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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.

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    $\begingroup$ Just by the way, this is why braking distances increase by the square of the speed. See for example qld.gov.au/transport/safety/road-safety/driving-safely/…. You'll notice that if you double the speed (say, from 40km/h to 80 km/h), the braking distance increases by a factor of 4. If they had extended the diagram to 120km/h, the braking distance at 120km/h would have been 9 times the braking distance at 40km/h. $\endgroup$ – Graham Oct 17 at 13:58
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    $\begingroup$ I think your intuition goes wrong because heat dissipation isn't the limiting factor in how long it takes to slow down at the speeds you are talking about. $\endgroup$ – chepner Oct 17 at 17:21
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    $\begingroup$ Other frictional forces (e.g. air resistance) are also greater at increased speeds. So the brakes themselves don't have to do all of the work that you calculated. $\endgroup$ – John Oct 18 at 1:31
  • $\begingroup$ One practical consequence of this phenomenon is that brake design depends on the typical speeds of the vehicle. High-performance sports cars have much larger brake discs than similar-sized family saloons. Formula One cars have massive brake discs, despite the cars being rather light, which glow red-hot when the brakes are applied at high speed. $\endgroup$ – Oscar Bravo Oct 18 at 7:32
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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.

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  • $\begingroup$ This makes sense. One more thing that popped into my mind - what about a rocket travelling through space? For it to accelerate from 900m/s to 1000m/s, does it use less rocket fuel than from 1000m/s to 1100m/s? I would assume not, because I think this is possibly a conservation of momentum question rather than a conservation of kinetic energy - (the rocket is 'throwing' rocket fuel out rather than using traditional work for propulsion. $\endgroup$ – Jethro McLean Oct 18 at 4:38
  • $\begingroup$ @JethroMcLean You are right that in rocketry, we are more concerned with momentum than energy. The major equations in rocketry are primarily concerned with velocities and momentums. The energy side of the balance is less important. Energy gets conserved, it just gets conserved in ways that we typically don't find important for rockets. As such, it is common to talk about a "delta-V" budget. $\endgroup$ – Cort Ammon Oct 18 at 5:07
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    $\begingroup$ @JethroMcLean It's still true for rockets, of course, but there's one important difference - in a car, you can largely ignore the amount of kinetic energy you impart on the Earth (remember Newton's laws?). In space, there's nothing to push against - you're throwing stuff (propellant) out the backside as fast as you can. The total momentum and energy of the "rocket + propellant" system is always the same, and the propellant is always expelled at the same speed, regardless of the rocket's speed relative to e.g. Earth. If you throw the propellant faster, you need more energy per unit mass. $\endgroup$ – Luaan Oct 18 at 7:30
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    $\begingroup$ @JethroMcLean The simplest rocket is "throw half of your mass backwards at 10 m/s". If you aren't moving and you weigh 100 kg, this requires 50 * (10m/s)^2 J = 5 kJ of energy (note: KE of both "ship" and "thrown part"). Now suppose we are already moving at 10m/s relative to the observer. Afterwards, you have 25 * (20m/s)^2 J = 10 kJ and before you had 50 * (10m/s)^2 J = 5 kJ for a change of ... 5 kJ. The "loss of fuel KE" from shooting it backwards balances the "gain of ship KE". Magically. Well, no, noetherly, but close enough. $\endgroup$ – Yakk Oct 18 at 15:36
  • $\begingroup$ @JethroMcLean I think the rocket case is tricky because energy is only conserved if you measure everything within one inertial reference frame. So the energy in the frame of a rocket going at speed x isn't the same as the energy in the frame of that same rocket after it accelerated. If you look at the two situations from the same reference frame, external to the rocket, you'd need to consider that the fuel itself has more kinetic energy in the second situation. $\endgroup$ – bdsl Oct 19 at 16:06
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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.

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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.

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    $\begingroup$ +1 for showing that force is actually the same, so that is force and not work what meets the OP's intuition. $\endgroup$ – Pere Oct 18 at 17:46
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To reach a new $E_k$ one must dissipate energy from the initial state, to the new final state:

$${\Delta}E_{k}=E_{ki}-E_{kf}$$

Where kinetic energy is defined as:

$$E_k=\frac{1}{2}mv^2$$

Do the brakes have to do more work (ignoring air resistance) slowing a vehicle from 10m/s to 8m/s than from 8m/s to 6m/s?

