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

[A.T.]

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.

 

[sophiecentaur]

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?

 

[A.T.]

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.

 

[sophiecentaur]

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.

 

[A.T.]

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:

There has to be non-zero relative motion.

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.

 

[sophiecentaur]

Which is wrong, as I explain in post #22.

More of an opinion than an explanation.

The trouble with your thesis is that it applies at every point in any mechanical arrangement (all the way along a rigid rod). So is it helpful? There is a direct analogue with electrical circuit methods where it's only nodes that are of any interest and, by the use of simple +/- signs, you can solve anything that's been specified sufficiently. But I already said it's just a matter of preference

I already addressed the point that there has to be a reaction force in order for work to be done.

Do you have an example where that doesn't apply? That's the basis of post #1.

 

[A.T.]

Do you have an example where that doesn't apply?

Linear inertial force in an accelerating frame or centrifugal inertial force in a rotating frame have no reaction force but can do work.

 

[sophiecentaur]

Linear inertial force in an accelerating frame or centrifugal inertial force in a rotating frame have no reaction force but can do work.

I was hoping for an actual example. You surely have a real world situation to quote.

 

[Akbar Karim]

Yes, it is accurate to say work is the result of motion against an opposing force, as defined in physics. Work is done when a force is applied to an object, causing it to move a certain distance against resistance.

 

[sophiecentaur]

Yes, it is accurate to say work is the result of motion against an opposing force, as defined in physics. Work is done when a force is applied to an object, causing it to move a certain distance against resistance.

I was waiting for @A.T. to quote a practical example where it doesn't apply. I've been trying to unpick his last post to make sense of it. Working in non-inertial frames can yield surprises so I wait to be enlightened / surprised.

 

[gmax137]

Work is done when a force is applied to an object, causing it to move a certain distance against resistance.

Isn't "work" equal to the integral

$W =\int \vec F \cdot \, d\vec S$

Where's the "resistance"?

 

[HAYZEN]

I GUESS it is fine because every action has an equal and opposite reaction so, let us consider like you are applying some force to cause the work on the object then the force you apply to change it state is equal and opposite to the force or resistance that the object gives to in back for not changing it state by inertia , that is how mechanical actions basically works

 

[HAYZEN]

I was waiting for @A.T. to quote a practical example where it doesn't apply. I've been trying to unpick his last post to make sense of it. Working in non-inertial frames can yield surprises so I wait to be enlightened / surprised.

can you explain why non inertial frame?

 

[jbriggs444]

can you explain why non inertial frame?

It gets back to the question of whether there is a reaction force.

If we are using an inertial frame then there are no inertial forces arising from our choice of frame. All of the forces are interaction forces. Real forces. Not fictitious forces. So they all obey Newton's third and have an equal and opposite reaction force.

But @A.T. asks the obvious next question. "What sort of forces have no third law partner"? The answer is the fictitious forces arising from the use of a non-inertial frame of reference.

In the accelerating frame we can have an object accelerating due to a fictitious force being applied to it. It may gain or lose kinetic energy (as assessed in the accelerating frame) due to this force.

It is reasonable to extend the notion of "work" to cover this case. Just as with an interaction force, the work energy theorem still applies. The object subject to the fictitious force gains (or loses) kinetic energy according to the "work" done by the fictitious force.

 

[sophiecentaur]

The answer is the fictitious forces arising from the use of a non-inertial frame of reference.

That certainly sounds like a good answer except that energy is relative. Is there not work needed to cause an energy change in a non-inertial frame. This is why I've been asking for actual practical instances so that we can find a 'loop hole' - or not.

 

[sophiecentaur]

Isn't "work" equal to the integral

$W =\int \vec F \cdot \, d\vec S$

Where's the "resistance"?

He probably meant reaction(?).

 

[jbriggs444]

That certainly sounds like a good answer except that energy is relative. Is there not work needed to cause an energy change in a non-inertial frame. This is why I've been asking for actual practical instances so that we can find a 'loop hole' - or not.

Yes, energy is relative. Relative to a reference frame.

