Work done on accelerating car is zero?

[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.