Work done on accelerating car is zero?

[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.

 

[Doc Al]

What? Where did I say that?

Right here:

... 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?

 

[OmCheeto]

Right here:

Oh...

Umm...

It never crossed my mind that west coast, internally powered hamster balls:

might be externally powered.

We call those:

elephant balls.

My bad.

 

[Doc Al]

It never crossed my mind that west coast, internally powered hamster balls:

might be externally powered.

Just because the energy source is internal does not remove the need for an external force to accelerate.

 

[OmCheeto]

Just because the energy source is internal does not remove the need for an external force to accelerate.

Hmmm... My boss and I were discussing "mathematical word problems" the other day, and I showed him that internet meme about aliens, hats, and purple. It was inspired by this thread.

If you are Hercules, positioned between Earth and BrainSalads car, and pushed really hard, in equal and opposite directions, what happens?

If you removed gravity, the car would go one way, and the Earth would move the other way.

Mathematically, the resulting momentum of the car and Earth would have the same magnitude, but the kinetic energy would go almost exclusively to the car.

If you're not willing to ignore gravity, then model this as a "Car in Spaaaaaace...", mechanically linked, but free to move, via fancy gears and such, to the midpoint of a million kilometer long, 6E24 kg, roadway. Floor the accelerator pedal for about a second.

Where now is this "external force" that you require?

--------------------------------
ps. Everything I've learned about momentum and impulse, I've learned in the last two days.

 

[BrainSalad]

Where now is this "external force" that you require?

--------------------------------
ps. Everything I've learned about momentum and impulse, I've learned in the last two days.

Taking the car to be the system, the external force is the static friction force of the road on the tires.

If you've really only learned about momentum the past two days, add to your knowledge that "external" and "internal" are relative terms. Any force exerted by an object outside the system in question, arbitrarily defined, is external. Thus, an external force is needed to change momentum (accelerate). Of course, if you change the size of the system, car+Earth for example, a previously external force may now be internal and cause no net change of momentum at all (Earth gains as much negative momentum as the car gains positive).

 

[OmCheeto]

Taking the car to be the system...

I learned long ago, to ignore everything past the first wrong premise.

 

[Doc Al]

If you're not willing to ignore gravity, then model this as a "Car in Spaaaaaace...", mechanically linked, but free to move, via fancy gears and such, to the midpoint of a million kilometer long, 6E24 kg, roadway. Floor the accelerator pedal for about a second.

Where now is this "external force" that you require?

You're kidding, right?

For the car to accelerate, the road must exert an external force on it. Same for your hamster wheel.

--------------------------------
ps. Everything I've learned about momentum and impulse, I've learned in the last two days.

So it seems.

 

[Doc Al]

Taking the car to be the system...

I learned long ago, to ignore everything past the first wrong premise.

I suggest that you reread this thread from the beginning.

 

[BrainSalad]

I learned long ago, to ignore everything past the first wrong premise.

What's wrong with the premise? We can't take the whole universe as the object in question.

You don't have any real physics education, do you? Every force is an internal force if you define the system large enough, but that isn't helpful a lot of the time.

 

[A.T.]

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

That is true in the rest frame of the objects in contact. But when you model the entrie car as one translating block, then there is no rest frame of both objects: ground and car.

I know, in reality there is a wheel that has a contact point at rest. But when you do calculations in physics, you do them based on your model, not on reality. And if you apply the F*d definition of work here, then the external force is doing work, because it is not static in this particular coarse model of reality.

 

[A.T.]

The correct answer is that the road performs 0 work on the car. Any method that disagrees with that is simply wrong.

Rather than "wrong", I would say inconsistent with the thermodynamic definition of work. I like this definition, and I used per default in post #2 to give a straight forward answer.

But still, I would like to know how/if the F*d definition can be applied when non-rigid bodies are modeled as rigid blocks. Or when such modelling is permissible. Therefore I would like to hear you thoughts on my question from post #55:

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.

 

[OmCheeto]

What's wrong with the premise? We can't take the whole universe as the object in question.

Yes we can. The universe is the system. The car is a part of the system.

 

[OmCheeto]

I suggest that you reread this thread from the beginning.

