B The work done by two objects on each other

[John Constantine]

TL;DR Summary

The work done by two objects on each other.

The work done by A on B as Wab, and the work done by B on A as Wba.

I have always thought that Wab = -Wba generally always holds true. Looking at it from the perspective of energy transfer:

1. "A do 10J of work on B."
2. "A transfer 10J of energy to B."
3. "B received 10J of energy from A."
4. "B do -10J of work on A."

If 1 is true, then 2 through 4 all hold true, making Wab = -Wba valid.

However, consider the case where two people of different masses push each other from rest. In this scenario, it is intuitively clear that Wab and Wba are not the same. The kinetic energy is not conserved in this case.

I asked about this on another physics forum, and the response was that sometimes it is true (e.g., in the case of a perfectly elastic collision) and sometimes it is false (e.g., when two objects of different masses attract each other due to gravity, or when two stationary objects of different masses push each other). In other words, the statement is sometimes correct and sometimes incorrect. What conditions are necessary for this statement to be true, and what conditions make it false?

 

[PeroK]

Newton's third law involves forces and momentum. In general, it's the force (hence total impulse = change in momentum) of A on B that is equal in magnitude and opposite in direction to the force (hence impulse) of B on A.

This is not true for energy. In general, $W_{ab} = - W_{ba}$ is not true.

 

[jbriggs444]

However, consider the case where two people of different masses push each other from rest. In this scenario, it is intuitively clear that Wab and Wba are not the same. The kinetic energy is not conserved in this case.

I asked about this on another physics forum, and the response was that sometimes it is true (e.g., in the case of a perfectly elastic collision) and sometimes it is false (e.g., when two objects of different masses attract each other due to gravity, or when two stationary objects of different masses push each other). In other words, the statement is sometimes correct and sometimes incorrect. What conditions are necessary for this statement to be true, and what conditions make it false?

One first needs to be clear about what sort of work is being considered.

Is it net work (net force times the [parallel] displacement of the center of mass)?

Is it mechanical work (individual force times the [parallel] displacement of the material of the target at the place where a particular force acts.

Based on the scenarios that you have called out, it appears that you are considering a particular interaction force times the [parallel] displacement of the center of mass. This is what I would call the "center of mass" work associated with a particular force.

For a force pair to do no "center of mass" work, the component of the relative velocity between the two centers of mass in the direction of the force pair must be zero.

Two balls springing apart have non-zero relative motion. Non-zero work is done by the repulsive force pair.

Two planets revolving about one another in a circular orbit have zero relative motion (along the force axis). The attractive force pair does zero work.

A hand on the end of a wrench has zero motion relative to the wrench. The force pair between hand and wrench does zero work.

A sanding block and a slab of walnut have non-zero relative motion in the direction of the frictional force between them. The frictional force pair between block and wood absorbs non-zero work.

A sanding block and a slab of walnet have zero relative motion in the direction of the normal force between them. The normal force pair between block and wood does zero work.

A man holding the center of mass of man+hand+sanding block steady as he moves a sanding block over the walnut has zero relative motion in the direction of the frictional force between block and a stationary slab of walnut. The frictional force pair between man+hand+block and wood does zero center of mass work.

A car driving along the highway has non-zero relative motion in the direction of the frictional force between car and road. The force pair between car and road does non-zero work.


Turning our attention to mechanical work...

For a force to do no "mechanical" work, the relative velocities of the parts of the two systems that exert force on one another must have zero relative motion in the direction of the force pair.

The contact patch on a drive wheel on an automobile has [near] zero relative motion relative to the road upon which it rides. The force pair between tire and road does [near] zero mechanical work

 

[A.T.]

I have always thought that Wab = -Wba generally always holds true.

A very basic counter example is kinetic friction, that dissipates some of the mechanical energy at the interface, so Wab + Wba < 0.

You can also have Wab + Wba > 0 when mechanical energy is generated from some other form of energy at the interface.

But note that this is very dependent the level of abstraction, or what you include in the interface between the two objects, and whether you consider "real work" or "center of mass work", as mentioned above by @jbriggs444.

 

[sophiecentaur]

TL;DR Summary: The work done by two objects on each other.

However, consider the case where two people of different masses push each other from rest. In this scenario, it is intuitively clear that Wab and Wba are not the same.

Why would it have to be the same? We have to resist looking for patterns that are not there. A will have a net Energy (relative to some frame) which will consist of its KE plus whatever fuel it has in its tank. As a result of an 'equal and opposite' force it will interact with B , which will also have a net energy (KE + fuel). Where the net energy ends up will depend on the details but there is no reason to assume how much energy A and B gain or lose will be equal.
If I do a standing jump on the ground, I will do (virtually ) no work on the Earth. When I start the jump I will have so much KE due to the leg work which will gradually change to PE (relative the Earth). 'All' the work was done "ON" me (if you really want some anthropomorphic thing to be responsible. But I supplied all the energy and, as I start to fall down, I steal back 'all' the Potential Energy (the Earth remains stationary). The only energy that the Earth gets back will be some strain energy and some heat. This could be significant if I land in something soft. Did I do the work 'on' myself and 'on' the landing spot? It's a meaningless question.

 

[john1954]

TL;DR Summary: The work done by two objects on each other.

The work done by A on B as Wab, and the work done by B on A as Wba.

I have always thought that Wab = -Wba generally always holds true. Looking at it from the perspective of energy transfer:

1. "A do 10J of work on B."
2. "A transfer 10J of energy to B."
3. "B received 10J of energy from A."
4. "B do -10J of work on A."

If 1 is true, then 2 through 4 all hold true, making Wab = -Wba valid.

However, consider the case where two people of different masses push each other from rest. In this scenario, it is intuitively clear that Wab and Wba are not the same. The kinetic energy is not conserved in this case.

