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I've had an interesting conversation a day ago and we came to a stop regarding one explanation.

Conversation went about how increasing brake rotors doesn't make your car stop faster. But we came to the part where I couldn't explain how can we calculate the work done by human foot pressing on brake pedal.

When we press on pedal we exert a force on it. And this pedal travels some distance, so there is work done. Through all of hydraulic system we get a normal force on brake caliper and through it we get friction on the rotor. But let's assume that the brake rotor isn't spinning. We can press on the pedal as much as we want, but there is no friction created.

One more thing that bothers me is also equivalent of work done when we want to stop the car. We know its kinetic energy and all of this kinetic energy must transfer to heat energy of brake rotors (if we are correct, some goes to tires). I don't know the physical correct explanation of work done by foot and all this energy that transfers to heat on brake rotors. Can we even make this equal?

I know that calculation of work done by friction is straightforward. We know the magnitude of force and distance. But how do we calculate work done by normal force, which is perpendicular on the path of friction so dot product is zero?

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    $\begingroup$ Not an expert of how cars work, but I think it is used the same principle of friction but exerted by the very brake rotors. So probably you should calculate the torque exerted by the dynamic friction force on the brake rotors which will reduce the angular momentum to zero (opposing the rotation). $W = \tau_{br. rot} \cdot \theta $ $\endgroup$ – Costantino Aug 27 '18 at 19:41
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    $\begingroup$ The work to stop the car is not done by the force your foot applies to the brake pedal. It is done by the friction force between the stationary brake pads and the moving brake disks. If the disks are not rotating, then the wheels are not rotating either, and the force that stops the car is the friction between the tires and the road. $\endgroup$ – alephzero Aug 27 '18 at 20:01
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    $\begingroup$ You are right that the normal force between the brake pad and the disk does no work, but the friction force is not normal to the brake disk. It acts in the direction along the surface of the pads and the disk, in the direction that oppose their relative motion. $\endgroup$ – alephzero Aug 27 '18 at 20:04
  • $\begingroup$ I didn't write this before, but as I know all the work is done with friction between the tire and the road. This is what stops the car. It is best if our tires don't lock, that's why we use ABS. But let's assume that we don't have it and we press the brake pedal as much as we can. The discs usually lock up, because the limiting factor is tire/road not disc/brake pad. But the car moves on with friction on discs doing no work. All the work is done by the friction on the tire/road. $\endgroup$ – Grega Aug 27 '18 at 20:16
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    $\begingroup$ Careful - the role of the tire/road interaction is very tricky. Since the point of contact between the tires and road is stationary, no energy is being transferred between them (unless the car is sliding). The engine does work on the tires to give them rotational kinetic energy, and the tires use the road to transfer that energy into the translational energy of the car. Of course, this is ignoring the entire ABS issue - idealized situation! $\endgroup$ – levitopher Aug 27 '18 at 20:33
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@levitopher's answer is good, but I just want to add that you are doing work with your foot when you press the brake pedal; however, that work is just going into compressing the springs which, in their default position, hold the brake calipers open and away from the rotor. When you release your foot, that work is returned to your foot by the springs - the work from your foot is not directly going into stopping the car.

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The comments are essentially all correct, I will try to summarize.

When you push on the brake, your foot travels some distance as the brake depresses. Since you feel some resistance, you are doing some amount of work on the pedal. However, that amount of work is absolutely insignificant for stopping the car. (BTW, that foot-on-pedal force is a normal force, but I don't think it's the normal force you are talking about).

The force that stops the car is the friction due to the brake pad, as you mentioned. There is a hydraulic system in the car which transfers and "multiplies" the force you push on the pedal with to the normal force between the brake pads and the brake disk (using something called Pascal's law, maybe a separate topic).

Although the normal force is very large, as you point out, it's acting perpendicularly to the motion of the wheels, so it does no work. However, the wheels are spinning, moving in the opposite direction as the applied force of friction from the pads - so this is the force which does work to stop the car. The amount of work done looks something like this:

$$W=F_f r \theta$$

where $F_f$ is the force of friction, $r$ is the distance from the center of the wheel the brake is applied, and $\theta$ is the angle that the wheel rotates while braking.

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    $\begingroup$ 's I Think taht the work enegry to bring the car from velocity $v_0$ to $0$ is $W=\frac{1}{2} m v_0^2$ $\endgroup$ – Eli Aug 28 '18 at 13:52
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    $\begingroup$ @Eli: what you're referring to is the content of the work-energy theorem: The work done on a system is equal to it's change in kinetic energy. But you can also write the work done by a particular force as I have, as a force times a distance (or in the case above, torque times an angular distance). If you know there is only one force acting on the system, you can set them equal, e.g. $F_f r \theta = \frac{1}{2}mv_0^2$ $\endgroup$ – levitopher Aug 28 '18 at 14:46

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