# 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.

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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 journey.

(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 answer ?

## 1 Answer

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(i) W = FS = – mg sin q × h = – 14·7 J is the W.D. by gravitational force in moving the body up the inclined plane.

W′ = FS = + mg sin θ × h = 14·7 J is the W.D. by gravitational force in moving the body down the inclined plane.

∴ Total W.D. round the trip, W1 = W + W′ = 0.

(ii) Force needed to move the body up the inclined plane,

F = mg sin θ + fk

= mg sin θ + µk

= mg sin θ + µk mg cos θ

∴ W.D. by force over the upward journey is

W2 = F × l = mg (sin θ + µk cos θ) l

= 18·5 J

(iii) W.D. by frictional force over the round trip,

W3 = – fk (l + l) = – 2fk

= – 2µk mg cos θ 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 θ – µk mg cos θ) l

= 10·9 J

⇒ K.E. of body = net W.D. on the body.

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