Is it accurate to say work is motion against an opposing force?

  • #1
zenterix
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TL;DR Summary
I saw the following quote in a thermodynamics book:

"The fundamental property in thermodynamics is work: work is done to achieve motion against an opposing force"

I'd like to know if this is accurate.
Is the following quote accurate:

"The fundamental property in thermodynamics is work: work is done to achieve motion against an opposing force"

Specifically, I am asking about the portion after the colon. I am a little confused by the notion of an opposing force.

Let's say we are in outer space and there is an object of mass ##m##. If we apply a force ##f## to this object it will accelerate, it will be displaced, and as far as I know, by definition we will be doing work. Of course, when we apply the force, there is a third law force in the opposite direction acting on us.

Similarly, if we have two charged particles, they each exert a force on the other of same magnitude and opposing directions. Thus, if we have particle A and particle B, particle A does work on particle B, and vice versa. Particle A is exerting a force and achieves motion in B. However, I don't see how this work is "against an opposing force" in an intuitive way. Sure there is an opposing force, but even in the absence of his opposing force, A would be doing work on B (because it would be exerting a force and there would be displacement of B).

Thus, I am confused that the definition of work used above is based on an opposing force.
 
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  • #2
How is the acceleration being achieved ? [edit: could be wrong, of course]
 
  • #3
zenterix said:
TL;DR Summary: I saw the following quote in a thermodynamics book:

"The fundamental property in thermodynamics is work: work is done to achieve motion against an opposing force"
Personally, I always ignore statements like this because they are too vague. Presumably the writer knows exactly what they are talking about, but it's too ambiguous to convey a precise understanding. In general, it's not true, IMO, as there is no need for an opposing force for a single force to do work. Gravity does work on an object in freefall, but there is no opposing force on the object.
 
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  • #4
zenterix said:
Is the following quote accurate:

"The fundamental property in thermodynamics is work: work is done to achieve motion against an opposing force"
It likely refers to the special case of motion at constant speed against an opposing force (doing negative work), which requires a propelling force (doing positive work) to maintain constant speed.

Or it really means the Newton's 3rd Law force by "opposing force" acting on whatever does work on the object, not on the object itself. But that would be rather confusing.

In either case, it's to vague to be of any use.
 
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  • #5
You have taken the quote out of context. Work in thermodynamics is not the same as work in Newtonian mechanics.

If you consider, for example, the expansion of a gas, if the expansion is against a force, the gas will do work, but if the expansion is in a vacuum, the work is zero.
 
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  • #6
Note, the wording of the quote does not match your title.
 
  • #7
zenterix said:
work is done to achieve motion against an opposing force"
To do work there has to be a reactive force - you can't 'push against nothing'. So the statement is fair enough and Work is equal to Force times Distance moved in the direction of the force. No Force or no Distance = no Work.
 
  • #8
The work does not need to "achieve" motion. The motion can already be there. Or can even be in the opposite direction, in which case, the work done on this object by this force on this object will be negative.

Work done is force times (parallel) displacement of the body acted upon. Regardless of what is responsible for the displacement.

There is room for a quibble about where the motion is measured. It can be motion of the target at the interface or motion of the target's center of mass. This gives rise to two distinct notions of work and two distinct versions of the work energy theorem.
 
  • #9
jbriggs444 said:
The work does not need to "achieve" motion. The motion can already be there.
There has to be non-zero relative motion. In some situations you need to be smart to recognise this. for instance in a slipping clutch in a moving car, work is done and it generates heat.
I reckon that just because you can't spot it, doesn't mean its not there.
 
  • #10
sophiecentaur said:
There has to be non-zero relative motion. In some situations you need to be smart to recognise this. for instance in a slipping clutch in a moving car, work is done and it generates heat.
I reckon that just because you can't spot it, doesn't mean its not there.
No non-zero relative motion is required.

Consider, for example, the meshing gears on a transmission. There is no relative motion. Yet work is transferred. Positive in one direction, negative in the other.

I agree that non-zero relative motion is required if one wants the sum of the work done in the one direction and the work done in the other direction to achieve a non-zero total.
 
