Work done on accelerating car is zero?

[BrainSalad]

The static friction force which provides the acceleration of a car does not move through a distance (the point of application is stationary with respect to the road at any instant). Isn't it that only external forces may change the translational kinetic energy of an object? If so, and if the only external force does no work, how does the car's kinetic energy change?

Obviously, there is no problem with energy conservation here (the chemical potential energy in the oxygen/gasoline mixture in the engine is converted to kinetic energy in the wheels and body of the car). But doesn't this mean that the kinetic energy of a system may change without net work being done?

 

[A.T.]

If so, and if the only external force does no work,

In the rest frame of the ground the ground has no kinetic energy. So it obviously cannot deliver energy to the car.

But doesn't this mean that the kinetic energy of a system may change without net work being done?

Sure. If you jump up, the ground reaction also does no work on you in the rest frame of the ground, and yet you accelerate and gain kinetic energy.

 

[russ_watters]

The static friction force which provides the acceleration of a car does not move through a distance (the point of application is stationary with respect to the road at any instant).

Untrue. The contact points are virtual and change over time (move) and the wheel is always rotating about the contact point, even though the individual points are not translating.

The effect of multiple contact points in different places is mathematically equal to one continuous force acting over a distance.

You can even switch your frame of reference to the car and see that the force is always applied at the same spot and moves with the car.

A similar example would be a drive chain or toothed belt pulled by a gear: the virtual contact point never moves, but the belt clearly does.

 

[jbriggs444]

The change in kinetic [including rotational] energy of a rigid body is equal to the sum of the work done by all of the external forces on that body. But a car is not a rigid body. The tires move with respect to the frame.

I disagree with Russ' response. The fact that the contact point is changing is irrelevant. The fact that the constact point is stationary is relevant. The fact that the contact point is not at rest with respect to the car is relevant.

 

[Doc Al]

BrainSalad and A.T. are correct. No work is being done by the road on the car, yet that force does accelerate the car.

 

[russ_watters]

I disagree with Russ' response. The fact that the contact point is changing is irrelevant. The fact that the constact point is stationary is relevant. The fact that the contact point is not at rest with respect to the car is relevant.

You have the last two backwards at least insofar as what my statements were, but I think also reality.

 

[russ_watters]

BrainSalad and A.T. are correct. No work is being done by the road on the car, yet that force does accelerate the car.

Is there a simplifying assumption there that the road is stationary? Doesn't in reality the car cause the road to accelerate backwards?

Or is my terminology just not quite right: car is doing work on the road and not the other way around?

 

[BrainSalad]

The change in kinetic [including rotational] energy of a rigid body is equal to the sum of the work done by all of the external forces on that body. But a car is not a rigid body. The tires move with respect to the frame.

So what does this mean for the Work-Energy theorem? Is it just a special case of the conservation of energy, applicable to rigid bodies? I find that hard to believe, since the change in translational kinetic energy of a car can be found with the theorem (treating it as a point mass subjected to the road force), even though no real work is done and the car is not a rigid body. What's really going on here?

 

[jbriggs444]

You have the last two backwards at least insofar as what my statements were, but I think also reality.

Well, let me re-read then. Likely we are having a communications failure rather than a disagreement about factual matters.

Untrue. The contact points are virtual and change over time (move) and the wheel is always rotating about the contact point, even though the individual points are not translating.

"The contact points are virtual and change over time (move)". The physical point which is in contact will not always be the same point. It changes over time. So the "contact point" is a virtual notion and is changing with respect to the physical points. Yes, I agree with that.

"The wheel is always rotating around the contact point". Yes, I agree with that.

"Even though the individual points are not translating". By this, I take it that you refer to the individual physical points that are momentarily in contact with the pavement. Those are not translating with respect to the pavement. Yes, I agree with that.

But the "untrue" bit is perplexing. What you are saying did not disagree with what OP said.

The effect of multiple contact points in different places is mathematically equal to one continuous force acting over a distance.