I asked about this on another physics forum, and the response was that sometimes it is true (e.g., in the case of a perfectly elastic collision) and sometimes it is false (e.g., when two objects of different masses attract each other due to gravity, or when two stationary objects of different masses push each other). In other words, the statement is sometimes correct and sometimes incorrect. What conditions are necessary for this statement to be true, and what conditions make it false?

Wab=-Wba is only true if the displacements of a and b (in the direction of their mutual force ) are identical.

 

[john1954]

TL;DR Summary: The work done by two objects on each other.

The work done by A on B as Wab, and the work done by B on A as Wba.

I have always thought that Wab = -Wba generally always holds true. Looking at it from the perspective of energy transfer:

1. "A do 10J of work on B."
2. "A transfer 10J of energy to B."
3. "B received 10J of energy from A."
4. "B do -10J of work on A."

If 1 is true, then 2 through 4 all hold true, making Wab = -Wba valid.

However, consider the case where two people of different masses push each other from rest. In this scenario, it is intuitively clear that Wab and Wba are not the same. The kinetic energy is not conserved in this case.

I asked about this on another physics forum, and the response was that sometimes it is true (e.g., in the case of a perfectly elastic collision) and sometimes it is false (e.g., when two objects of different masses attract each other due to gravity, or when two stationary objects of different masses push each other). In other words, the statement is sometimes correct and sometimes incorrect. What conditions are necessary for this statement to be true, and what conditions make it false?

The forces between two objects are always equal and opposite (Newton III). Thus we can always ignore internal forces when doing a force/momentum analysis. However, the displacements of the two equal and opposite forces may not be the same - so the work done may not be equal and opposite and the work done by internal forces may not cancel. For perfectly rigid bodies internal work cancels. when gas explodes in the cylinder of your car the explosion exerts equal and opposite force on piston and cylinder head. But because the cylinder head does not move while the piston does you can extract useful work.

 

[sophiecentaur]

Wab=-Wba is only true if the displacements of a and b (in the direction of their mutual force ) are identical.

Something that is making this all more difficult than it need be is the apparent (almost emotional) need to look for 'conservstion' when it need not apply. When two isolated objects interact, momentum is transferred from one to the other and conservation of momentum is obviously conserved. But that's as far as it goes. What actually happens in practice will depend on the nature of the interaction. The energies before and after the interaction will depend on the 'losses' in a collision. This loss will reduce the available mechanical energy for sharing between the objects. Work can be done in sliding / distortion during the collision. In that situation how would one describe whether work was done 'by or on' the objects (forces times distances) when a lot of the work would be done internally on the squashy bits. What's important is the share of the remaining energy, after the lossy collision. Both objects are required to be there for this energy to be lost. Who's doing the work? If you are scrupulous about your arrows and signs in any calculation then you will get the right answer without worrying about the by or on question.

 

[john1954]

Something that is making this all more difficult than it need be is the apparent (almost emotional) need to look for 'conservstion' when it need not apply. When two isolated objects interact, momentum is transferred from one to the other and conservation of momentum is obviously conserved. But that's as far as it goes. What actually happens in practice will depend on the nature of the interaction. The energies before and after the interaction will depend on the 'losses' in a collision. This loss will reduce the available mechanical energy for sharing between the objects. Work can be done in sliding / distortion during the collision. In that situation how would one describe whether work was done 'by or on' the objects (forces times distances) when a lot of the work would be done internally on the squashy bits. What's important is the share of the remaining energy, after the lossy collision. Both objects are required to be there for this energy to be lost. Who's doing the work? If you are scrupulous about your arrows and signs in any calculation then you will get the right answer without worrying about the by or on question.

I always told my students to assume that internal forces do do net work , and then to justify situations where they do not do net work.

 

[A.T.]

If you are scrupulous about your arrows and signs in any calculation then you will get the right answer without worrying about the by or on question.

If you are not clear about what is doing the work on what for the work quantities you define, then you are not being scrupulous.

 

[sophiecentaur]

I always told my students to assume that internal forces do do net work , and then to justify situations where they do not do net work.

An ideal spring between two approaching objects would be the latter and a collision would be elastic.

If you are not clear about what is doing the work on what for the work quantities you define, then you are not being scrupulous.

What you have written implies you would have to know the answer before doing the calculation but I'm sure you don't mean that. In a complicated system, you could not predict this.

 

[A.T.]

What you have written implies you would have to know the answer before doing the calculation

No it doesn't. You can define Wab as the work done by a on b, without knowing it's value or sign. You have to be clear about the 'by a on b' part in order to interpret the sign of Wab resulting from the calculation correctly. If Wab turns out te be negative, then it represents a loss of mechanical energy by b due to interaction with a.

 

[sophiecentaur]

No it doesn't. You can define Wab as the work done by a on b, without knowing it's value or sign. You have to be clear about the 'by a on b' part in order to interpret the sign of Wab resulting from the calculation correctly. If Wab turns out te be negative, then it represents a loss of mechanical energy by b due to interaction with a.

So why use 'on or by'? The work is already determined by the sign you have chosen and that will determine the behaviour (acceleration / extension etc.) and that would be the energy gain / loss. I think you ignore the fact that not everyone is confident with your (the common) way of approaching those problems because they feel a need to 'know' if it's 'on or by'. You have already acknowledged that the sign and not the word is the final indicator. When the problem gets complicated, with several sources of energy and several objects, it's even more confusing.

Of course, there is always a problem with choosing the direction of arrows on diagrams, in any case. Using words in addition really doesn't help people (until they reach the competence and confidence levels that you clearly have). If it really were as straightforward as you claim then why are there so many PF questions where the confusion is present? A steady favourite is on the subject of car tyres on roads.

 

[jbriggs444]

So why use 'on or by'? The work is already determined by the sign ...

You have already debunked yourself by using the definite article, "the".

We use "on" and "by" because for every interaction force there are [at least] two works. The one done on object A and the one done on object B.

