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I am solving work and energy objective from HC Verma.I got a question in which I have a basic doubt.The question is total work done on a particle is equal to change in kinetic energy always.But,I think it was wrong bcz if a body falls from a inclined plane and its final velocity is 'v' and it moves through height 'h' ,then total work done =mv 2 /2+mgh.But,in the question it is mention that total work done is only equal to change in kinetic energy.So,the work done must only be equal to mv 2 /2,butfrom thisway I can't get the answer.Please correct me where I was wrong.

I am solving work and energy objective from HC Verma.I got a question in which I have a basic doubt.The question is total work done on a particle is equal to change in kinetic energy always.But,I think it was wrong bcz if a body falls from a inclined plane and its final velocity is 'v' and it moves through height 'h' ,then total work done =mv2/2+mgh.But,in the question it is mention that total work done is only equal to change in kinetic energy.So,the work done must only be equal to mv2/2,butfrom thisway I can't get the answer.Please correct me where I was wrong.

Grade:12th pass

1 Answers

Kapil Khare
80 Points
5 years ago
The total work done by both external and internal forces is equal to change in kinetic energy.
But the Total work done only by external forces is equal to change in total energy.
In case of inclined plane,
Work done by gravity = mv2/2 + mgh
Work done by internal forces = change in potential energy = -mgh
So, in this case also total work done = Wext + Wint = mv2/2 + mgh – mgh
Total work done = mv2/2 = Change in kinetic energy

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