Measured against a non-inertial frame, energy is varying without an interaction force to explain it. The work done by the fictitious force explains the discrepancy.

For a practical example, consider a rocket ship with uniform acceleration. There is a "gravitational" force which can do work on falling objects.

 

[gmax137]

Getting back to the original post:

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.

I think @DrClaude has it right here:

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.

If you have a gas confined to a portion of a box, and then slide the partition out of the way, the gas will expand into the entire box without doing any work. If the scenario is gas in a cylinder with a piston, with the pressure sufficient to result in a net force on the piston, then the expansion does work on the piston.

I'm still a little unsure about the tie in between this, and the Newtonian
$W =\int \vec F \cdot \, d\vec S$

Maybe @Chestermiller could chime in on this.

 

[A.T.]

Work is done when a force is applied to an object, causing it to move a certain distance against resistance.

Work is done when a force acts on an object along a certain distance. Resistance is not required.

 

[Chestermiller]

Getting back to the original post:
I think @DrClaude has it right here:
If you have a gas confined to a portion of a box, and then slide the partition out of the way, the gas will expand into the entire box without doing any work. If the scenario is gas in a cylinder with a piston, with the pressure sufficient to result in a net force on the piston, then the expansion does work on the piston.

I'm still a little unsure about the tie in between this, and the Newtonian
$W =\int \vec F \cdot \, d\vec S$

Maybe @Chestermiller could chime in on this.

Let x be the displacement of the piston and A be the piston area. Then the force the gas exerts on the piston is PA, and the work done is $W=(PA)x = P(Ax)=P(\Delta V)$

 

[A.T.]

Isn't "work" equal to the integral

$W =\int \vec F \cdot \, d\vec S$

Where's the "resistance"?

He probably meant reaction(?).

OK, where's the "reaction"?

 

[sophiecentaur]

OK, where's the "reaction"?

The friction force and any force due to acceleration. Your equation contains an F; that's the one.

 

[A.T.]

Your equation contains an F; that's the one.

F is the force doing the work. Where is the "reaction" to F in the equation?

 

[sophiecentaur]

F is the force doing the work. Where is the "reaction" to F in the equation?

I have already asked you for a reactionless force (a concrete practical example). Nothing so far.

 

[A.T.]

I have already asked you for a reactionless force (a concrete practical example). Nothing so far.

See post #47

 

[sophiecentaur]

????

For a practical example, consider a rocket ship with uniform acceleration. There is a "gravitational" force which can do work on falling objects.

Sorry but I just don't get that.
1. A gravitational force is experienced by an object on the floor of the ship and the object gains KE.
2. Alternatively, the work done in expelling the ejecta is the force on ship and/or ejecta times distance moved.
3. Or you could be saying that the work done with the engine is changing the relative velocity of a free object before it hits the floor. But work is still being done on the ship in a conventional way as it accelerates to 'overtake' the free object.
How are those three changes in energy not examples of N3?

There is KE 'in abeyance' as the object and ship have relative motion change but that applies to any pair of objects in the Universe that are under different forces. So everything is involved in some sort of work translation with respect to everything else? I see how that's a possible view but is it of any use? You may be able to convince me more. Try.

 

[jbriggs444]

????

Sorry but I just don't get that.
1. A gravitational force is experienced by an object on the floor of the ship and the object gains KE.

No. In the frame of the ship, the object on the floor is just sitting there. It gains no KE.

It is subject to an interaction force from the floor of the ship. But that force is applied over zero displacement, so it does zero work. It is subject to the fictitious force from the acceleration of the ship. But that force is also applied over zero displacement so it also does zero work. KE is unchanged. Which is correct.

2. Alternatively, the work done in expelling the ejecta is the force on ship and/or ejecta times distance moved.

The [nearly] instantaneous action of expelling a particular bit of ejecta does the same amount of work in the accelerating frame as it does in the inertial frame. There is no time for acceleration to make a significant difference.

However, a bit of ejecta seen by the accelerating rocket will be accelerating away, gaining kinetic energy as it does so. This is explained by the fictious force from the acceleration of the frame.

