Where Does the Work Go When Pushing a Heavy Box Against Friction?

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Discussion Overview

The discussion revolves around the concept of work done when pushing a heavy box against friction, particularly focusing on scenarios where the box does not move and where it moves at a constant velocity. Participants explore the implications of work, energy conversion, and the role of friction in these contexts.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants assert that if there is no motion, then no work is performed on the box, and any energy expended by the person is converted to heat, suggesting a zero efficiency in energy conversion.
  • Others propose that when the box moves at a constant velocity, the work done is still significant, as it involves balancing the applied force and friction, even though the kinetic energy of the box does not increase.
  • A participant introduces the idea that work done can be viewed as the electromagnetic atomic reactions involved in the deformations of the system (ground-person-box circuit).
  • There is a discussion about the concept of "work done against friction," with some participants questioning its meaning in the context of energy dissipation as heat.

Areas of Agreement / Disagreement

Participants generally agree that when the box does not move, the work done on it is zero, but there is disagreement regarding the implications of work done when the box is in motion at constant velocity and how energy is transformed in these scenarios.

Contextual Notes

Participants express uncertainty about the efficiency of energy conversion and the definitions of work done against friction, indicating that these concepts may depend on specific interpretations and assumptions.

chingcx
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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.
 
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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.
 
russ_watters said:
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?
 
chingcx said:
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.
 
chingcx said:
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...
 
Loren Booda said:
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.
 
russ_watters said:
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?
 
chingcx said:
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!
 

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