You'll have to spell out this mathematical equivalence. It seems obvious that it is also mathematically equivalent to one continuous force acting over zero distance.

You can even switch your frame of reference to the car and see that the force is always applied at the same spot and moves with the car.

That's not the definition of work. It is the motion of the material that is subject to the applied force that matters. Not the motion of the contact point.

Let us take an extreme example... A rocking chair with a fairly flat rocker. The point of contact of this rocking chair on the floor will move dramatically in response to small changes in the orientation of the chair.

Push the chair across the floor with a one Newton force moving it by one centimeter while rocking it so that the contact point moves by one meter. Have you done 1 Joule of work? or 0.01 Joules?

I think we can agree that the answer is 0.01 Joules and has to do with how far the chair moved rather than with how far the contact point moved.

In response to a later message...

Is there a simplifying assumption there that the road is stationary? Doesn't in reality the car cause the road to accelerate backwards?

Yes, there is an assumption that the road is stationary.

The notion that there is an underlying reality that is different is jarring. There is no such thing as an underlying reality that makes a particular choice of frame of reference "real" or "unreal".

If one adopts a frame of reference in which the car is stationary then the car does positive work on the road and the road does negative work on the car -- the contact point between road and car is moving backwards under a forward force from the road and under a backward force from the car.

 

[jbriggs444]

So what does this mean for the Work-Energy theorem? Is it just a special case of the conservation of energy, applicable to rigid bodies? I find that hard to believe, since the change in translational kinetic energy of a car can be found with the theorem (treating it as a point mass subjected to the road force), even though no real work is done and the car is not a rigid body. What's really going on here?

For a non-rigid body, exernal forces need not result in changes in kinetic energy. The energy can go into deformation and heat (e.g. if you crash your car into a wall), potential energy (e.g. when you stretch a rubber band), electrical energy (e.g. if you crank a generator) or other forms.

So it is clear that the Work-Energy theorm need not apply per se. But, as you point out, it gives the right answer -- how can it be wrong?!

My answer is that if you abstract away the details of the tires, the drive train, the engine, the fuel, the oxidizer and the exhaust then you are left with a rigid block sliding across a frictionless surface subject to an external force. That external force is applied to a moving object. The work-energy theorem applies just fine with that understanding.

 

[BrainSalad]

My answer is that if you abstract away the details of the tires, the drive train, the engine, the fuel, the oxidizer and the exhaust then you are left with a rigid block sliding across a frictionless surface subject to an external force. That external force is applied to a moving object. The work-energy theorem applies just fine with that understanding.

Understood. But, in light of this, why do we still define work to be the dot product between a force and the displacement of the point of contact, when in reality it need not be the point of contact that is displaced?

It amazes me, when thinking about things like this, that idealized models of phenomena can accurately predict reality, even though the two are very different (an accelerating car has no work done on it it reality, but can be treated like something that has in order to reach correct results).

 

[Doc Al]

Is there a simplifying assumption there that the road is stationary? Doesn't in reality the car cause the road to accelerate backwards?

I think we can safely ignore the movement of the road!

Or is my terminology just not quite right: car is doing work on the road and not the other way around?

No work is being done by the road. It is not a source of energy.

 

[Doc Al]

So what does this mean for the Work-Energy theorem? Is it just a special case of the conservation of energy, applicable to rigid bodies? I find that hard to believe, since the change in translational kinetic energy of a car can be found with the theorem (treating it as a point mass subjected to the road force), even though no real work is done and the car is not a rigid body. What's really going on here?

The so-called 'work'-Energy theorem is really an application of Newton's 2nd law, not a statement about work in general. Only in the special case of a point mass (or rigid body) is that "work" term really a work (in the conservation of energy sense).

If you take a net force acting on an object (like friction) and multiply it by the displacement of the object's center of mass, you get a quantity that looks like a work term but is better called pseudowork (or "center of mass" work)--what it determines is not the real work done on the object, but the change in the KE of the center of mass of the object. This is usually called the "Work-Energy" theorem:
F_{net}\Delta x_{cm}=\Delta (\frac{1}{2}m v_{cm}^2)
Despite the name, this is really a consequence of Newton's 2nd law, not a statement of energy conservation.

