# Work done by friction when car is breaking

1. Jul 22, 2013

### bksree

Hi all
Please clarify this doubt on work done by friction. The text referred is from the book 'Introduction to classical physics by Morin' - Chap 5, sec 5.1
He is referring to the work-energy theorem : Wexternal = ΔK + ΔV + ΔKinternal
"Consider a car that is braking (not skidding). The friction force from the ground on the tires is what causes the car to slow down. But this force does no work on the car, because the ground isn't moving; the force acts over zero distance. So the external work on the left side (of the above equation) is zero. The right side is therefore also zero. That is the total energy of the car doesn't change. .... ΔK = - ΔKinternal ..."

My doubt is regarding the line But this force does no work on the car, because the ground isn't moving; the force acts over zero distance
The frictional force (rolling friction since the car isn't skidding) by the ground ON the car is F = μkN where N = wt of car. This acts opposite to the direction of motion and isn't work done by friction given by ( F*dist travelled during braking)

TIA

2. Jul 22, 2013

### Staff: Mentor

The friction is static friction (not rolling friction, which is something else), since the tires are not slipping. The instantaneous speed of the tire patch in contact with the ground is zero, thus no work is done by that friction force. Morin is correct.

You can certainly calculate F*distance traveled, which will give you the change in translational KE. But the distance is not the distance through which the force acts; it is the distance traveled by the center of mass. So it is not really the work done by that force. Sometimes that quantity is called "center of mass" work or pseudo-work.

3. Jul 22, 2013

### Staff: Mentor

This situation is a classic example of why the usual F.d definition of work is difficult to apply in some scenarios. If everything is rigid then the distance, d, is clear and unambiguous, but as soon as you are dealing with non-rigid motion d becomes difficult to pin down. Is d the distance travelled during braking or is it 0? Either answer seems reasonable.

A more general and usable definition of work is that "work is the transfer of energy by means other than heat". In this case, it is clear that the work done by friction is 0 since the energy of the car is not changing. All that is happening is that kinetic energy of the car is being converted internally within the car into thermal energy in the brake pads and shoes. Since there is no transfer of energy from the car to the road there is no work.

4. Jul 22, 2013

### A.T.

Does there have to be an absolute answer to this? It seems to me, that just as work done by some force can depend on the reference frame, it can also depend on the level of detail of your free body diagram:
- If you model the entire car as one rigid body moving linearly, then work is done on the car by the ground in the ground frame.
- If you model the non slipping wheel as a separate object, then no work is done on the wheel by the ground in the ground frame.

Consider a block sliding over ground, slowing down due to friction. Usually we call this kinetic friction which does negative work on the block. But we actually don't know (and usually don't care) what exactly is going on at the interface between block and ground. There could be some static forces at the interface, combined with deformation (and thus energy dissipation) within the block, just like the brakes dissipate energy within the car. In the case of the block we almost always abstract the complex interaction away, and express it in terms of a friction coefficient. In the case of a car we can too abstract away the interface (wheels, brake pads, etc), if we are not interested in those details. Then the force of the ground becomes kinetic friction, which does negative work in the ground frame.

Last edited: Jul 23, 2013
5. Jul 22, 2013

### rcgldr

It doesn't matter that the ground doesn't move. The point of application of force does move with respect to the ground, so you have a decelerating force exerted by the pavement onto the tires over the distance it takes for the car to stop, and that equals the work done.

6. Jul 22, 2013

### Staff: Mentor

Except that the car's energy doesn't change. So if there is work done and no change in energy then you have violated the conservation of energy and can make a PMM. For conservation of energy, the work must be 0.

Last edited: Jul 22, 2013
7. Jul 22, 2013

### Staff: Mentor

The ground does move... Not much, but if we're to conserve momentum of the car+earth system as the car slows and stops relative to the earth, the earth has to acquire some small velocity in the direction of the car's travel.

A substantial fraction of the work done happens not at the tire/road contact patch, but at the boundary between brake pads and rotors. There the frictional force is quite high, and it acts across many meters every second as the rotor moves under the pads.

8. Jul 22, 2013

### Staff: Mentor

We end up with car's kinetic energy going to zero while the brakes become hot (A few good laps on a racetrack and the rotors will glow red hot - even a drive to the grocery store will get them to where they will spit and sizzle if you splash water on them) and a tiny change in the earth's kinetic energy as we transfer momentum from the decelerating car to the earth.

9. Jul 23, 2013

### rcgldr

The car's energy changes, but the car is part of a larger system (earth and car), as mentioned by Nugatory.

The kinetic friction converts some of the mechanical energy into heat, so that the total mechanical energy of the earth + car is reduced by the losses.

If the car used regenerative braking charging a secondary battery, and assuming an ideal case where no energy is lost in the conversion, then the total energy of earth + car + battery potential energy remains constant.

10. Jul 23, 2013

### Staff: Mentor

Yes, exactly.