  • #11
jbriggs444 said:
Consider, for example, the meshing gears on a transmission. There is no relative motion. Yet work is transferred. Positive in one direction, negative in the other.
I think you are demonstrating my point that it may be hard to identify the motion that's counted. There is a distinction between work transferred and work 'done'. Some work is 'done' in meshing gear teeth, even the best designed ones.
 
  • #12
DrClaude said:
You have taken the quote out of context. Work in thermodynamics is not the same as work in Newtonian mechanics.

If you consider, for example, the expansion of a gas, if the expansion is against a force, the gas will do work, but if the expansion is in a vacuum, the work is zero.
I have a hard time agreeing that the two concepts are different. Work in thermodynamics is defined starting with the definition of work in mechanics: a dot product of force and displacement vectors. In thermodynamics, the magnitude of force is replaced by pressure times area.

I suspect that when I study thermodynamics further, an in particular statistical mechanics, I will gain insight related to this whole topic. We have gas particles with some velocity hitting a container internal surface and the velocity changes: this is a change in momentum, and thus there was an impulse and hence a force applied on the surface. If the surface has zero mass, then the applied force is zero and so no work is done.

I don't know if this is true but it is my guess.

I just can't understand that the claim that work in thermodynamics is different from work in newtonian mechanics can be true.
russ_watters said:
Note, the wording of the quote does not match your title.
Sure but for some reason this website only allows very small titles so it was not possible to get the whole quote in.
 
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  • #13
sophiecentaur said:
To do work there has to be a reactive force - you can't 'push against nothing'. So the statement is fair enough and Work is equal to Force times Distance moved in the direction of the force. No Force or no Distance = no Work.
What about gravity? If the Newton's third law were not true for gravity, and it were the case that earth attracted an object but not vice-versa, then the earth would do work on the object without a reactive force.

But even with the third law being true, the gravitational force of earth on an object does work. The gravitational force of the object on earth seems unrelated, for purposes of work, to the gravitational force of the earth on the object.
 
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  • #14
zenterix said:
the gravitational force of earth on an object does work.
Not if there is no change oh height of a mass relative to ground.
 
  • #15
sophiecentaur said:
Not if there is no change oh height of a mass relative to ground.
Obviously. But we're discussing your claim: that to do work there has to be a reactive force. So I assumed a situation where work is being done.
 
  • #16
zenterix said:
What about gravity? If the Newton's third law were not true for gravity, and it were the case that earth attracted an object but not vice-versa, then the earth would do work on the object without a reactive force.
We don't live in that universe.

zenterix said:
But even with the third law being true, the gravitational force of earth on an object does work. The gravitational force of the object on earth seems unrelated, for purposes of work, to the gravitational force of the earth on the object.
It seems like your concern/complaint is that they are picking the "wrong" force in an N3 pair? "Work done on" is the same as "work done by" in a universe that obeys Newton's 3rd law, so I don't understand why this would be a problem. Or is the issue here that you don't believe N3 is universally true?
 
  • #17
russ_watters said:
We don't live in that universe.It seems like your concern/complaint is that they are picking the "wrong" force in an N3 pair? "Work done on" is the same as "work done by" in a universe that obeys Newton's 3rd law, so I don't understand why this would be a problem. Or is the issue here that you don't believe N3 is universally true?
Doesn't matter what universe we live in. We're talking about definitions of concepts here in an analytical sense. What universe we apply the concepts to is a separate question that relates to the correctness or usefulness of the concepts in making accurate predictions.

I am definitely not questioning Newton's third law. I am just arguing against the argument that for there to be work there has to be a reactive force, given the definition of work in physics.
 
  • #18
zenterix said:
Doesn't matter what universe we live in. We're talking about definitions of concepts here in an analytical sense. What universe we apply the concepts to is a separate question that relates to the correctness or usefulness of the concepts in making accurate predictions.
Well:
1. Definitions can be whatever we want them to be....as long as they satisfy #2. And by "we" I mean the physics community that created this definition. In that sense, it isn't really up for debate by us here.
2. The definition does need to/does work to make accurate predictions.
zenterix said:
I am definitely not questioning Newton's third law. I am just arguing against the argument that for there to be work there has to be a reactive force, given the definition of work in physics.
Again, work done on vs work done by. One is the applied force, the other is the reactive force. The choice is arbitrary and the result is the same (aside from direction). N3 says "there has to be" both. So this point of logic doesn't appear to me to be meaningful.