Nor is the sign relevant. The two works are not always equal and opposite, as has been pointed out ad nauseum.

 

[A.T.]

...by the sign you have chosen

Did you even read what I wrote? The sign is determined by the calculation, not chosen:

... the sign of Wab resulting from the calculation ...

 

[Herman Trivilino]

TL;DR Summary: The work done by two objects on each other.

1. "A do 10J of work on B."
2. "A transfer 10J of energy to B."

No. The first statement is a purely dynamical relation, that is, it's based on forces, momentum, and displacement. The second statement is far more general. It includes thermodynamics. It took researchers a century or so to realize that heat is a form of energy. And that form of energy is not included in a purely dynamical relation.

Anther way of looking at this is the conservation laws. Conservation of momentum is a purely dynamical relation and does not include thermal interactions. Conservation of energy is more general, in that it also includes thermal interactions. It took researchers a century to be able to place dynamic and thermal interactions under one umbrella and thus formulate the principle of the Conservation of Energy.

 

[sophiecentaur]

The sign is determined by the calculation, not chosen:

But you haven't done the calculation yet. At some stage you will have to draw / assume the direction of your arrows before you do the calculation. Until you have some values of masses and signs of forces (i.e. a usable formula) then any choice will be arbitrary and could easily be the same for all arrows.

The two works are not always equal and opposite,

Of course but how does that affect my comments?

Many of you guys don't seem to be aware of the problems that are caused by established terminology. You 'all get it' and assume that beginners just automatically get it. Responses (I've seen countless threads) just don't accept that repeating the same explanation may not provide a solution. @jbriggs444 demonstrated this excellently with a single line of thought / explanation which produced a lightbulb. See thread post #21. When answers are 'appropriate' confusion can so often be resolved and everyone can be happy. (and educated)

 

[A.T.]

But you haven't done the calculation yet.

What do you mean by "but" here? This exactly what I'm talking about:

You can define Wab as the work done by a on b, without knowing it's value or sign.

Again, do you even read what we write?

 

[jbriggs444]

Of course but how does that affect my comments?

You seem to have a problem with using the words "on" and "by" when speaking of the work done by a force. As if the distinguishing between the work done by A on B and the work done by B on A just sows confusion among the uninitiated. As if it is nothing but a sign convention anyway.

To me, failing to include "on" and "by" encourages wooly thinking.

Let me go off on a tangent and pontificate for a bit. I've never been a physics teacher. Nor have I ever been a poor student. So maybe I will be talking nonsense here.

I've often imagined that there is a sort of physics student who just does not get it. He skates through Physics 101 by using a pattern matching strategy, playing a game called "find the equation". Give him a radius and a velocity and ask him for a force and he will give you $F = \frac{mv^2}{r}$. To him, the fact that the situation involves a dinner plate sliding across a hockey rink is irrelevant.

We do such students no favors by speaking loosely and providing ambiguous definitions. If we want to encourage sharp thinking, we need to speak precisely.

 

[weirdoguy]

Many of you guys don't seem to be aware of the problems that are caused by established terminology. You 'all get it' and assume that beginners just automatically get it.

You don't seem to be aware that throwing out established terminology will make it even worse. They will not get it even more. Physics is hard. Don't make it harder.

 

[sophiecentaur]

You don't seem to be aware that throwing out established terminology will make it even worse. They will not get it even more. Physics is hard. Don't make it harder.

I realise that a risk is there but the risk of countless people having serious problems due to the presentation of the topic is real and present (just look at the wording of many queries on PF). It would be easy to establish a principle of forces always being assumed into a system, for instance. Far less of a struggle than getting involved with the flow of energy in a system.

In ore than 60 years of observing and experiencing Physics Education I have seen a significant number of changes in terminology and methods in many topics. Changes are often welcomed.

To me, failing to include "on" and "by" encourages wooly thinking.

I feel the opposite is true. Insisting on on and by can involve confusion from the very start. Work is defined by the forces and the distances and you can start from there without having to commit to a totally arbitrary choice of sign. I obviously realise that there is a similar difficulty in choice of the directions of forces but the context there and the words used in setting a problem - like pulling / pushing and gravity- usually gives you a clue for the choice of force vector direction in most cases.

Maybe the scalar nature of energy could account for the problem.

 

[jbriggs444]

I feel the opposite is true. Insisting on on and by can involve confusion from the very start. Work is defined by the forces and the distances and you can start from there without having to commit to a totally arbitrary choice of sign.

We can agree that choosing badly between the force of A on B and the force of B on A has the same effect as muffing a sign convention. Getting it wrong will invert the sign of the result.

But that is not what we are discussing. We are talking about the distinction between the work done by A on B and the work done by B on A. Getting those wrong is far worse than muffing a sign convention. We can get the wrong answer by orders of magnitude.

In reality, there is no choice of sign to be made. We are computing the dot product of a force and a displacement. No signs anywhere. The dot product does not depend on coordinates or the orientation of a coordinate system.

We do need to identify which force we are talking about. In a third law interaction there are two to choose from. We also need to identify which displacement we are talking about. In a two body interaction there are [at least] two to choose between. [I like to distinguish between four]

To be completely clear, the choices for force are:

1. The force exerted by A on B
2. The force exerted by B on A

Meanwhile, the choices for displacement are:

1. The displacement of the center of mass of A
2. The displacement of A at the point where the force from B is applied.
3. The displacement of the center of mass of B
4. The displacement of B at the point where the force from A is applied.

There are four useful combinations.

If we multiply (dot product) the force exerted by A on B by the displacement of the center of mass of B then we get the "center of mass" work done by A on B. This work contribute to the bulk kinetic energy associated with the linear motion of B: $\Delta E = \frac{1}{2}m_b v_{\text{com}}^2$

If we multiply (dot product) the force exerted by A on B by the displacement of B where the force is applied we get the mechanical work done by A on B. This work will contribute to various forms of mechanical energy in B including rotation, vibration and other internal motion along with bulk kinetic energy.