Think conceptually, what are you really asking? You are asking, "is kinetic energy dissipation linear or non-linear"? Well, by looking at the equation above, we clearly see that the energy is a function of velocity squared, ie a non-linear function.

So, yes, your math is correct, and conceptually it follows as kinetic energy is a non-linear equation.

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    $\begingroup$ And just why is it non-linear? Because it's force applied over a distance -- and while it takes the same time to dispose of a certain ${\Delta}v$ with a given acceleration, the car has traveled farther in that time. So you need (a) more energy because, well, you have more ${\Delta}v$ to shed to begin with, and (b) you cover all that additional ground due to higher speeds, and that addition is proportional to the average speed, which is proportional to the speed when you start to brake, hence that speed multiplies twice. $\endgroup$ – Peter - Reinstate Monica Oct 17 at 17:55
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Lets conduct a thought experiment. Build a special car.

The car is special because its wheels are perfect omni wheels. There are wheels that rotate sliplessly when the car moves along $X$, but they slide without friction when the car moves along $Y$. And there are wheels that rotate sliplessly when the car moves along $Y$, but they slide without friction when the car moves along $X$. The two directions are horizontal and perpendicular to each other.

Each set of wheels has a brake, so there are two brakes. One brake allows you to bring the $X$ component of the car velocity to $0$, the other allows you to reduce the $Y$ component to $0$. Each brake affects one component only, the brakes are in this sense "orthogonal".

Now imagine you drive the car and the velocity components are $1\frac m s$ along $X$ and $1\frac m s$ along $Y$ – but you don't know it yet because the windows are deliberately covered. Your task is to measure your initial speed with respect to the room somehow, knowing how the car works and how much it weights.

Your idea: engage the first brake and measure all the heat you will get. Because the braking wheels will still slide freely along $Y$, the velocity component along $Y$ neither will interfere with this process nor itself change. You will get the energy associated with the car's movement along $X$ only.

You do this, perform calculations and the answer is $1\frac m s$ along $X$. You repeat the procedure with the other brake and the answer is $1\frac m s$ along $Y$. Both brakes are engaged, now the car is at rest.

You got the heat corresponding to $1\frac m s$ twice. Your intuition says it's the same amount of heat you'd get by braking from $2\frac m s$ to $1\frac m s$ and then to $0$. You declare your initial speed was $2\frac m s$.

Pythagoras strongly disagrees. He says your initial speed was $\sqrt 2\frac m s$. After a bit of thinking you know he's right and you change your answer. Then you realize this means that decelerating from $\sqrt 2\frac m s$ to $1\frac m s$ would give you the same amount of heat as decelerating from $1\frac m s$ to $0$. And because you expect you can get some heat by decelerating from $2\frac m s$ to $\sqrt 2\frac m s$, then you have to admit that braking from $2\frac m s$ to $1\frac m s$ would convert more kinetic energy into heat than braking from $1\frac m s$ to $0$.


Your intuition would like to see the kinetic energy proportional to the speed (absolute value of velocity). Assume the intuition is right and imagine the initial velocity as the hypotenuse of some right triangle, where two other sides are along $X$ and $Y$ (velocity components). A traditional car could convert the length of the hypotenuse to heat by just braking. Our car with omni wheels could convert the length of one component with one brake, the length of the other component with the other brake. In total we would get more energy as heat. Different directions of velocity would give us different amounts of heat, each time at least as much as the traditional car would get. And each time we would say the final kinetic energy is 0, we converted all the kinetic energy there was.

In fact (and you know it) the kinetic energy is proportional to the speed squared. A traditional car converts the squared length of the hypotenuse to heat. Our special car converts the sum of squared lengths of two other sides. By the Pythagorean theorem these values are equal. The velocity direction doesn't matter.


To connect our thought experiment to the values in question, let's imagine you want to experimentally measure how much heat you get by braking from $10\frac m s$ to $8\frac m s$; and separately from $8\frac m s$ to $6\frac m s$. You use your special car for this.

One inconvenience though: once a brake is applied, it cannot be released until the car totally stops and you unblock things from the outside.

So you cannot just accelerate to $10\frac m s$ along $X$. If you did, you would be able to decelerate to $0$, not to the desired value of $8\frac m s$.