So if you're saying "work done by" is about applied force not reaction force so it doesn't depend on reaction force, ok that's true. But what has really changed in our accurate description of the universe?

And this gets even messier when you try to decide which is which.
 
  • #19
The object of discussion is a specific statement of a definition of work present in a specific book I read. The debate here isn't to try to change physics. The debate is about trying to obtain a precise english-language statement of the definition of work that is better than the one in the book. Or at least to write out the meaning of the book's definition in more words and in a more didactical manner.

So, not sure what you're getting at with your 1.
 
  • #20
zenterix said:
I have a hard time agreeing that the two concepts are different. Work in thermodynamics is defined starting with the definition of work in mechanics: a dot product of force and displacement vectors. In thermodynamics, the magnitude of force is replaced by pressure times area.

I suspect that when I study thermodynamics further, an in particular statistical mechanics, I will gain insight related to this whole topic. We have gas particles with some velocity hitting a container internal surface and the velocity changes: this is a change in momentum, and thus there was an impulse and hence a force applied on the surface. If the surface has zero mass, then the applied force is zero and so no work is done.
In thermodynamics, work is essentially in used in opposition to heat. Conservation of energy is written as
$$
\Delta U = q - w
$$
(sign may be different depending on the convention), the idea being simply to distinguish between energy transferred as heat (i.e., because of a temperature difference) and the rest. Work is important because it is the only form of energy that it useful, that can be used to achieve some other goal.

When considering gases, work is mostly mechanical, but there are other forms of work in thermodynamics, such as chemical and electrical.
 
  • #21
sophiecentaur said:
There has to be non-zero relative motion.
An elevator going up at constant is doing work on the person inside, without relative motion between them.

In a frame that moves downward at constant speed, a table is doing work on the book resting on it, without relative motion between them.
 
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  • #22
zenterix said:
I am definitely not questioning Newton's third law. I am just arguing against the argument that for there to be work there has to be a reactive force, given the definition of work in physics.
In non-inertial frames of reference there are indeed inertial forces, which do not obey Newton's 3rd Law (they have no equal but opposite force), but sill can do work.
 
  • #23
A.T. said:
An elevator going up at constant is doing work on the person inside, without relative motion between them.
I frequently point out the problem in identifying who 'does the work' and to whom. We accept that work is done. The elevator is just an intermediate mechanism and no work needs to be done at the contact surface nor within any of the parts of the elevator (ignore possible hysteresis in stretching cables or in slippage / friction) All the diagrams that we draw when discussing mechanisms ignore the intermediate forces in rigid structures because there's no relative internal movement.
It's sufficient to talk of "work done" and the resulting amount.
 
  • #24
sophiecentaur said:
I frequently point out the problem in identifying who 'does the work' and to whom. We accept that work is done. The elevator is just an intermediate mechanism and no work needs to be done at the contact surface nor within any of the parts of the elevator (ignore possible hysteresis in stretching cables or in slippage / friction) All the diagrams that we draw when discussing mechanisms ignore the intermediate forces in rigid structures because there's no relative internal movement.
It's sufficient to talk of "work done" and the resulting amount.
You speak of "does the work" to denote a transfer of mechanical energy to a system.

You speak of "work is done" to denote the creation or destruction of mechanical energy when the work that A does on B plus the work that B does on A sums to something non-zero.

We agree that there is a distinction to be made between the two. But switching from the active voice to the passive voice seems like a poor choice of terminology to make the meaning clear.
 
  • #25
This is along the same lines as using ‘acceleration’ and ‘deceleration’. Most of us on PF would prefer Acceleration and a sign. That makes sense, imo and avoids confusion due to preconceptions about a situation.
Work plus a sign would be consistent with this. That avoids confusion and the problem “how can brakes do work?”. There’s an answer, of course but why should it be needed?
But these things are matters of preference as much as anything.
 
  • #26
sophiecentaur said:
The elevator is just an intermediate mechanism and no work needs to be done at the contact surface
The normal contact force at the feet is doing work on the person. This follows from the definition of work and energy conservation, since the potential energy of the person increases.
 
