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When you apply the brakes to a car and it slows down, there are two kinds of friction and I am getting confused.

On the one hand, there is friction between tyres and road. This friction

  • a) Is the only external force on the car, so ultimately according to Newton's law this must be equal to mass x acceleration and should be responsible for slowing down.

  • b) Is static and performs no work. This is weird. How can this force slow down the car but produce no work? How about the kinetic energy theorem?

On the other hand, there is friction in the brakes. This friction

  • a) Dissipates mechanical energy into heat, so you might say that actually it is this force that slows down the car.

  • b) But, all interaction at the brakes is internal to the car-system, so cannot produce any center-of-mass acceleration.

So it seems one force is responsible for the acceleration and another force is responsible for dissipating the energy. I could accept this, but it sounds weird. Should I just leave it at that or is that wrong?

EDIT: This question is not a duplicate. I understand different types of friction and the forces present in the problem. The answers to other questions do not answer this one. This one can be summarized as "How can static friction with the road produce acceleration but no work, and kinetic friction in the brakes produce work but no acceleration?" Other questions are different.

Qmechanic
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thedude
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2 Answers2

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First, when the car is braking, besides the static friction, there is a rolling friction and possibly, a sliding friction between the tires and the road. These two friction components slow down the car and at least partially dissipate its kinetic energy.

In an extreme case, when brakes stop the wheels instantly, the sliding friction between the tires and the road will be solely (ignoring air resistance) responsible for the car deceleration and for the dissipation of its kinetic energy.

In an more common case, the brakes stop the wheels very gradually, so that the wheels keep spinning, with no sliding, until the car stops. In this scenario, the kinetic energy of the car will be passed, through the axles, to the wheels and will be dissipated by the brakes, i.e., if we ignore the rolling friction and the friction in the bearings, the brakes will be solely responsible for the deceleration of the (center of mass of the) car and for the burning of all its kinetic energy.

There is nothing contradictory here, since this is done through the mediation of the road (through the static friction), i.e., an external object is involved in the change of the car’s center of gravity. Similarly, the engine speeds up the car or a person walks.

If the brakes are not applied at all (coasting), then the car will be stopped due to the friction forces we were ignoring before: rolling friction of the tires, friction in the bearings, air resistance.

So, we can say, that in a general case, there are a number of friction components slowing down the car, involving brakes, tires (rolling and sliding), bearings and air resistance, and, for each component, the force responsible for the deceleration will be also responsible for the dissipation of the corresponding portion of the total initial kinetic energy of the car, which, technically, includes the initial kinetic energy of the rotating wheels.

V.F.
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  • This answer approaches the question, but is not complete. "There is nothing contradictory here, since this is done through the mediation of the road" - this is a bit vague. "the force responsible for the deceleration will be also responsible for the dissipation" - it doesn't look that way at first sight, that was exactly the point of the question. – thedude Apr 10 '18 at 12:30
  • Let's ignore rolling and sliding fric. and focus on the brakes and static fr. btw tires and road which "mediates" the action of the brakes. When a frog leaps forward, the forces and enrgy originate inside the frog, but the "mediation" for the st. fric. of the grnd is needed to convert these forces and enrgies into a jmp. Similarly, the static fric. of the road is needed to convert the braking forces to car deceleration. The static fric. did not perform any work (reactive force), since there is no relative movement involved: all the work and power dissipation happened inside frog and brakes. – V.F. Apr 10 '18 at 18:29
  • So we can think of brakes as producing a force that pushes the ground (in the direction of motion) and due to Newton's third law the reaction forces from static friction decelarates the car? That is, the friction helps to have the condition of rolling without slipping? – Antonios Sarikas Apr 14 '22 at 18:08
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In a moving car there is only static friction between the tyres and the road. When you apply brakes you usually try to stop the rotation of the wheel. For the wheel the friction from the brake is an external force. If the wheel suddenly stops rotating, as the theory of rolling and kinetic friction says, there will be two surfaces (the tyres and the road) in relative motion. Now kinetic friction will try to stop this relative motion. At last the car comes to rest. But you should not try this if you are at a high speed otherwise your inertia will throw you outside the car...XDXDXD

Sayantan Das
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