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I'm little confused here. Work done on the body when we lift it and put it on the table is zero, because according to work energy theorem, change in kinetic energy of the body is zero. So, the net work done is zero.

Fine, but now the object has $mgh$ (where $m=$mass of body; $h=$height of body) amount of potential energy stored in it. If the net work done is zero then who has increased its energy?

I'm confused and unable to think it through kindly help.

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""according to work energy theorem, change in kinetic energy of the body is zero. So, the net work done is zero. "" Where did You get that from? –  Georg Oct 29 '11 at 15:49
@Georg, read it carefully. The work done on the body is zero because the change in kinetic energy of the body is zero, according to the work energy theorem $W=\Delta K$. –  Andyk Oct 29 '11 at 16:00
"Work done" is all work done, if not stated otherwise. (My opinion) –  Georg Oct 29 '11 at 16:07
you, the lifter has supplied the energy :) –  user1355 Oct 29 '11 at 16:29
Related: physics.stackexchange.com/q/9089/2451 –  Qmechanic Oct 29 '11 at 16:43

3 Answers 3

up vote 1 down vote accepted

You did net work on the body. Gravity did negative net work on the body. The over all work done was zero. The original confusion arose because the work-energy theorem demands we calculate the change in kinetic energy using the net force on the body, but your question considered only the force exerted by you, and ignored that exerted by gravity.

If the body has mass $m$, you were putting a force $mg$ on it to raise it at constant speed. The work-energy theorem says that if you had done this when there were no other forces on the body, the body would have gained kinetic energy $mgh$ as you moved it from the floor to the table.

That analysis ignores gravity, though. Gravity pulled down on the box with force $-mg$. This means the work done by gravity was $-mgh$, and so the total work done on the box was zero. This makes sense because if you lift the box at constant speed, the net force on the box is zero by $F = ma$.

If the box starts by sitting stationary on the floor, there would have to be some small net work done on the box to get it started going up. There would have to be some small negative net work on it to get it to stop.

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Don't complicate things for youself.Net work done is a simple term , which means the transfer in energy to do certain work(where energy transfers , not transform into its original form) .In the above case energy is transferred and then transform into its original form (another e.g is "SPRING").

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There is work because the box transfer from one place to another which means there is a change in position.

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This does not follow in general: free motion with non-zero velocity can move an object with no work done. –  dmckee Jun 25 '13 at 15:38
It all depends on what you include and exclude in work and energy. In the view analyzed by the question, both the work done by the person and the negative work done by gravity count, with a total of zero. However, unless gravity gets turned off there is effectively a potential energy not obviously accounted for. The way that is accounted for is that a situation has been set up such if the table suddenly goes away, then gravity will do work on the object. Perhaps the object will then do work on the person's toe, returning to a net total of zero. –  Chris Stratton Jun 25 '13 at 17:11

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