  • #27
zenterix said:
The object of discussion is a specific statement of a definition of work present in a specific book I read. The debate here isn't to try to change physics. The debate is about trying to obtain a precise english-language statement of the definition of work that is better than the one in the book. Or at least to write out the meaning of the book's definition in more words and in a more didactical manner.

So, not sure what you're getting at with your 1.
English is not the language of physics, so I don't see a lot of value in trying to word-smith this as long as people can use it to correctly apply the math. There is no one right/best way to say something in English.
 
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  • #28
A.T. said:
The normal contact force at the feet is doing work on the person.
Mais oui! But, if there is whole chain of forces and mechanisms involved, the "by's" and "on's" can mount up and the 'flow' of energy may not be self evident. The same could be said about + and - signs but that's how we normally do things in other cases. Maths usually works best.

But each to his own.
 
  • #29
sophiecentaur said:
But, if there is whole chain of forces and mechanisms involved, the "by's" and "on's" can mount up and the 'flow' of energy may not be self evident.
You don't have to consider the whole chain to apply the definition of work to one particular interaction within it.

sophiecentaur said:
The same could be said about + and - signs but that's how we normally do things in other cases.
The sign follows from the definition via the dot product of force and displacement.
 
  • #30
A.T. said:
You don't have to consider the whole chain to apply the definition of work to one particular interaction within it.
OK, riddle me this: you have a complicated mechanical, arrangement with more than one motor and a number of sources of friction - even slippy clutches. You will produce a number of simultaneous equations which have just enough to get a solution. Where would you put the "by's" and "on's" in those equations? You would be dragged, kicking and screaming into world of mathematical signs if you wanted a solution.
To be realistic, I believe you'd just stick with a sign convention at each 'node' and you wouldn't actually need to know the by or the on at each point. You are a careful sort of guy so it would all come out in the wash and you'd get the right answer.
 
  • #31
sophiecentaur said:
OK, riddle me this: you have a complicated mechanical, arrangement with more than one motor and a number of sources of friction - even slippy clutches. You will produce a number of simultaneous equations which have just enough to get a solution. Where would you put the "by's" and "on's" in those equations? You would be dragged, kicking and screaming into world of mathematical signs if you wanted a solution.
To be realistic, I believe you'd just stick with a sign convention at each 'node' and you wouldn't actually need to know the by or the on at each point. You are a careful sort of guy so it would all come out in the wash and you'd get the right answer.
Sorry, I don't really understand your point. The more complex a problem is, the more important it is to apply the formal definitions and rules systematically, and not trying to take short cuts and hoping the right answer comes out in the wash.
 
  • #32
A.T. said:
Sorry, I don't really understand your point. The more complex a problem is, the more important it is to apply the formal definitions and rules systematically, and not trying to take short cuts and hoping the right answer comes out in the wash.
Are you saying that, in a complex problem, you would need to specify, at each node, which way “on” points? How would circuit analysis calculations work if you had to know the direction of power flow before you could start? You choose arbitrary directions for currents and the answer falls out but you don’t assume power flow.
I think you must have mis-understood the point I’m trying to make. What are you defending?
 
  • #33
sophiecentaur said:
I think you must have mis-understood the point I’m trying to make.
Is your point relevant to the OPs question? It is about the general definition of work, not about practical approaches to solve problems.
 
  • #34
A.T. said:
Is your point relevant to the OPs question? It is about the general definition of work, not about practical approaches to solve problems.
Very much. I already addressed the point that there has to be a reaction force in order for work to be done. I have also made the point that requiring to know whether the work is done on or by can only add to the difficulty.

I notice you are bailing out just when I challenge you about 'on or by' in complicated situations because it is often impossible to decide which way round it actually is until you've solved the equations.
 
  • #35
sophiecentaur said:
I already addressed the point that there has to be a reaction force in order for work to be done.
Which is wrong, as I explain in post #22. And the below is also wrong, as explained in post #21:
sophiecentaur said:
There has to be non-zero relative motion.

sophiecentaur said:
I notice you are bailing out just when I challenge you about 'on or by'
I don't understand your challenge, and what it has to do with my objections to your statements above.
 

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