A body of mass 0.3 kg is taken up an inclined plane to length 10 m and height 5 m and then allowwed to slide down to the bottom again.The coefficient of friction between the body and the plane is 0.15. what is the (i) work done by the gravitational force over the round trip. (ii) work done by the applied force over the upward journey. (iii) work done by frictional force over the round trip. (iv) Kinetic energ of the body at the end of the trip. How is the answer to (iv) related to the first thre answer?

A body of mass 0.3 kg is taken up an inclined plane to length 10 m and

1.	A body of mass 0.3 kg is taken up an inclined plane to length 10 m and height 5 m and then allowwed to slide down to the bottom again.The coefficient of friction between the body and the plane is 0.15. what is the (i) work done by the gravitational force over the round trip. (ii) work done by the applied force over the upward journey. (iii) work done by frictional force over the round trip. (iv) Kinetic energ of the body at the end of the trip. How is the answer to (iv) related to the first thre answer?
Answer» Solution : `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 thetrip, `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.