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I apply a force on an object. The object applies an equal and opposite force on me. I realize the forces are on two distinct bodies and do not cancel each other out. Accordingly, the work done by me and the work done by the object are being applied on and by different objects.

However, the work done by me on the object is $F\times{}d$ and the work done by the object on me is the same. If a body does $x$ joules of work, and has $x$ joules of work done on it, there should be no change in the total energy of the object. The work done by it and on it is equal. How can this be possible? Has the object not gained kinetic energy?

Not a duplicate of Newton's Third Law and Work Done on a Body. My question is about the perspective of a single box rather than of the whole system.

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    $\begingroup$ This question is similar to: Given Newton's third law, why are things capable of moving?. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. $\endgroup$
    – Ruffolo
    Commented Nov 12 at 10:26
  • $\begingroup$ @Ruffolo I have clarified in the first few sentences that the force on the object and the force from the object are acting on different bodies, which I believe was the source of confusion in the linked question. Is there anything else in that question relevant to mine that I may be missing? $\endgroup$
    – stickynotememo
    Commented Nov 12 at 10:36

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You are absolutely right. You do work on the object, and the object does an equal amount of (negative) work on you. It cancels out and this system of you-and-object doesn't gain any energy and hence doesn't move as a whole.

But if you only look at the object and only draw a free-body diagram for the object, then there is no force from the object on anything. There is only a force on the object and thus only work done on the object. This work is unbalanced, and manifests as kinetic energy and thus motion of the object.

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  • $\begingroup$ Wouldn't forces applied by the object on something else not show up on a free body diagram? AFAIK the free body diagram only shows forces on the object, not forces it's applying $\endgroup$
    – stickynotememo
    Commented Nov 12 at 10:35
  • $\begingroup$ @stickynotememo Yes, exactly. $\endgroup$
    – Steeven
    Commented Nov 12 at 10:37
  • $\begingroup$ You state that there is no force from the object on anything (2nd paragraph), but there should be the N3L reaction force. This would mean the object is doing work on me, which should cancel out the work done on it. If no work is done there should be no change in energy - at least according to my understanding. $\endgroup$
    – stickynotememo
    Commented Nov 12 at 10:41
  • $\begingroup$ @stickynotememo In your first comment here you wrote: "AFAIK the free body diagram only shows forces on the object, not forces it's applying". That is exactly right. And because of that, if you only look at the object and its free-body diagram, then you will calculate a non-zero net work done on the object. So, this explains the energy gain of the object. But you are correct, that if you include yourself in a free-body diagram, then the object does do work on you simultaneously, and that is precisely why the system of you-and-object doesn't gain any energy. $\endgroup$
    – Steeven
    Commented Nov 12 at 10:48
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    $\begingroup$ @stickynotememo Remember that only forces do work. Not objects, forces. So, a better phrasing than "the object does work on you" would be: "the force exerted by the object does work on you". If you do not include this force in a scenario, then the energy transfered by means of this force will not be involved either. $\endgroup$
    – Steeven
    Commented Nov 12 at 10:54

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