If we multiply (dot product) the force exerted by B on A by the displacement of the center of mass of A then we get the "center of mass" work done by B on A.

If we multiply (dot product) the force exerted by B on A by the displacement of A where the force is applied we get the mechanical work done by B on A.

Perhaps we need to set a real problem for which we can work toward a real solution. You show how it is unimportant to decide which force and which displacement is relevant and how only the sign convention matters.

Here goes...

A ball massing 1 kg is suspended 10 m above the floor of a gymnasium. It is released from rest and falls to the floor below.

Question 1: What relevant forces exist in this scenario?
Question 2: For each of those forces, can you compute the work done?

You may assume that the local acceleration of gravity is 9.8 m/s² and that the mass of the earth is 6 x 10²⁴ kg. You are expected to work from a reference frame where everything is initially at rest.

 

[Herman Trivilino]

I've often imagined that there is a sort of physics student who just does not get it. He skates through Physics 101 by using a pattern matching strategy, playing a game called "find the equation".

Well, I have been a full-time college-level physics instructor for 34 consecutive years and what you imagine is true for the vast majority of students. It takes skill and effort by an instructor to make this strategy unsuccessful. Many if not most instructors never figure out how, or perhaps are never motivated, to address this issue.

By the way, I learned and chose to speak only of the work done by an object when teaching thermodynamics. Suppose the work done on A by B is $W$ and the work done on B by A is $−W$. There is no need to speak of the work done on anything. I write the First Law as $ΔU=Q−W$. Of course one can instead choose to speak only of the work done on objects and write the First Law as $ΔU=Q+W$. There are advantages and disadvantages with either choice. The issue, among others, is being consistent (or in my case not) between the $W$ used in dynamics and the $W$ used in thermodynamics.

This is all a very confusing pedantic issue. This very discussion exemplifies some of this confusion.

 

[sophiecentaur]

We can agree that choosing badly between the force of A on B and the force of B on A has the same effect as muffing a sign convention. Getting it wrong will invert the sign of the result.

But that is not what we are discussing. We are talking about the distinction between the work done by A on B and the work done by B on A. Getting those wrong is far worse than muffing a sign convention. We can get the wrong answer by orders of magnitude.

In reality, there is no choice of sign to be made. We are computing the dot product of a force and a displacement. No signs anywhere. The dot product does not depend on coordinates or the orientation of a coordinate system.

We do need to identify which force we are talking about. In a third law interaction there are two to choose from. We also need to identify which displacement we are talking about. In a two body interaction there are [at least] two to choose between. [I like to distinguish between four]

To be completely clear, the choices for force are:

1. The force exerted by A on B
2. The force exerted by B on A

Meanwhile, the choices for displacement are:

1. The displacement of the center of mass of A
2. The displacement of A at the point where the force from B is applied.
3. The displacement of the center of mass of B
4. The displacement of B at the point where the force from A is applied.

There are four useful combinations.

If we multiply (dot product) the force exerted by A on B by the displacement of the center of mass of B then we get the "center of mass" work done by A on B. This work contribute to the bulk kinetic energy associated with the linear motion of B: $\Delta E = \frac{1}{2}m_b v_{\text{com}}^2$

If we multiply (dot product) the force exerted by A on B by the displacement of B where the force is applied we get the mechanical work done by A on B. This work will contribute to various forms of mechanical energy in B including rotation, vibration and other internal motion along with bulk kinetic energy.

If we multiply (dot product) the force exerted by B on A by the displacement of the center of mass of A then we get the "center of mass" work done by B on A.

If we multiply (dot product) the force exerted by B on A by the displacement of A where the force is applied we get the mechanical work done by B on A.

Perhaps we need to set a real problem for which we can work toward a real solution. You show how it is unimportant to decide which force and which displacement is relevant and how only the sign convention matters.

Here goes...

A ball massing 1 kg is suspended 10 m above the floor of a gymnasium. It is released from rest and falls to the floor below.

Question 1: What relevant forces exist in this scenario?
Question 2: For each of those forces, can you compute the work done?

You may assume that the local acceleration of gravity is 9.8 m/s² and that the mass of the earth is 6 x 10²⁴ kg. You are expected to work from a reference frame where everything is initially at rest.

That's a nice straightforward situation and the answer confirms the 'intuitive' feel. But the situation in a car with the front wheels driving over a rough road produces many questions like "is the car doing work on the road?" or "is there work done by the road surface on the car / wheel". The problem here is that people expect to know the by or on before the problems's been solved because it usually works for straightforward problems (as above).

what you imagine is true for the vast majority of students.

I agree but those students are not all destined to a life of flipping burgers and a teacher who does a bit of 'thinking for them' can help with this. sort of learning topic. I'd say that dealing with the work done should be after the forces have been dealt with. The total energy flow / work / dissipation / storage can be dealt with at the end.

 

[jbriggs444]

That's a nice straightforward situation and the answer confirms the 'intuitive' feel. But the situation in a car with the front wheels driving over a rough road produces many questions like "is the car doing work on the road?" or "is there work done by the road surface on the car / wheel". The problem here is that people expect to know the by or on before the problems's been solved because it usually works for straightforward problems (as above).

That is a problem with correctly choosing which displacement one is talking about. I advocate being explicit and clear on this point. My own teachers were not clear and explicit about that.

The example I chose above involved rigid and non-rotating objects. As a consequence of this, there is no ambiguity due to a choice of which displacement to use. The displacement of those objects at the points of application of the respective forces is identical to the displacement of the object's centers of mass.

I could have chosen a more complicated example, but it is good to start with something simple on which we can agree before proceeding to something more complicated.

So you wish now to speak of a 1000 kg car proceeding at 10 m/s down a straight section of highway. There is a frictional force of 1000 N between tires and road propelling the car forward and and air resistance of 1000 N tending to slow the car down. There is a 1 m/s headwind.