 

[BrainSalad]

The so-called 'work'-Energy theorem is really an application of Newton's 2nd law, not a statement about work in general. Only in the special case of a point mass (or rigid body) is that "work" term really a work (in the conservation of energy sense).

If you take a net force acting on an object (like friction) and multiply it by the displacement of the object's center of mass, you get a quantity that looks like a work term but is better called pseudowork (or "center of mass" work)--what it determines is not the real work done on the object, but the change in the KE of the center of mass of the object. This is usually called the "Work-Energy" theorem:
F_{net}\Delta x_{cm}=\Delta (\frac{1}{2}m v_{cm}^2)
Despite the name, this is really a consequence of Newton's 2nd law, not a statement of energy conservation.

Okay, I'll list what I think I've learned from this discussion. Correct me or add something.

1.) Real work need not be done on an object for the object's kinetic energy to change. Rather, the energy transformation can be totally internal, from chemical potential energy, etc.
2.) Only objects which possesses energy have the ability to do work, or transfer kinetic energy.
3.) When dealing with rigid bodies or point masses subjected to external forces, the Work-Energy Theorem is effective in determining the change in translational kinetic energy of the objects, but does not necessarily reflect real work done on any object involved.

 

[Dale]

I think that the problem is that there are multiple definitions of work. The most straightforward one is the thermodynamic definition, which basically says that work is any transfer of energy besides heat. By that definition the ground does no work since no energy is transferred from the ground to the car.

The mechanical definition is that work is a force applied over a distance. If you are dealing with a rigid body in the absence of heat transfer then it is easy to show that the two definitions are equivalent (work energy theorem), but if you have a non-rigid body then the mechanical definition is no longer straightforward to use. The applicable distance becomes challenging to define.

In such circumstances my inclination is to use the thermodynamic definition since it is clear and easy to apply. I don't think that use of the mechanical definition is wrong, but it becomes confusing.

I recommend this page for a good reference on the topic:
http://www.lightandmatter.com/html_books/me/ch13/ch13.html

 

[rcgldr]

I had the impression that the distance part of the force x distance formula refers to the point of application of a force (not the instantaneous point of contact between tire and road), and the point of application of force is moving if the car is moving.

For constant acceleration, I think it would be easier to start with impulse = force x time = m Δv (change in momentum), combined with force = m a, and v = a Δt to derive force x distance = 1/2 m (Δv)^2.

f Δt = m Δv
f = m a = m Δv / Δt
a = f/m = Δv / Δt
Δv = a Δt = f/m Δt
vavg = 1/2 a Δt
d = vavg Δt = 1/2 a Δt^2 = 1/2 (Δv / Δt) Δt^2 = 1/2 Δv Δt
f d = (m Δv / Δt) (1/2 Δv Δt) = 1/2 m Δv^2

For a more generic case, integrals could be used.

 

[Doc Al]

I had the impression that the distance part of the force x distance formula refers to the point of application of a force (not the instantaneous point of contact between tire and road), and the point of application of force is moving if the car is moving.

The point of application of the force is the instantaneous point of contact between tire and road.

 

[jbriggs444]

I had the impression that the distance part of the force x distance formula refers to the point of application of a force (not the instantaneous point of contact between tire and road), and the point of application of force is moving if the car is moving.

The motion of the point of where the force is applied is unimportant. The motion of the material where the force is applied is what counts. The movement of the contact point has about as much physical significance as the movement of the point where the blades of a pair of scissors intersect.

 

[sophiecentaur]

I had the impression that the distance part of the force x distance formula refers to the point of application of a force (not the instantaneous point of contact between tire and road), and the point of application of force is moving if the car is moving.

Perhaps if, for a start, we replaced the (theoretical) wheel with a belt, as on a tracked vehicle. Then it may be easier to picture a force being moved back along the track, against the road to produce non-instantaneous force moving a finite distance.