Worry not! Our previous experiments revealed that kinetic energy (amount of heat you can get from it) doesn't depend on direction of movement. So you accelerate the car to $8\frac m s$ along $X$ and to $6\frac m s$ along $Y$. Now your speed is $10\frac m s$ and you can reduce it to $8\frac m s$ by applying one of the brakes for good. You do this and measure the heat. The movement direction has changed but it's OK, the only thing that matters is you're traveling $8\frac m s$ now.

On the second run you accelerate the car to $6\frac m s$ along $X$ and to $2\sqrt 7\frac m s$ along $Y$. The speed is $8\frac m s$. You already expect you will get less heat than in the previous run because $2\sqrt 7 < 2\sqrt 9 = 6$. You apply the proper brake, reduce the $Y$ component to $0$ and measure the heat. It's indeed less than before.

Note your current speed is $6\frac m s$ now (along $X$ only). Aren't we lucky? You can masure the heat from the other brake when you come to a total stop and confirm it's equal to the result from the previous run, when you lost the $6\frac m s$ component.

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About the difference between decelerating from 10 units of velocity to 8 units, versus from 8 units to 6. (This doesn't address you question yet, but I want to get this out of the way first.)

Thought experiment:
Construct a "road" on a chain of freight train cars, so that a car can travel the length of the train. You have the train going along at 2 units of velocity, you have the car going at 8 units of velocity relative to the train.

Then you decelerate the car, from 8 units to 6 units (relative to the train).

As we know, the surface that the tires are gripping to decelerate the car is the train, so that is what counts. The car was slowed down from 10 units to 8 units relative to the earth, but the tires aren't gripping the Earth, so that doesn't count.

Now to the core of your question:

What is kinetic energy?

I like to think of it in terms of the following thought experiment.

Set up a long row of easily tearable paper screens, and shoot a ball through. (Let's say a marble is used.)
Each time the marble tears through a screen it loses a bit if its velocity. Let it be set up in such a way that the succesion of of tearing through the screens is to a good approximation a constant deceleration.

Let me define:
'Total time': the total time for the marble from the first tearing to coming to a stop.
'Total distance': the total distance traveled from the first tearing to coming to a stop.

As we know:
When an object decelerates at a constant rate the change of velocity is proportional to the time. That is: the marble loses half of its velocity in the first half of the total time, and it loses the other half of its velocity in the remaining half of the total time.

But we also know:
In the first half of the total time the marble covers 3/4 of the total distance.
The (averaged) force causes a constant deceleration. As we know: with a constant acceleration/deceleration the distance travelled is proportional to the square of the time.

It's because of that quadratic relation that the marble tearing through the series op paper screens covers 3/4 of the total distance in the first half of the time. Of all the tearing that the marble did 3/4 happened in the first half of the time.

So we can see the kinetic energy of an object as a measure of how much damage that object will do per unit of time.

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You have the physics. Your issue is with the intuition.

If you need to feed your intuition, get in your car. Accelerate the car from 0-60 kph. That should be easy, yes? Now accelerate it from 60-120 kph. Three times harder.

"But aerodynamics," you say. OK, try it from 0-20 and 20-40 kph. In a car this will occur too fast to measure, so do it on a bicycle (making full use of 10-21 gears).

Now venture in the world of the freight railways, where traction drive is electric and masses are enormous, up to 10,000 or 20,000 tonne on North American, Russian or Australian roads. This is still colored by the need to not overcurrent motors or tear out drawbars, but once the traction current meter is out of the red zone, train behavior can be easily observed to conform to theory. Throttle notch position will correspond tightly to KW to wheels. Rolling resistance is essentially nil, especially on concrete ties and welded rail, and aero drag doesn't start stacking until over 40 kph.

Anyway, this is why the old chestnut of slowing down your car is so darn useful. When you reduce speed, you are quadratically reducing the amount of kinetic energy you're "bringing to the party" i.e. that the tires must redirect in a turn, or that combines with other forces like wind, or that the guardrails and your car's energy absorption structure must dissipate in a crash. Cut speed 29.3% and you've halved KE.

If you ever see me pass you, and then we come onto a tall bridge or wind hazard and I slow down notably, that's me reducing KE to improve stability and handling.

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