3. Or you could be saying that the work done with the engine is changing the relative velocity of a free object before it hits the floor. But work is still being done on the ship in a conventional way as it accelerates to 'overtake' the free object.

This one.

A falling object is accelerating as viewed by an observer standing on the rocket floor. It is gaining kinetic energy. This is explained by the work done by the fictious force arising from the acceleration of the reference frame.

How are those three changes in energy not examples of N3?

The first one (object on the floor) did not involve a change in kinetic energy at all. If you think it did, you are frame jumping.

The second one (expelling the ejecta) is an ordinary interaction force. Of course it involves N3.

The second one (watching the ejecta after ejection) does not involve a N3 interaction. The gain in KE is from a fictitious force. Fictitious forces lack a third law partner.

The third one (watching a falling object) does not involve a N3 interaction. The gain in KE is from a fictitious force. Fictitious forces lack a third law partner.

There is KE 'in abeyance' as the object and ship have relative motion change but that applies to any pair of objects in the Universe that are under different forces. So everything is involved in some sort of work translation with respect to everything else? I see how that's a possible view but is it of any use? You may be able to convince me more. Try.

The only notion of "KE in abeyance" that I know of is potential energy. Some accelerating frames allow a potential to be defined. A uniformly accelerating frame is one example. A rotating frame is another.

In a uniformly accelerating frame, the potential is the familiar $\text{PE} = mgh$.

In a rotating frame, I believe that the centrifugal potential is $\text{PE} = -\frac{1}{2}m\omega^2r^2$

 

[sophiecentaur]

No. In the frame of the ship, the object on the floor is just sitting there. It gains no KE.

I like the description 'frame jumping'. It needs good discipline to avoid that unconsciously. it's as if the quantity 'Work' should only be restricted to inertial frames or else the KE in abeyance needs to be taken into account. Work is done to accelerate the ejecta of the ship and there you have an N3 pair and those GJ contained in the propellant are responsible for the Work.

Here's a thing. The object is suspended from the ceiling of the ship and 'weighs' ma. You let go of the object and the ship will (?) accelerate a bit faster. (in the observer frame) the object will gain KE and lose PE on the way to the floor. Can that energy be identified as Work? An accelerometer on the ship will record an 'upwards' acceleration as the crew notice the object accelerating 'downwards'. Doesn't that represent a third law pair in the same way that gravitational force between the Earth and a dropped object works on both bodies? Don't dismiss that as ravings until you've positively destroyed the idea because it gives work a chance to be the same in all situations. I would be much happier if it could be.

I have a feeling that there's a similar situation in a rotating frame where, as radius changes the centrifugal force changes for both the object and the central attractor (as you vary the string length).

 

[jbriggs444]

You let go of the object and the ship will (?) accelerate a bit faster.

Since it is specified that the rocket accelerates uniformly, this declaration is false.

If one discards the uniformly accelerating frame and uses the ship's rest frame instead then it is true that the object accelerates a bit faster. The same way that a fat man accelerates a bit faster toward the gravitating earth [in the Earth's rest frame] than a thin man. Yet we do not normally discard the notion of a fixed value for $g$.

You proceed to focus on this approximation error and miss the point entirely.

Since a not-uniformly-accelerating frame has a time-varying acceleration, the resulting potential field will be time varying. It would not be proper to define a single static potential for such a field. The notion of potential energy could not be usefully defined except, perhaps, as an approximation.

 

[hmmm27]

Since a not-uniformly-accelerating frame has a time-varying acceleration, the resulting potential field will be time varying. It would not be proper to define a single static potential for such a field. The notion of potential energy could not be usefully defined except, perhaps, as an approximation.

IMO, assuming $a$ is otherwise constant, and not including the impulses(right term?) at "string-cut" and "thud", the "in-air" segment of the journey can be seen as totally discontinuous from the rest, and PE&KE (object) and $a$(ship) are constant during that time.

(unless of course I skipped a post or ten upstream and that's not what you're talking about)