 

[russ_watters]

But the "untrue" bit is perplexing. What you are saying did not disagree with what OP said.

The OP said "stationary" and I don't agree. The point on the wheel where it touches the ground is not translating, but it is rotating. Rotation is motion too.

You'll have to spell out this mathematical equivalence. It seems obvious that it is also mathematically equivalent to one continuous force acting over zero distance.

I'm not quite sure how to spell it out: the numbers just are what they are. The distance is the distance and the force is the force in both cases. That's where my confusion or perhaps disagreement as an engineer comes from. If the force is 1N and the distance is 1M, that's 1x1=1J. The ground applies a force and the car moves, so the ground did work.

That's not the definition of work. It is the motion of the material that is subject to the applied force that matters. Not the motion of the contact point.

Agreed! The ground applied a force (or, rather, the ground and wheel applied a force to each other) and the car moved.

Let us take an extreme example... A rocking chair with a fairly flat rocker. The point of contact of this rocking chair on the floor will move dramatically in response to small changes in the orientation of the chair.

Push the chair across the floor with a one Newton force moving it by one centimeter while rocking it so that the contact point moves by one meter. Have you done 1 Joule of work? or 0.01 Joules?

This is a really bad example because it combines motions in a way that is unclear by adding oscillation to the mix, making the distances traveled unequal and the force not constant, none of which apply to the car.

Lets try this one:
Lock the brakes of the car and push it, making it slide. The contact point on the road is "stationary" in that we define the road stationary and "moves" in that it is a different point on the road over time - both exactly as it does when the car is rolling. Regardless of which way you look at it, the car is moving and the road applies a force to it, applying work (just in the opposite direction from when the car is accelerated by its motor).

Yes, there is an assumption that the road is stationary.

The notion that there is an underlying reality that is different is jarring. There is no such thing as an underlying reality that makes a particular choice of frame of reference "real" or "unreal".

I didn't say that it made the frames "real" or "unreal" - frames are completely arbitrary. But a simplifying assumption is something that is assumed that isn't actually true. In this case, assuming the Earth doesn't move results in a contradiction of conservation of momentum.

In any case, this is more on point:

If one adopts a frame of reference in which the car is stationary then the car does positive work on the road and the road does negative work on the car -- the contact point between road and car is moving backwards under a forward force from the road and under a backward force from the car.

Agreed. So isn't there a contradiction here? The OP is arguing (asking) that if we adopt a frame of reference where the road is stationary, it does no work but I agree with you that if in a frame of reference where the car is stationary, both do work. Isn't it a contradiction for the change of frame to result in no work being done on/by the road? My argument is that it is doing work either way; the choice of reference frame shouldn't change that.

 

[russ_watters]

I had the impression that the distance part of the force x distance formula refers to the point of application of a force (not the instantaneous point of contact between tire and road), and the point of application of force is moving if the car is moving.

The point of application of the force is the instantaneous point of contact between tire and road.

Right, but Doc, the question wasn't where is it, the question is is it moving? My answer was yes: Yes, there is a force applied to the car by the road at the contact point and yes that point is moving.

It appears to me that the fact that the half of the contact point that is on the tire is not translating (but is rotating - and yes, I know points can't have halves) causes people to say that the contact point is not moving.

There are other variations of this we could do:
-Flip the car over and set a long board on the wheels. The car is stationary WRT Earth and the contact point is stationary WRT the car, but it throws the board (gaining the board kinetic energy).
-Put the car on a dyno. Again, the car and contact point are stationary WRT earth, but energy is transferred to the dyno (and dissipated as heat).

 

[russ_watters]

Okay, I'll list what I think I've learned from this discussion. Correct me or add something.