Can you now identify the various forces, the bodies upon which they act and the various amounts of work that are done over a one second duration.

[There are a lot of more focused questions that could be asked to probe your knowledge]

For convenience, you may adopt a reference frame in which the highway is at rest. For this exercise you may feel free to ignore any recoil by the earth.

If you want extra credit, consider that the wheel assemblies are 15 kg each with a 0.5 meter radius. You may assume a uniform mass distribution. Break down the works done on the wheel assemblies and by the wheel assemblies on the entities with which they mate.

Which flavors of work have you used? Which one identifies an energy flow from the engine?

 

[sophiecentaur]

My own teachers were not clear and explicit about that.

As for most of us. You obviously survived and you would have been able to deal with or without the by or on.

When you get down to it, the energy in the tank ends up as friction losses and accelerating the car. The work done is largely of interest inside the cylinders (@Mister T ) and against the road. Thermodynamics one end and conservation of momentum at the other. At the road, work done (energy) is shared by wheel and road / Earth and distorting / warming - as with an inelastic collision. I really can't see how the by or on is actually relevant. Yet the question still bugs people in their hundreds. Wouldn't it be worth while helping them?

 

[jbriggs444]

As for most of us. You obviously survived and you would have been able to deal with or without the by or on.

Without using "on" or "by", do you have a concrete proposal to distinguish between the work done by A on B and the work done by B on A?

When you get down to it, the energy in the tank ends up as friction losses and accelerating the car. The work done is largely of interest inside the cylinders (@Mister T ) and against the road.

Perhaps so. But if we are going to address the question of whether the power that drives the car comes from the road or from the engine, we are going to need precise terminology rather than a hand wave.

Thermodynamics one end and conservation of momentum at the other.

And two or three relevant flavors of work. The thermodynamics end needs mechanical or generalized work. The momentum end needs "center of mass" work.

At the road, work done (energy) is shared by wheel and road

1. How much work is done at the road interface? Can you answer that question without knowing which sort of work you are being asked about?

2. If we invoke the work energy theorem, where does that energy go? Or come from?

3. Once you have answered that question, do you still think that the energy is shared between wheel and road?

Let me help you out by answering those questions.

1...

In the rest frame of the road, zero mechanical work is done by tire on road. Also, zero mechanical work is done by road on tire. The mating surfaces are both at rest.

In the rest frame of the road, the center of mass work done by car on road is also zero. The road does not move, so zero work is done on the earth.

However, in the rest frame of the road, the car is moving. The center of mass work done by road on car across displacement $d$ is $Fd$.

If we shift to a different reference frame, these answers can change. Energy is not invariant. Displacement over a non-zero duration is also not invariant.

We might ask about the [invariant] total mechanical work done across the tire/road interface. Regardless of reference frame, this is zero since the mating surfaces are not in relative motion. That is to say that as long as slippage and rolling resistance are negligible, zero mechanical energy is lost at the tire/road interface.

2...

For mechanical work, the work energy theorem is about energy conservation. Zero mechanical work is done, so there is no energy flow to account for.

For center of mass work, the work energy theorem is about kinematics rather than energy conservation. To the extent that the question is meaningful, the resulting kinetic energy of the car comes from its interaction with the road. At the level of abstraction associated with center of mass work, there is an abstract interaction between car and road involving a net force and relative motion. That interaction is the source of any excess kinetic energy in the bulk motion of the interacting entities. If you want to dig deeper into that interaction, you will want to use mechanical work or generalized work.

3....

No the energy is not shared. In the case of mechanical work, there is no energy to share. In the case of center of mass work, all of the energy goes into the car and none into the road.

To me, physics is about problem solving. Finding numerical answers to physical problems. Having the terminology available to precisely express the question being asked is essential to getting a precise and correct numerical answer.

 

[A.T.]

As for most of us. You obviously survived and you would have been able to deal with or without the by or on.

Just because we survived some bad teachers, doesn't mean we have to propagate bad teaching.

 

[Herman Trivilino]

I agree but those students are not all destined to a life of flipping burgers

Certainly. In fact some of them are masterful at it and go on to excel in chemistry, physics, engineering, or math. Others succeed in lots of other fields, such as medicine, but are left thinking their physics course was a waste of time, used only to weed out the unworthy.

Sadly, many go on thinking they are too stupid to learn physics. And it's in large part a failure of the instructor who never learned how to teach the subject properly.

 

[Herman Trivilino]

Without using "on" or "by", do you have a concrete proposal to distinguish between the work done by A on B and the work done by B on A?

One would be a positive quantity and the other negative.

 

[jbriggs444]

One would be a positive quantity and the other negative.

Not necessarily. @PeroK had pointed this out in post #2 in the thread.

It holds true for two objects (or interacting surfaces) with zero relative motion in the direction of the force between them. It does not hold in general.

For many situations, there is no relative motion in the direction of the interaction force. This would include ropes on pulleys, levers on loads and hands on boxes.

For other situations, there is relative motion. This would include kinetic friction, gravity and the electrostatic force.

For some situations there is ambiguity. It can depend on how closely we look at the interaction. There is relative motion between car and road. But not between tire treads and road. There is relative motion between the bulk of the combustion gasses in the cylinder and the piston. There is zero relative motion between the gas right at the piston surface and the directly adjacent piston surface.

 

[sophiecentaur]

Just because we survived some bad teachers, doesn't mean we have to propagate bad teaching.

Do you have actual evidence that removing the frequent use of by and on has worse results? I haven't suggested anything revolutionary; all I am saying is that you can avoid the confusion by changing the emphasis and the way through mechanics. You are ignoring the problems that arise with the standard approach because you are bright enough to cope with all that cognitive dissonance.

 

[sophiecentaur]

which sort of work

That's a new concept for me. Isn't work always a dot product of Force and Displacement?