This tiny change, while present, is irrelevant to the question in the OP. We are likewise ignoring energy lost to air resistance, rolling resistance, acoustic emissions, and thermal emissions, etc. all of which are very small compared to the work supposedly being done.

11. Jul 23, 2013

### Staff: Mentor

This is a very good example. Using conservation of momentum and conservation of energy you can calculate the change in energy for each. The work done by the tires on the road is equal to the energy transferred to the earth. You can compare that to your suggestion in post 5.

With respect to the road there is no difference between regenerative braking and normal braking. The only difference is that the cars internal energy distribution is shifted between KE and chemical PE in one case and between KE and thermal energy in the other.

12. Jul 23, 2013

### Staff: Mentor

The surfaces exerting the forces against each other--the ground and the contact patch of the tire--have zero relative velocity. The force of friction does no work on the car. (Similar to jumping in the air. The ground pushes you, but does no work.)

As I said before, you can certainly calculate Fnet*Δxcm to determine the change in KE. But that's not work.

13. Jul 23, 2013

### rcgldr

Getting back to the original post:

If the car had lossless regenerative braking charging a battery, then the total energy of the car doesn't change (the mechanical energy is converted into electrical potential energy).

If the car is using regular brakes, then mechanical energy is being converted into heat, and I would consider heat as a loss instead of part of the car's internal energy.

ΔK = - ΔKheat

14. Jul 23, 2013

### cjl

The heat is still internal to the car though - it goes into the brake disks, which are definitely part of the car. The energy is then lost as the disks cool off, so the mechanism of energy transfer is the brake disks heating the air. The act of braking does not change the car's internal energy, it's the post-braking brake cooling (that's an awkward statement if I ever typed on...) that causes the car's energy to change. If your brake disks were perfectly insulated from their surroundings, they would contain the full energy that the car had prior to stopping, and you could even possibly extract some of it with a heat engine (though that's getting a bit rube-goldbergish).

15. Jul 23, 2013

### rcgldr

Ignoring the issue of whether heat is an internal energy or a loss, the main point of the original post is correct, static friction doesn't perform work. The work is being performed by something else that is responsible for the static friction force.

As Doc Al mentioned, I was confusing the ability to calculate the change in kinetic energy as friction force times distance to brake as work, but static friction isn't the source of this work, it's the brakes.

16. Jul 23, 2013

### Staff: Mentor

I could have been more clear about why I mentioned that particular tiny effect...
The transfer of this tiny amount of kinetic energy is what reduces to zero the car's momentum relative to the earth - the pad-rotor friction inside the brakes cannot affect the car's linear momentum. So despite its small size, it's essential to understanding the dynamics of the situation; if it were zero we'd see very different behavior.

The car does slow, and momentum and kinetic energy are transferred from the car to the earth, so some force must be acting between the road and the tire contact patch. It's not dynamic friction (we're all agreed about rolling without slipping here).

Imagine a block resting on a frictionless surface; I put my hand flat on the top of the block and move it around, thereby doing work on the block. The $W=Fd$ calculation works just fine here, but the force cannot exceed that allowed by the coefficient of static friction (if that were zero my hand would slide around the top of the block without doing any work at all). The car interacts with the earth in the same way, although the value of $d$ is very small.

I gotta say that I completely agree with you when you said:
Such a transfer does happen or the car wouldn't stop, it is mediated through forces between the tire contact patch and the road surface, and I find Morin's description in terms of frictional forces to be less than lucid.

Last edited: Jul 23, 2013
17. Jul 23, 2013

### Staff: Mentor

Thermal energy is part of a system's energy. You cannot just neglect it nor exclude it. The entire field of thermodynamics requires it. What you can do is to define the brakes to be a separate system from the rest of the car, but you cannot just wish away the thermal energy.

If you consider the brakes to be a separate system from the rest of the car then there is a transfer of energy from the car to the brakes resulting in a decrease in the car's kinetic energy and an increase of the brakes thermal energy. Thus the car does work on the brakes.

There is no significant transfer of energy between the road and the car or brakes, and therefore no work done by or on the road.

18. Jul 23, 2013

### Naty1

bksree: All this is not at all obvious when first exposed....Just keep in mind kinetic [sliding] friction, rolling friction, and static friction are all a bit different. Which one to utilize follows from the problem description.

19. Jul 23, 2013

### Staff: Mentor

Well, static friction doesn't perform work in this case, but it can perform work in other cases. For example, consider a box on the back of an accelerating truck. Energy is being transfered to the box so the static friction force is doing work in that case.

20. Jul 23, 2013

### Staff: Mentor

There always exists a frame where the change in the earth's energy is 0 despite the fact that the car comes to rest relative to the earth. This is the frame where the earth is at rest halfway through the braking. So the change of energy is not in fact what changes the momentum.

In the usual reference frame I see its only value as being for the conservation of momentum, which is not of interest here. For energy questions it only becomes "essential" in other reference frames (in my opinion).

EDIT: I should mention that in those other reference frames it can be quite large, which is why it becomes essential in those frames.

Last edited: Jul 23, 2013