# Work done?

If I push a heavy box on the ground but it doesn't move at all (due to friction), the work done on the box is of course zero. But where does the work I done go? From a website, it says the work is done on myself (in the muscle whatsoever), which I find weird because once it moves, the work is done against friction? Or in both cases it is done against friction?

Any help is appreciated. Thank you.

russ_watters
Mentor
If there is no motion, there is no work performed on the box, period. Your body, no the other hand, is constantly using chemical energy (equivalent to work), whether you are pushing on the box or not. When you push on the box and fail to move it, what can really be said, then is that the chemical to mechanical energy conversion efficiency is zero and all of that energy is converted to heat.

If there is no motion, there is no work performed on the box, period. Your body, no the other hand, is constantly using chemical energy (equivalent to work), whether you are pushing on the box or not. When you push on the box and fail to move it, what can really be said, then is that the chemical to mechanical energy conversion efficiency is zero and all of that energy is converted to heat.

ya, thanks, I get your meaning, but if the box is moving, for example, at a constant velocity. It is evident that the friction balances the applied force, but on the other hand the box's KE is not increasing. So the work done converts to heat energy again? But in this case, the box is really moving, the work done should not be zero?

If I push a heavy box on the ground but it doesn't move at all (due to friction), the work done on the box is of course zero. But where does the work I done go? From a website, it says the work is done on myself (in the muscle whatsoever), which I find weird because once it moves, the work is done against friction? Or in both cases it is done against friction?

Any help is appreciated. Thank you.

The way I see it, the work done in your example equals the electromagnetic atomic reactions integrated over related deformations of the ground-person-box circuit.

russ_watters
Mentor
ya, thanks, I get your meaning, but if the box is moving, for example, at a constant velocity. It is evident that the friction balances the applied force, but on the other hand the box's KE is not increasing. So the work done converts to heat energy again? But in this case, the box is really moving, the work done should not be zero?
Yes, on both counts. Ultimately, pretty much all mechanical energy (on earth, at least) ends up as heat. One exception would be carrying a weight up a hill and leaving it there...

russ_watters
Mentor
The way I see it, the work done in your example equals the electromagnetic atomic reactions integrated over related deformations of the ground-person-box circuit.
"The ground-person-box circuit" defoms once, but the body will continue to release energy indefinitely.

Yes, on both counts. Ultimately, pretty much all mechanical energy (on earth, at least) ends up as heat. One exception would be carrying a weight up a hill and leaving it there...

so you mean work done on the box = Fs
work done by friction = -fs
so no net work done on the box? All work done (both by me and the friction) dissipated as heat energy?

it sounds reasonable to me.

so what does "work done against friction" mean?

russ_watters
Mentor
so you mean work done on the box = Fs
work done by friction = -fs
so no net work done on the box? All work done (both by me and the friction) dissipated as heat energy?
Correct.
so what does "work done against friction" mean?
Not sure what you are still confused about - you explained it well enough above!