1.) Real work need not be done on an object for the object's kinetic energy to change. Rather, the energy transformation can be totally internal, from chemical potential energy, etc.
2.) Only objects which possesses energy have the ability to do work, or transfer kinetic energy.
3.) When dealing with rigid bodies or point masses subjected to external forces, the Work-Energy Theorem is effective in determining the change in translational kinetic energy of the objects, but does not necessarily reflect real work done on any object involved.

It looks to me like the first two are wrong, but I'm the engineer here, not the physicist.

1. This one implies reactionless motion which isn't possible.
2. Energy is exchanged between objects sharing a force and motion: one object gains and the other loses -- but which is which is a matter of choice of coordinates. If I push a car, it gains energy and I lose it, if I define the starting point to be at rest. If it is already rolling and I push it to stop it, the math is identical except for the reversed signs.

 

[russ_watters]

I apologize if I'm making a mess of this, but I think I figured something out:

If you have internal energy (such as is provided by a motor or spring), the total energy of the system must be constant and the kinetic energies of the objects involved sum to equal it.

Take two blocks and a spring, for example: the spring pushes the blocks apart and expends a certain amount of energy. If the blocks are of equal mass, they each get half (the spring expands and pushes each an equal distance from the starting point). If one block is made much more massive, it moves a shorter distance and the less massive one moves a larger distance: one gets more energy than the other and the total remains the same. Momentum is conserved in this situation, but it doesn't mean that the kinetic energies of the two objects have to be equal. Make the large object much, much larger and its absorbed kinetic energy becomes close to zero.

 

[BrainSalad]

If one block is made much more massive, it moves a shorter distance and the less massive one moves a larger distance: one gets more energy than the other and the total remains the same. Momentum is conserved in this situation, but it doesn't mean that the kinetic energies of the two objects have to be equal. Make the large object much, much larger and its absorbed kinetic energy becomes close to zero.

I think this is actually a pretty good model of the original scenario. In your model, no work need be done by the very large block (Earth in the original case). It transfers no energy, but it allows the spring energy (chemical energy in gasoline) to be converted, almost entirely, into the kinetic energy of the small block (car).

 

[rcgldr]

If we replaced the (theoretical) wheel with a belt ...

Which is the real world situation, since real objects deform due to force, so there will always be a non-zero length contact patch between a wheel and the surface the wheel rolls on.

 

[jbriggs444]

The OP said "stationary" and I don't agree. The point on the wheel where it touches the ground is not translating, but it is rotating. Rotation is motion too.

Points do not rotate.

I'm not quite sure how to spell it out: the numbers just are what they are. The distance is the distance and the force is the force in both cases.

The distance can be measured in a number of ways. The integral of the distance moved by the material at the location of the contact patch with respect to the road over the duration of a one meter roll is zero.

That's where my confusion or perhaps disagreement as an engineer comes from. If the force is 1N and the distance is 1M, that's 1x1=1J. The ground applies a force and the car moves, so the ground did work.

The material at the contact patch was always stationary with respect to the ground.

Agreed! The ground applied a force (or, rather, the ground and wheel applied a force to each other) and the car moved.

Agreed!

Lets try this one:
Lock the brakes of the car and push it, making it slide. The contact point on the road is "stationary" in that we define the road stationary and "moves" in that it is a different point on the road over time - both exactly as it does when the car is rolling.

We are examining the work done by the road on the car still? In that case the [tire material at the] contact patch is moving. As the car skids to a stop the road is doing negative work on it.

This is different than when the car was rolling. The difference is the motion of the material at the contact patch. My notion of work done has to do with the interface between two objects, not with their bulk motion with respect to one another.

Regardless of which way you look at it, the car is moving and the road applies a force to it, applying work (just in the opposite direction from when the car is accelerated by its motor).

Agreed!

Agreed. So isn't there a contradiction here? The OP is arguing (asking) that if we adopt a frame of reference where the road is stationary, it does no work but I agree with you that if in a frame of reference where the car is stationary, both do work. Isn't it a contradiction for the change of frame to result in no work being done on/by the road? My argument is that it is doing work either way; the choice of reference frame shouldn't change that.

