta-mu costheta)=(tantheta+mu)/(tantheta-mu)`
If the displacement along the inlined plane is s and the time taken to move up and to come down are `t_(1)` and `t_(2)` respectively then ,
`s=(1)/(2)a_(1)t_(1)^(2)=(1)/(2)a_(2)t_(2)^(2)" ":." "a_(1)=(2s)/(t_(1)^(2))" ""and"" "a_(2)= (2s)/(t_(2)^(2))`
`:. " "(a_(1))/(a_(2))=(t_(2)^(2))/(t_(1)^(2))=((t_(2))/(t_(1)))^(2)=2^(2)=4" " [because t_(1)=t_(2)/(2)]`
`:." " (tantheta+mu)/(tan-mu)=4 " ""or",mu= (3)/(5)tantheta = 0.6 tan30^(@)= 0.346`