A body of mass 0.3 kg is taken up an inclined plane to length 10 m and height 5 m 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 (i) work done by the gravitational force over the round trip? (ii) work done by the applied force over the upward joumey? (iii) work done by frictional force over the round trip? (iv) kinetic energy of the body at the end of the trip? How is the answer to (iv) related to the first three answers?

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 allowed 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 joumey? (iii) work done by frictional force over the round trip? (iv) kinetic energy of the body at the end of the trip? How is the answer to (iv) related to the first three answers?
Answer» Solution :`sin theta = (CB)/(CA) = 0.5` `therefore =30^@` (i) W= FS =- mg `sin theta xx H = -14.7 J `is the W.D. by gravitational FORCE in moving the body up the inclined plane. W. = FS + mg `sin theta xx h = 14.7`. Is the W.D. by gravitational force in moving the body down the inclined plane. `therefore` Total W.D. round the trip, `W_1 = W + Q. = 0 ` (ii) Force needed to move the body up the inclined plane, `F = mg sin theta + f_k = mg sin theta + mu_k R = mg sin theta + mu_k mg cos theta` `therefore` W.D. by force over the upward journey is `W_2 = F xx l = mg (sin theta + mu_k cos theta) l = 18.5 J` (iii) W.D. by frictional force over the round trip, `W_3 = - fk(l+l) = - 2 fkl = - 2 mu_k mg cos theta l = - 7.6 J ` (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 = 10.9 J `IMPLIES ` K.E. of body = net W.D. on the body