I've read it at least 6 times. I will let things digest, and reread it next weekend, as the thread's title still makes me want to poke someone in the eye.

 

[BrainSalad]

Yes we can. The universe is the system. The car is a part of the system.

Newton's laws would be a good place to start, my friend.

 

[Dale]

. Therefore I would like to hear you thoughts on my question from post #55:

I'm still thinking about it. It made me reconsider my stance, and I am still not sure where I will wind up.

 

[OmCheeto]

You're kidding, right?

No.

For the car to accelerate, the road must exert an external force on it. Same for your hamster wheel.

Are you referring to the normal force? A.T. mentioned that in post #2 as I recall.

So it seems.

I didn't chose my name lightly.

 

[sophiecentaur]

This question and the way it is being discussed seems totally crazy to me. If I jump up in the air, before I actually leave the ground, my foot is in contact all the time and there is no movement. Are you bothered that you can't say that the ground is doing work on me - or that I am doing work on the ground? Why do you choose to be arguing about the 'wheel' question rather than the 'jumping' question? They are, in essence, exactly the same in that you need to look at the whole mechanism and not just the contact point for the answer. The only difference is that the contact point on the wheel keeps changing but your foot is on the same spot until you actually leave the ground by jumping.

 

[BrainSalad]

No.Are you referring to the normal force?

No. It's a static friction force in the forward direction that drives the car forward. It could not accelerate from rest without that force, as per Newton's 1st law. A normal force always acts perpendicular to the surface.

 

[catellus]

You're all making this more complicated than necessary, and oversimplified at the same time.

The OP started with a very simple model of a car as a point mass and a force being applied to it somehow, but then asked a question about a more complex system (one with rotating wheels and a road). That simple model doesn't capture enough about the real world to answer that question.

If you want to model what happens on a car with rotating tires, you need a more complex model. Let's take a point mass car with a single wheel. The important part of the wheel is the part from the stationary contact point to the axle, which forms a lever, rotating about the contact point. The torque of this lever, supplied by the drive train, applies a force to the axle, where we will assume the point mass of the car is located. Now you have a force applied to the car over a distance and hence doing work.

p.s. Yes, the distance involved in the motion of this lever can only be small, but when you consider the rotating wheel, you always have that same lever applying the same force. And I changed "patch" to "point' to keep the lever model simple. And the wheel has inertia if you're looking for a numerical answer. If you want to look at other aspects of the car, expand your model to include them.

Catellus

 

[OmCheeto]

No. It's a static friction force in the forward direction that drives the car forward. It could not accelerate from rest without that force, as per Newton's 1st law. A normal force always acts perpendicular to the surface.

Thank you for pointing out that both forces ultimately point in the same direction.

I really like this thread.

 

[BrainSalad]

If you want to model what happens on a car with rotating tires, you need a more complex model. Let's take a point mass car with a single wheel. The important part of the wheel is the part from the stationary contact point to the axle, which forms a lever, rotating about the contact point. The torque of this lever, supplied by the drive train, applies a force to the axle, where we will assume the point mass of the car is located. Now you have a force applied to the car over a distance and hence doing work.Catellus

I like this idea, and have thought about it before, but it's wrong. The drive train does just as much positive work on the axle as the axle does negative work on the drive train, so no net work is done (equal, opposite forces exerted over same distance). This is the very reason that only external forces may do work on an object. Since the only external force acting on the car does no work, no work is done on the car. All this means is that no kinetic energy is transferred to the car via outside sources. All its energy comes from the engine.

 

[rcgldr]

This is the very reason that only external forces may do work on an object.

As a counter example to this, take the case of a spinning ice staker that pulls their arms inwards. Although the work done pulling the arms inwards is internal, the angular kinetic energy of the spinning ice skater increases.

 

[Dale]

That isn't a counter example. The skater has exchanged internal chemical potential energy for kinetic energy. No transfer of energy has happened and therefore no work.

 

[BrainSalad]

As a counter example to this, take the case of a spinning ice staker that pulls their arms inwards. Although the work done pulling the arms inwards is internal, the angular kinetic energy of the spinning ice skater increases.

No work is done on the ice skater. NO internal forces can do work, because they transfer no energy from outside the system.