The work done by a force on an object is indeed a frame dependent quantity. This makes sense since the energy of an object is also frame dependent.

Take for example the case of a car that is driving westward and slams on the brakes, skidding to a stop relative to the Earth. There was work done by the road on the car. In a frame of reference in which the Earth is stationary this was negative work. It slowed the car to a stop. In an Earth-centered inertial frame within which the Earth is rotating this was positive work. The force actually sped the car up.

 

[A.T.]

The work done by a force on an object is indeed a frame dependent quantity.

It also depends on how you define your objects. If you model the entire car as one rigid translating block called "car", then you abstract away all the rotating parts and lump them together as the interface between "car" and "ground". In such a coarse model the external force is indeed doing work on the "car", while the equal opposite force is doing no work on the stationary "ground". Energy is conserved because the interface generates/dissipates kinetic energy from/to other energy forms.

Whether such a coarse model, that obfuscates much of the function, is still useful is another question and depends on the application. It might be, that you don't care where the KE of the car is exactly dissipated during braking (brakes or tires). Then you can model your "car" as a sliding block, and the interface to the ground (representing wheels, brakes etc) dissipates the cars KE. The force pair at this interface does negative work on the "car", but no positive work on the stationary "ground".

 

[rcgldr]

I've often read here that static friction does no work. In my opinion, the power source of a car is its engine, which after losses and gearing, generates a torque times angular velocity at the driven wheels. The force equals this torque divided by the radius of the wheels. There are forces involved at the contact patch, and the car is accelerated in the direction of force from the pavement of the Newton third law pair of forces at the contact patch (tire backwards on pavement, pavement forwards on tire), but this doesn't mean the ground is performing work.

skidding - this is a different situations. Kinetic friction consumes energy converting it to heat. Either the brakes or the tires are skidding during braking (unless regenerative braking is being used).

 

[sophiecentaur]

This worry about 'who does the work' is all a bit too anthropomorphic for me. As far as I'm concerned, it's usually pretty clear where the energy for the work comes from and the work is just a force times a displacement.
It seems like some of the posters on this thread are just seeking a place in the system where they can find that 'the laws' are broken. In general, this will be at a discontinuity of some sort in the idealised version of the system. Somewhere there is a 0/0 situation and that upsets them. I think that's where you have to accept that the Mathematical model needs the 'appropriate interpretation'. (This happens throughout Science and it is not a problem).

 

[A.T.]

Either the brakes or the tires are skidding during braking

On a very coarse abstraction level you might not care where the dissipation actually happens. When you model the entire car as one translating rigid body, then you don't have bodies like wheels and brakes in your free body diagram.

When modeling an actual sliding block, you also don't differentiate between dissipation at the actual contact vs. dissipation deeper inside the block due to deformation. You approximate the block as perfectly rigid, and associate all the dissipation with the sliding at the contact.

 

[BrainSalad]

This worry about 'who does the work' is all a bit too anthropomorphic for me. As far as I'm concerned, it's usually pretty clear where the energy for the work comes from and the work is just a force times a displacement.
It seems like some of the posters on this thread are just seeking a place in the system where they can find that 'the laws' are broken. In general, this will be at a discontinuity of some sort in the idealised version of the system. Somewhere there is a 0/0 situation and that upsets them. I think that's where you have to accept that the Mathematical model needs the 'appropriate interpretation'. (This happens throughout Science and it is not a problem).

I don't know what you mean. Analyzing a situation at greater depth is not an offense to the gods of physics, it is the reason we have laws in the first place to work with. They aren't the end, they are the means. The scenarios that the principles are introduced with are often so simplified that they have no physical meaning in themselves, but can be used to idealize real world phenomena.

 

[Dale]

-Flip the car over and set a long board on the wheels. The car is stationary WRT Earth and the contact point is stationary WRT the car, but it throws the board (gaining the board kinetic energy).
-Put the car on a dyno. Again, the car and contact point are stationary WRT earth, but energy is transferred to the dyno (and dissipated as heat).