In your post, you have quoted a very simple scenario of car wheels on roads but it's far too simplified to add the sort of confusion I'm talking about. You have to consider the 'sharing' of the energy delivered to the wheels into useful kinetic energy and losses in the tyres and the tarmac; slip too. The rim of the wheel definitely does work because the car engine is the only source of energy (a level road, here) and any student can understand that. After the losses, there will be a force on the road which will have a reaction which accelerates the car. You could also say that the axle 'does work' by accelerating the car forward. Your ('my moderately bright ') student would not find that confusing.

But questions are always asked like "how does friction make the car go forward when friction slows things down?". You must have been asked to deal with that one. In that chain of interactions between components there are wheel - tyre - road surface / the Earth and the definition of Work could be applied in there. Easy for you because you have confidence that your system works reliably but if 'the student' has to decide who does what to whom, that's a needless barrier with a perceived risk. You can just use the forces and avoid another arbitrary choice of signs for intermediate values of Work. I can see only problems and not any significant advantages.

 

[jbriggs444]

That's a new concept for me. Isn't work always a dot product of Force and Displacement?

Which displacement? Perhaps you should read what I have written in this thread.

And no. Generalized work includes other forms of energy transfer.

In your post, you have quoted a very simple scenario of car wheels on roads but it's far too simplified to add the sort of confusion I'm talking about.

Yet you are unwilling to address it at the requisite level of detail. You have to show an ability to do the work. A level of understanding sufficient to proceed with an intelligent conversation.

You have to consider the 'sharing' of the energy delivered to the wheels into useful kinetic energy and losses in the tyres and the tarmac; slip too.

We could consider those. We need not. If dissipative losses are negligible then we can neglect them.

Dissipative losses in kinetic friction are a key reason that the distinction between the work done by road on treads and treads on road is more than just a sign change.

The rim of the wheel definitely does work because the car engine is the only source of energy (a level road, here) and any student can understand that.

Yet you fail to understand that. Because it is false. The rim of the wheel does zero mechanical work on the road.

Time for you to work the exercise.


Let me work through the situation for the wheel using the notion of mechanical work. Recall that mechanical work is defined as the dot product of the force applied on an object and the motion of the object at the point of application of the force.

We begin by identifying the relevant interfaces across which mechanical work may be done.

1. Static friction between tread and road.
2. Linear force between hub and wheel bearings.
3. Torsion between drive axle and wheel.

1...

The static friction between tread and road is (per the givens of the exercise) 1000 N. This is a forward force on the tire. The tread at the contact patch has zero velocity. So zero total displacement over the duration of the scenario. The work done by road on wheel is zero.

The road is also motionless. The work done by wheel on road is zero.

2...

For convenience, we assume that air resistance on the tires is negligible. The 1000 N of retarding air resistance is entirely on the body of the car. Accordingly, we can conclude that there is a forward force of 1000 N (total across the four wheels) from hub on wheel bearings. The car is moving forward at 10 m/s over a duration of 1 s for a total displacement of 10 meters forward. So we have $F \cdot d = +1000 \cdot +10 = +10000 \text{ J}$ of work done by wheel on car.

There is no relative motion in the forward direction. So the work done by car on wheel through this interface is the opposite: -10000 J.

3...

To simplify the analysis, I will pretend that only one wheel is driven. We could total across four driven wheels with various loads, but that would complicate matters with no benefit to understanding.

We are given that the radius of the wheel assembly is 0.5 meters. The required torque ($\tau$) to achieve 1000 N at the rim is 500 N m. In order to traverse 10 meters in one second, we are going to need 20 radians of rotation ($\Delta \theta$). The work done by torque is given by $\tau \cdot \Delta \theta = 500 \text{ Nm}\cdot 20 \text{ rad} = 10000 \text{ J}$. For drive axle on wheel the torque is in the same direction as the rotation of the wheel. So +10000 J of mechanical work is done.

[If you like, I can walk you through a derivation of $W = \tau \cdot \Delta \theta$ from first principles -- integration and the definition of mechanical work]

For wheel on drive axle, the torque is opposite and -10000 J of mechanical work is done.

In summary, the mechanical energy flows over the duration of the scenario are 10000 J from drive axle to wheel and 10000 J from wheel to car body.

 

[Herman Trivilino]

Not necessarily. @PeroK had pointed this out in post #2 in the thread.

That seems like a different point to me.

When you refer to the work done "on" something, or the work done "by" something aren't you always referring to a positive quantity?

My point is that if a positive amount of work $W$ is done on A in some interaction with B, then a negative amount of work $-W$ is done by A in the same interaction.

 

[sophiecentaur]

If dissipative losses are negligible then we can neglect them.

That's so true but answering questions about brakes, skidding and slipping and the dreaded word "friction" need answers involving more than the ideal.

Yet you fail to understand that. Because it is false. The rim of the wheel does zero mechanical work on the road.

Here's more confusion. In the frame of the car (we can choose?) the work is done on the road But you are right; the Earth doesn't move so that's the best reference frame. bI was being sloppy about the rim doing work; It's the couple (wheel rim and axel) that does work

I appreciate that the workings of the transmission system is analysed in terms of torque. The work done in the cylinder starts off as linear motion.

In summary, the mechanical energy flows over the duration of the scenario are 10000 J from drive axle to wheel and 10000 J from wheel to car body.

You put that well.

If dissipative losses are negligible then we can neglect them.

Now here's a bit of a problem. Your idealised model with no losses is fine and has much fewer apparent paradoxes but, I'm sure you will remember countless questions about friction (slipping) and internal losses in tyres and tarmac. The reason I got started on all this was because of the confused phrases about friction doing the work etc. etc.. I guess I'll just have to get over it but, when you read the next post that's confused about work being done by friction and "how is that possible?", you'll notice the cognitive dissonance and understand that there real difficulties because, basically, by the' by or on' question.

