# How does kinetic energy work in braking a vehicle?

The work is basically the amount of energy that is used to make something move. So first some math to gain insight how work works:

In the case of constant force work is defined as $$W=F s,$$ where $W$ is work, $F$ is the applied force and $s$ is the distance the object traveled in the direction of the force. The force is defined as $$F=m a,$$ where $m$ is the mass of the object and $a$ its acceleration. For constant force we have constant acceleration, which can be computed as $$a=\frac{v_2-v_1}{t},$$ where $v_2$ is the end velocity, $v_1$ is the starting velocity and $t$ is the time that passed during slowing down from $v_1$ to $v_2$. We also need the distance that the object traveled, which is: $$s=v_1 t +\frac{at^2}{2}=v_1 t +\frac{v_2-v_1}{2}t=\frac{v_2+v_1}{2}t,$$ where we plugged in our formula for acceleration. Now to put it all together we get: $$W=m\frac{v_2-v_1}{t}\frac{v_2+v_1}{2}t=m\frac{v_2^2-v_1^2}{2}=E_2-E_1,$$ where $E_2$ is end kinetic energy and $E_1$ is starting kinetic energy of the object.

So why is this not proportional to velocity difference but to velocity squared distance? That is simply because the force applied is proportional to the velocity difference through the acceleration being proportional to the velocity difference. That makes sense doesn't it? To slow down your car your force need to be bigger the bigger the velocity difference is, if your are to take it the same amount of time.

But this force you need to multiply by the distance traveled and that distance depends on your initial velocity. The bigger your initial velocity, the bigger the distance you travel to slow down by the same amount of speed with the same acceleration, which seems pretty intuitive to me. So once you multiply the force, that is proportional to the velocity difference, by something that is bigger the bigger your initial velocity is, your resulting work must be bigger the bigger the initial velocity is, if your are to have the same velocity difference. Just as your computation suggests.

It looks like you know how to work through the formulas, but your intuition isn't on board. So any answer that just explains why it follows from the formula for kinetic energy might not be satisfying.

Here is something that might help your intuition. For the moment, think about speeding things up rather than slowing them down, since the energy involved is the same. Have you ever helped someone get started riding a bike? Let's imagine they're just working on their balance, and not pedaling. When you start to push, it's easy enough to stay with them and push hard on their back. But as they get going faster, you have to work harder to keep the same amount of force at their back.

It's the same thing with pushing someone on a swing. When they're moving fast, you have to move your arm fast to apply as much force, and that involves more energy.

If that isn't helpful, consider a more physically precise approach. Suppose, instead of regular brakes, you have a weight on a pulley. The cable goes from the weight straight up over the pulley, straight back down to another pulley on the floor, and then horizontally to a hook that can snag your car's bumper. And just for safety, assume the weight is pre-accelerated so the hook matches the speed of the car as you snag it. Some mechanism tows the hook and then releases it just as it snags your car. Then all the force of the weight goes to slowing the car down.

If you snag the hook at 100 kph, that weight will exert the same force, and hence the same deceleration, as if you snag the hook at 10kph. The same deceleration means you slow down the same amount in the same time. But obviously the weight is going to go up a lot farther in one second if you're going 100 kph than if you're going 10 kph. That means it's going to gain that much more potential energy.

**Work is force times distance**.

Assuming that your brakes apply the same force in each deceleration, it takes the same amount of time to go from `10m/s`

to `8m/s`

as it does to go from `8m/s`

to `6m/s`

. However, **the vehicle is slower in the second deceleration, so it doesn't travel as far**. As such, force is the same, but distance is smaller, and less work is done. Exactly what you expect from differencing the kinetic energies.

To see that traveled distance is actually important, just consider the ground that supports you. It constantly applies quite some force on you, but it does exactly zero work because it does not move up/down with you on top. A lift, however, needs to put in energy to get you to the top of a building: It pushes on you with the same force as the ground does, but it also moves upwards in the direction of the force, and thus transfers energy to you. The work done by the lift is exactly your gravitational force times the vertical distance you traveled.