A body of mass `0.3kg` is taken up an inclined plane of length `10m` and height `5m,` and then allowed to slide down to the bottom again. The coefficient of friction between the body and the plane is `0.15`.
What is the
(a) work done by the gravitational force over the round trip ?
(b) work done by the applied force on the upward journey ?
(c) work done by the frictional force over the round trip ,
(d)kinetic energy of the body at the end of the trip ?

Explanation

`sin theta=(CB)/(CA)=0.5`
`:. theta=30^(@)`
(i)`W=FS=-mg sin qxxh=-14.7J` is the W.D. by gravitational force in moving the body up the inclined plane.
`W'=FS=+mg sin thetaxxh=14.7J` is the W.D. by gravitationa fore in moving the body down the inclined plane.
`:.` Total W.D. round the trip, `W_(1)=W+W'=0`
(ii) Force needed to over the body up the inclined plane
`F=mgsin theta+f_(k)`
`=mg sin theta+mu_(k)R`
`=mg sin theta+mu_(k)mg cos theta`
`:.` W.D. by force over the upward journey is
`W_(2)=Fxxl=mg(sin theta+mu_(k)cos theta)l`
`=18.5J`
(iii) W.D. by frictional over the round trip,
`W_(3)=-f_(k)(l+l)=-2f_(k)l`
`=-2mu_(k)mg cos theta l=-7.6J`
(iv) K.E. of the body at the end of round trip
`=` W.D by net force in moving the body down the inclined plane
`=(mg sin theta-mu_(k) mg cos theta)l`
`=10.9J`
`implies` K.E. of body `=` net W.D on the body.