Cheers

 

[jbriggs444]

That's so true but answering questions about brakes, skidding and slipping and the dreaded word "friction" need answers involving more than the ideal.

We are still working toward a meeting of the minds. Your insistence on adding complications at every turn is not helpful in achieving that goal.

Here's more confusion. In the frame of the car (we can choose?)

There you go again. Another complication.

Yes, we can choose to work in a different reference frame. I was trying to avoid that complexity until we had some agreement on the basics.

If we shift to the rest frame of the car, the wheel assembly does zero work on the car body. But now the wheel assembly does positive work on the retreating road. The energy balance through the wheel assembly remains intact, of course. 10000 J in and 10000 J out.

We could shift to the Earth centered inertial frame and consider an eastbound car at the equator. Now we wheel is doing 470000 J on the car body and -46000 J on the Earth. The energy balance is still intact with 10000 J in and 10000 J out.

A westbound wheel would do -450000 J on the car body and +460000 on the Earth. Still for a total of 10000 J in and 10000 J out.

the work is done on the road But you are right; the Earth doesn't move so that's the best reference frame. I was being sloppy about the rim doing work; It's the couple (wheel rim and axel) that does work

First of all, it is not correct to say that the Earth does not move. Its movement is conditional on the reference frame that one chooses. A key lesson from Newtonian mechanics is that there is no such thing as absolute rest.

In addition, the wheel, rim and axle are a rigid assembly. No dissipation. No energy production. Zero total work done on the assembly [in this case anyway].

If you want to talk about a couple, you will need to be clear about which couple you are talking about.

The obvious place to look is the expanding gasses in the combustion chamber. Add up the work done on the piston and the work done on the cylinder and one can see that positive total work is being done by those gasses. The rest of the drive train has [ideally] zero losses. Mechanical work flows down the line to the drive axle and the wheel assembly.

The interface between car body and atmosphere involves dissipative friction. That interface is a mechanical energy sink. Work goes in and mostly thermal energy comes out.

In the exercise I posed, there was a headwind. In addition to the mechanical energy dissipated into thermal energy, there is a loss in total atmospheric kinetic energy. The atmosphere slows down as the car plows through it.

But since you not yet worked through the exercise, you may not have discovered that fine point in the energy balance.

I appreciate that the workings of the transmission system is analyzed in terms of torque. The work done in the cylinder starts off as linear motion.

Good. So we do not have to worry about a disagreement about the work associated with torque and rotation.

Now here's a bit of a problem. Your idealised model with no losses is fine and has much fewer apparent paradoxes but, I'm sure you will remember countless questions about friction (slipping) and internal losses in tyres and tarmac. The reason I got started on all this was because of the confused phrases about friction doing the work etc. etc.. I guess I'll just have to get over it but, when you read the next post that's confused about work being done by friction and "how is that possible?", you'll notice the cognitive dissonance and understand that there real difficulties

Real difficulties. Yes. I recognize that. Perhaps you recall the discussion here around the Brennan torpedo and where work is being done in that system. That was a lively discussion.

you'll notice the cognitive dissonance and understand that there real difficulties because, basically, by the' by or on' question.

Here we disagree. I believe that problem with student's understandings of work have little to do with the words "on" and "by". If anything, I see most difficulties associated with a failure to properly identify the relevant displacement.

One can trace confusion over kinetic friction and work to that difficulty. The displacements of the two interacting objects are different.

One can trace confusion about work done on cars to that difficulty. The displacements of the treads and of the car are different.

 

[jbriggs444]

That seems like a different point to me.

When you refer to the work done "on" something, or the work done "by" something aren't you always referring to a positive quantity?

As I use the terminology, the result can be negative.

Perhaps I should explain the way I use the terms. I use both terms in a single phrase to characterize one side of a particular interaction.

If I exert a retarding force of 1000 N on an object displacing forward through 10 m then I have done -10000 J of work on the object. This is the work done "by" my hand "on" the object.

If this 1000 N is instead the result of kinetic friction from a long piece of sandpaper then the mechanical work done "by" the sandpaper "on" the object is still -10000 J.

However, if I speak about the work done "by" the object "on" the sandpaper then the result is 0 J since the sandpaper is not moving. In this case, the total of the two works is -10000 J. This discrepancy will manifest as thermal energy.

My point is that if a positive amount of work $W$ is done on A in some interaction with B, then a negative amount of work $-W$ is done by A in the same interaction.

No, that is not correct.

 

[A.T.]

all I am saying is that you can avoid the confusion

Avoiding confusion by being less explicit about what you mean? Is this a joke?

 

[Herman Trivilino]

If I exert a retarding force of 1000 N on an object displacing forward through 10 m then I have done -10000 J of work on the object. This is the work done "by" my hand "on" the object.

And the work done by the object on your hand would be +10 000 J.

If this 1000 N is instead the result of kinetic friction from a long piece of sandpaper then the mechanical work done "by" the sandpaper "on" the object is still -10000 J.

However, if I speak about the work done "by" the object "on" the sandpaper then the result is 0 J since the sandpaper is not moving. In this case, the total of the two works is -10000 J. This discrepancy will manifest as thermal energy.

But note that that thermal energy is shared by both the object and the sandpaper, the total being 10 000 J, but with no way of knowing how much of that thermal energy is in the sandpaper and how much is in the object.

So if we look at the work done on the object you claim it's -10 000 J, yet the object doesn't lose a total of 10 000 J of energy. It looses 10 000 J of mechanical energy, gains an unknown amount of thermal energy, and therefore looses an unknown amount of energy.

If this is a result of the way you are defining work, there is nothing wrong with that as a perfectly valid dynamical relation. But it is not the work as we define it in the 1st Law of Thermodynamics.

 

[jbriggs444]

And the work done by the object on your hand would be +10 000 J.

Yes. In this case the force pair is not dissipative. The two works are equal and opposite.