These are good examples. Although the basic way that mechanical work is taught is W=F.d, I find that in ambiguous situations I usually get the right answer more easily if I instead think of power, P=F.v where v is the velocity of the material at the point of contact.

In the case of the right-side-up car that v is 0 which gives the correct amount of work. In the case of the upside down car that v is not 0 which is also correct.

 

[Dale]

I've often read here that static friction does no work.

Hmm, that is not a general rule. There are many cases where static friction does work.

 

[sophiecentaur]

I don't know what you mean. Analyzing a situation at greater depth is not an offense to the gods of physics, it is the reason we have laws in the first place to work with. They aren't the end, they are the means. The scenarios that the principles are introduced with are often so simplified that they have no physical meaning in themselves, but can be used to idealize real world phenomena.

It depends what you mean by "in greater depth". It seems to me that you are really looking for some 'familiar' standpoint to view the problem and not looking any deeper at all. You want a force and you want a distance - you've got them both - except when you want to look just at the point of contact, which is a discontinuity, and refers to a strictly theoretical model. If you allow an infinitesimal angle for the wheel to rotate, with the same portion of the tyre in contact (which is totally what happens in practice) then you have a force and you have a distance.
I am not discounting the "physical meaning" when there is one. You are taking a mathematical abstraction (point contact etc.) and demanding that the problem can be taken from there and then given a physical explanation. That seems unreasonable to me. You seem to want to have your cake and eat it.
If you allow for a polygonal wheel then the problem goes away so do what calculus does all the time and take the limit as the number of sides approaches infinity.

 

[BrainSalad]

It depends what you mean by "in greater depth". It seems to me that you are really looking for some 'familiar' standpoint to view the problem and not looking any deeper at all. You want a force and you want a distance - you've got them both - except when you want to look just at the point of contact, which is a discontinuity, and refers to a strictly theoretical model

The fact remains that no work is actually done on the car. The road does no work on the car because it has no energy to give. However, the change in kinetic energy of the car can be calculated by pretending that work has been done by an outside source of energy. There's no problem with that. In reality though, the energy comes from the gasoline in the engine.

 

[rcgldr]

The fact remains that no work is actually done on the car.

IMO, there is work performed on the car, equal to the force at the contact patch times the distance that the force is applied. However, the source of the energy is the car's engine (after losses), not the road or the tires.

 

[BrainSalad]

IMO, there is work performed on the car, equal to the force at the contact patch times the distance that the force is applied. However, the source of the energy is the car's engine (after losses), not the road or the tires.

The force is not applied through any distance though. The COM moves through a distance, and that's useful for calculation. But, work (the transfer of energy from one object to another) is not done by the road.

 

[Dale]

The correct answer is that the road performs 0 work on the car. Any method that disagrees with that is simply wrong. If the road does work on the car then the road must lose energy and the car must gain it. You cannot have the road doing work on the car but the car's energy not increase and the road's energy not decrease.

 

[A.T.]

If the road does work on the car then the road must lose energy and the car must gain it.

Is the reverse also true? If the road does negative work on the car then the road must gain energy and the car must lose it?

 

[sophiecentaur]

Is the reverse also true? If the road does negative work on the car then the road must gain energy and the car must lose it?

That happens in a skid - the road gets a bit warmer from the work done 'on' the surface material.

But why is there this preoccupation about who does what? It's just another of these daft 'what really happens?' questions that are only there for entertainment. There is always some answer available; it just needs to be very convoluted, sometimes.

Is this question any different from the question about whether or not the Fulcrum of a lever does any work on the Load? If it doesn't move at all then it cannot - but how does knowing that improve the sum total of human knowledge? A real fulcrum will deflect a bit so there is work done on it and then by it (if you really feel those concepts are of any consequence) as the forces are first applied and later removed.

 

[A.T.]

If the road does negative work on the car then the road must gain energy and the car must lose it?

That happens in a skid - the road gets a bit warmer from the work done 'on' the surface material.

