# Work done on accelerating car is zero?

1. Jan 22, 2014

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?

2. Jan 22, 2014

### A.T.

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

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.

3. Jan 22, 2014

### Staff: Mentor

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.

4. Jan 22, 2014

### 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.

5. Jan 22, 2014

### Staff: Mentor

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

6. Jan 22, 2014

### Staff: Mentor

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

7. Jan 22, 2014

### Staff: Mentor

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?

Last edited: Jan 22, 2014
8. Jan 22, 2014

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?

9. Jan 22, 2014

### jbriggs444

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

"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.

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.

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...

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.

10. Jan 22, 2014

### jbriggs444

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.

11. Jan 22, 2014

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).

12. Jan 22, 2014

### Staff: Mentor

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

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

13. Jan 22, 2014

### Staff: Mentor

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.

14. Jan 22, 2014

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 possess 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.

Last edited: Jan 22, 2014
15. Jan 22, 2014

### Staff: Mentor

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.

http://www.lightandmatter.com/html_books/me/ch13/ch13.html

16. Jan 22, 2014

### 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.

17. Jan 22, 2014

### Staff: Mentor

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

18. Jan 22, 2014

### jbriggs444

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.

19. Jan 22, 2014

### sophiecentaur

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.

20. Jan 22, 2014

### Staff: Mentor

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