But note that that thermal energy is shared by both the object and the sandpaper, the total being 10 000 J, but with no way of knowing how much of that thermal energy is in the sandpaper and how much is in the object.

Yes. Agreed.

So if we look at the work done on the object you claim it's -10 000 J, yet the object doesn't lose a total of 10 000 J of energy. It looses 10 000 J of mechanical energy, gains an unknown amount of thermal energy, and therefore looses an unknown amount of energy.

Yes. Agreed.

If this is a result of the way you are defining work, there is nothing wrong with that as a perfectly valid dynamical relation.

Good. Then we have no fundamental disagreement. We are just nattering about terminology.

But it is not the work as we define it in the 1st Law of Thermodynamics.

You are apparently talking about thermodynamic work while I am talking about mechanical work.

Can you define "thermodynamic work" precisely and explain how it relates to kinetic friction?

If it were me, I would invent an abstract system to embody the interface across which the kinetic friction acts. This abstract system has a net surplus of "thermodynamic work" supplied from its two sides. It supplies an equivalent amount of heat which egresses (in an unknown ratio, as you point out) out the two sides.

 

[Herman Trivilino]

You are apparently talking about thermodynamic work while I am talking about mechanical work.

I didn't know that you were making a distinction.

 

[sophiecentaur]

Avoiding confusion by being less explicit about what you mean? Is this a joke?

I think you have been too interested in getting across the theory (and no-one is disputing that it's right) and have not grasped that using Work as a way into understanding makes the whole thing harder. What do you mean by "explicit"? These sort of basics are very often best taught in steps and avoiding the difficulty which Work can initially introduce would be an advantage. I have never suggested that Work shouldn't;t be taught.

If you only see intellectual problem from your personal viewpoint then you will find it hard to understand how many other people see things. From a perfectly logical point of view it can be said that people who want to learn should just accept being taught the way you were taught and that should be enough.
I am always coming across the stumbling block that arises when Work is introduced too early in learning some topics. If you just tell me that they should get over it and approach it 'your way' from the start, then it's basically 'end of conversation'.

Yes. In this case the force pair is not dissipative. The two works are equal and opposite.

the force pair is not dissipative if the displacements are the same amplitude. (That, at least is true for mechanical interactions.

 

[sophiecentaur]

You are apparently talking about thermodynamic work while I am talking about mechanical work.

Can you define "thermodynamic work" precisely and explain how it relates to kinetic friction?

I didn't know that you were making a distinction.

This sub-conversation rather makes my point for me. How would someone who is confused when trying to appreciatewhat goes on in a driven car and what happens to the Energy, be in a position to appreciate the meaning of those two contributions? It only clouds the basic issue of the thread which it the order in which all these concepts should best be explained.

There's a difference between vectors, which 'cancel' and scalars which merely have signs but don't actually cancel in the same way.

 

[jbriggs444]

This sub-conversation rather makes my point for me. How would someone who is confused when trying to appreciate what goes on in a driven car and what happens to the Energy, be in a position to appreciate the meaning of those two contributions? It only clouds the basic issue of the thread which it the order in which all these concepts should best be explained.

As @A.T. suggests, the way to combat confusion is with clarity. Not vague handwaving.

There's a difference between vectors, which 'cancel' and scalars which merely have signs but don't actually cancel in the same way.

More handwaving. Show me a scalar and I'll show you a vector.

 

[jbriggs444]

the force pair is not dissipative if the displacements are the same amplitude. (That, at least is true for mechanical interactions.

What does "amplitude" mean in this context?

 

[jbriggs444]

I didn't know that you were making a distinction.

I'd already distinguished between "center of mass" work and mechanical work. I'd alluded to something else called generalized work. I think that generalized work (a transfer of mechanical energy without a transfer of mass) is the same thing as thermodynamic work, but am still awaiting your definition.

 

[sophiecentaur]

What does "amplitude" mean in this context?

I was trying for a way of avoiding the word 'equal' because then you'd say they're vectors. Isn't it true that non-dissipative implies no relative motion?

Edit: but this is all well above the intended level of the thread and is another reason to try to avoid Work, early on. This is a very common approach to initial learning / teaching many topics in Physics, actually.

 

[Herman Trivilino]

This sub-conversation rather makes my point for me. How would someone who is confused when trying to appreciatewhat goes on in a driven car and what happens to the Energy, be in a position to appreciate the meaning of those two contributions?

There have been efforts in the last 15-20 years in many of the mainstream college-level introductory physics textbooks to make the definition of work the same in both dynamics and thermodynamics. It's an uphill battle, mainly because many of us who learned different formalisms have difficulty re-learning the newer formalism. Some people can be very stubborn and close-minded when it comes to re-learning something that took them years or even decades to master.

 

[jbriggs444]

I was trying for a way of avoiding the word 'equal' because then you'd say they're vectors. Isn't it true that non-dissipative implies no relative motion?

No relative motion in the direction of the force. You want the dot product of the relative motion and the force to be zero.

Edit: but this is all well above the intended level of the thread and is another reason to try to avoid Work, early on.

The original intent of this thread is to discuss the work done by two objects on each other. A detailed examination of the situation is directly on point.

A digression into the utility of the words "on" and "by" in physics pedagogy is arguably beyond the scope of this thread.

This is a very common approach to initial learning / teaching many topics in Physics, actually.

Certainly, we often paint a simple picture to begin with and fill in complexities and details later. There is nothing wrong with that. But if a student comes to us and asks why the work done by the road on the car can be different from the work done by the car on the road [as in the original post in this thread], how are we to answer if we are forbidden from discussing the matter in detail?

For me, I want to be able to reconcile a formalism with the physical reality. To do that, I need a formalism rather than a handwave. I'd be open to a formalism that focuses on the interaction rather than on the participants in the interaction. But then one loses the ability to talk about the work done by fictitious forces.