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Nov 15 at 6:19 vote accept stickynotememo
Nov 12 at 10:54 comment added Steeven @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.
Nov 12 at 10:49 comment added stickynotememo So are we not subtracting the work done by the object from the energy of the object? If we did we should have zero work done on the object. Sorry if I'm not being clear enough.
Nov 12 at 10:48 comment added Steeven @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.
Nov 12 at 10:41 comment added stickynotememo 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.
Nov 12 at 10:37 comment added Steeven @stickynotememo Yes, exactly.
Nov 12 at 10:35 comment added stickynotememo 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
Nov 12 at 10:34 history answered Steeven CC BY-SA 4.0