Let me pose the question quantitatively: Is the energy transferred from the car to the road, always equal to the negative work done on the car by the road?

 

[Dale]

Is the reverse also true? If the road does negative work on the car then the road must gain energy and the car must lose it?

Yes.

 

[Dale]

But why is there this preoccupation about who does what? It's just another of these daft 'what really happens?' questions that are only there for entertainment.

You know that I despise such questions. This isn't a philosophical question, it is a question about the correct use of the model.

In our model energy is transferred through work and heat. That is a rule of the model. Any approach which violates the rules is not using the model correctly.

 

[A.T.]

Yes.

What about the quantitative version: Is the energy transferred from the car to the road, always equal to the negative work done on the car by the road?

 

[sophiecentaur]

You know that I despise such questions. This isn't a philosophical question, it is a question about the correct use of the model.

In our model energy is transferred through work and heat. That is a rule of the model. Any approach which violates the rules is not using the model correctly.

I think that is too literal an interpretation of the question and you are getting inveigled into treating it as such. I see it precisely as a philosophical question (in as far as it assumes that it actually has meaning and can have an answer, within the gamut of the model). Imo, such questions are aimed at finding 'flaws' and loopholes in these basic models, rather than spotting where the model is being mis - applied. You can't apply the modhich is what you say here.
The only causal arrow that can be drawn is from engine to car motion. No energy comes from any other source and the 'by and to' ideas don't have to apply everywhere in the system. You may as well ask the same question about the bearings and chassis. Force is 'transmitted' through them but what has that to do with work?

 

[Dale]

I think that is too literal an interpretation of the question and you are getting inveigled into treating it as such. I see it precisely as a philosophical question (in as far as it assumes that it actually has meaning and can have an answer, within the gamut of the model).

I am not sure what your position is. I am both too literal and too philosophical?

Work isn't a philosophical term. It is a defined term in the model. This isn't an "interpretation" question, it is simply identifying correct use of the model and its defined terms. If work is done then energy is transferred. If you make a calculation where work is done without energy transfer then the calculation is wrong. I don't know what about that you can possibly find objectionable.

 

[Dale]

What about the quantitative version: Is the energy transferred from the car to the road, always equal to the negative work done on the car by the road?

Yes (neglecting heat).

Heat can be important in slipping. Consider a car skidding to a stop. F.v is negative for the car so negative mechanical work is being done on the car and the mechanical energy decreases. F.v is zero for the Earth so there is no mechanical work being done on the Earth and the mechanical energy does not change. The quantitative difference in mechanical energy goes to heat.

 

[sophiecentaur]

I am not sure what your position is. I am both too literal and too philosophical?

Work isn't a philosophical term. It is a defined term in the model. This isn't an "interpretation" question, it is simply identifying correct use of the model and its defined terms. If work is done then energy is transferred. If you make a calculation where work is done without energy transfer then the calculation is wrong. I don't know what about that you can possibly find objectionable.

They're not mutually exclusive, imo.
I am not finding anything "objectionable" - I am trying to dissuade you from getting tied up in something where your answer is bound to be in quadrature with the question, as I read it. It can only result in frustration for you.
Work is not a philosophical term, of course, because it is defined. But, insisting that every point in a machine has to have work done 'to or by ' something is philosophical.You seem to be trying to answer that philosophical (implied) question with reference to the 'model'.
We are not disagreeing about the model - or where it applies. I don't imagine you would feel it necessary to bring up the 'to or by' question, if it hadn't been slipped in under the radar.

 

[BrainSalad]

Yes (neglecting heat).

Right. In a situation like braking, the breaks keep the wheels from turning, allowing the road to apply a force which slows down the car. The kinetic energy of the car is turned into heat in the break pads and wheels. The road still does no work though. Skidding is about the only time significant energy is lost to the road, I think. Other than normal heat losses from the tires, which we've ignored so far.

 

[BrainSalad]

Before sophiecentaur casts this discussion into the depths of argument for no reason, thanks to everyone for the replies. I learned something.