Do the brakes have to do more work (ignoring air resistance) slowing a vehicle from $10\ \mathrm{m/s}$ to $8\ \mathrm{m/s}$ than from $8\ \mathrm{m/s}$ to $6\ \mathrm{m/s}$?
Say a $1000\ \mathrm{kg}$ vehicle is moving at $10\ \mathrm{m/s}$, it has a kinetic energy of
$$\frac12\times1000\ \mathrm{kg}\times(10\ \mathrm{m/s})^2=50\,000\ \mathrm J$$
Then the brakes are applied, and it slows to $8\ \mathrm{m/s}$, so now has a kinetic energy of
$$\frac12\times1000\ \mathrm{kg}\times(8\ \mathrm{m/s})^2=32\,000\ \mathrm J$$
The brakes are now applied again, and it slows to $6\ \mathrm{m/s}$, now the kinetic energy is
$$\frac12\times1000\ \mathrm{kg}\times(6\ \mathrm{m/s})^2=18\,000\ \mathrm J$$
So in the first braking instance, $50\,000\ \mathrm J – 32\,000\ \mathrm J = 18\,000\ \mathrm J$ of kinetic energy were converted into heat by the brakes.
In the second braking instance, $32\,000\ \mathrm J – 18\,000\ \mathrm J = 14\,000\ \mathrm J$ of kinetic energy was converted into heat by the brakes.
Doesn't seem intuitively right to me, I would imagine the work required from brakes would be equal to the amount velocity was reduced, regardless of the start velocity.
Best Answer
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.