Work done on accelerating car is zero?

[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.

 

[rcgldr]

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?

No. It's a closed system, momentum is conserved, so the momentum change in the Earth is equal in magnitude but opposite in direction to the momentum change of the car. As mentioned earlier, since the Earth is so much more massive than the car, almost all of the energy from the engine goes into the car.

 

[BrainSalad]

No. It's a closed system, momentum is conserved, so the momentum change in the Earth is equal in magnitude but opposite in direction to the momentum change of the car. As mentioned earlier, since the Earth is so much more massive than the car, almost all of the energy from the engine goes into the car.

I think he was referencing the skidding scenario, in which negative work is done on the car and the road and tires are heated up.

 

[Dale]

But, insisting that every point in a machine has to have work done 'to or by ' something is philosophical.

That is the model. In the model, energy conservation doesn't happen by energy just magically disappearing from one system and appearing at another. It is transferred from system to system via work and heat only. This isn't an optional part of the model.

 

[Dale]

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.

This is a correct analysis. The road does not do work in this situation because the energy of the car is unchanged. The car loses KE and gains thermal energy at the brake pads and shoes. No energy is transferred.

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.

I agree.

 

[A.T.]

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.

Right, so in respect to Newtons 3rd Law force pairs and work we we can say:

If kinetic energy is converted into other energy forms at the interface, then the positive work done by one force can be less than the negative work done by the other force.


What about the opposite situation?

If kinetic energy is generated from other energy forms at the interface, then the positive work done by one force can be more than the negative work done by the other force.


The later would apply to an accelerating car modeled as a single rigid body, with the propulsive system seen as part of the interface to the road. The force of the ground could do positive work on the car, despite the opposite force doing no negative work on the road. Note, that if you model the car as one translating body, then the force from the ground is not static any more. The "car" in the model is a sliding block. Just that the interface is not opposing the sliding, but propelling it.

I know that this sounds like a stretch, but the point of my question is to figure out how far you can abstract things, and still have a valid model. Is it permissible at all, to model a machine with moving parts as one rigid body, if those parts are interacting with other bodies?

 

[BrainSalad]

I know that this sounds like a stretch, but the point of my question is to figure out how far you can abstract things, and still have a valid model. Is it permissible at all, to model a machine with moving parts as one rigid body, if those parts are interacting with other bodies?

I think that depends on how the parts are interacting and which model you apply. Newtons 1st and 2nd, for instance, are valid for extended bodies and systems of particles as long as the velocities and accelerations considered are those of the COM. Work, however cannot be done without the exchange of energy across the interface(s). But, sometimes we can pretend it's done by modeling an object as a point particle or something like that and using the work energy theorem for the COM.

 

[OmCheeto]

What a confusing thread. Let me start from the beginning.

... Isn't it that only external forces may change the translational kinetic energy of an object?

If that were true, then how do hamster balls get around? I suppose you could say that gravity is the external force pulling on the hamster. But I don't think gravity pushed the hamster forward.

If so, and if the only external force does no work, how does the car's kinetic energy change?

The hamster moved.

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?

No. The hamster added potential energy to the system, and since he's in a sphere, the torque he created caused the sphere to rotate.

Points do not rotate.

They do if you model them as tidally locked geohubstationary tennis shoes.

I can quite obviously see that each point/tennis shoe is going to rotate once for each full orbit around the hub.

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.

I don't quite understand this answer.

Although... I think I got it.

Your answer is "No, as heat balances the equation.".

Yes?

Although... hmmm...

Can a car accelerating be modeled as the null velocity point in an elastic collision problem?

 

[Doc Al]

If that were true, then how do hamster balls get around? I suppose you could say that gravity is the external force pulling on the hamster. But I don't think gravity pushed the hamster forward.

Are you saying that a hamster ball can start moving without there being an external force acting on it? Better rethink that one.

 

[OmCheeto]

... I suppose you could say that gravity is the external force pulling on the hamster.
...

Are you saying that a hamster ball can start moving without there being an external force acting on it? Better rethink that one.

What? Where did I say that?

 

[BrainSalad]

Are you saying that a hamster ball can start moving without there being an external force acting on it? Better rethink that one.

This is the way I thought about it originally. It's net momentum can't change without an outside force, but that has no bearing on the kinetic energy. An bomb for instance, converts chemical energy into kinetic and heat energy, without outside forces. The net momentum of all its exploded pieces is the same as it was initially though. Kinetic energy also can't be added from outside sources without work being done. A car can move from rest as a single unit because it has an internal energy source and an external force acting on it, but that force does no work. If the car was in space, you push the gas and all the energy goes into wheel rotation. Add the ground and gravity and the car moves forward under the influence of an external force.