First, I’ve never taken a physics class or attended highschool before, hence the ELI5. I’ve read many explanations but it doesn’t really make intuitive sense to me. For example (assuming there’s no air resistance / drag), let’s say I was traveling in a car going 120 mph and I wanted to decelerate to 90 mph. This would take four times as much energy than going from 30 mph to 0.
But let’s say there were two cars traveling at 120 mph. The car next to me decelerates to 90 mph, but I’m still going 120. From my point of view, the car next to me just started going 30 mph in the opposite direction. Why would this require 4 times as much energy than if both cars were just stationary, and the car next to me actually started going 30 mph in the opposite direction?
And, let’s say we’re both standing on earth. One person at the north pole and one at the equator. Both of us throw a ball, but the ball at the equator is already traveling at something like 1,000 mph due to the earth’s rotation. Shouldn’t throwing a ball eastward then require way more energy to go from 1,000 to say 1,020 mph, than the person throwing the ball at the north pole who just has to accelerate it from 0 mph to 20 mph?
In: 3
Let’s chose as reference the a point on a road. And let’s accelerate a car. I assume you know the equations. Let them assume we use the car’s engine to generate the force, and this engine is a combustion one, and we never shift gear, fix gear ratio.
We start the reasoning with: Speed = time * force / mass.
It happens that the car will travel faster and faster while accelerating, covering more and more distance per second. Distance=time squared * force / mass.
I will keep using force/mass as a substitute for acceleration.
Now, let’s say you push ed the car with 1000force, for 1000 distance, it will get to 10speed. To get the same car to the speed of 20, it will take twice the time.
But how much distance it took? To get to 20 speed instead of 10 speed, we will need twice the time. Again, distance=time squared*force/mass. So we will cover 4 times more distance as 2time squared is gives 4time as a result.
Breaking it down to the combustion in the engine, while the car gets faster and faster, the engine spins faster and faster, burning fuel faster and faster. There will be a fix ratio between engine and wheel, for each revolution of the wheel there will be a fix number of combustion cycles, all burning the same fuel. Pause it at any point of the travel: you can see for each meter of road, you burn a fix amount of fuel.
If getting to twice the speed covers four time more road, therefore burning four time more fuel.
This means the car has received 4 times more energy to get to 20speed compared to the energy it takes to get to 10speed.
From energy pov, the time it took doesn’t matter, what matters is what you burn to get there and for how much distance you had to burn.
The car’s kinetic energy is nothing else that the burned fuel energy. And the amount is determined by how much force for how many meters It traveled under that force.
To recap: kinetic energy=“fuel used” to get the object to that speed.
(speed=force * time /mass, transform this you have:
force= speed * mass / time.
Distance= speed * time)
SO HERE WE ARE: Kinetic energy = “fuel used” = force * distance = (speed * mass / time) * (speed * time) = speed * speed * mass / ~~time / time~~= mass * speed squared
Hope this example helped visualize it. The rest of your question is answered just by changing the reference point. Law applies the same.
I also had this problem. It seems unintuitive that somehow going from 50 to 60 mph adds a different amount of oomph to an object than going from 60 to 70 mph.
I’m going to base my answer on [this](https://physics.stackexchange.com/a/14752) physics stackexchange answer. I’ll rephrase it to make the mathematical steps more clear, but I recommend reading that answer as well if you know high school algebra well.
Hopefully you have the intuition that energy is some property that objects have which can cause internal transformations in matter. Whenever an object collides and as a result deforms or heats up, that’s kinetic energy acting on the object.
So let’s look at a situation that makes it easy to account for all the kinetic energy – a clay ball hits a wall. It will splatter, and the size of this deformation can be used to calculate the energy. If two clay balls of equal size hit a wall and splatter to the same extent, they had the same amount of energy. Let’s say the balls were moving with speed **v**, and had an amount of energy **E**.
Our first step is to change this experiment. Instead of throwing the balls at the wall, we throw them at each other. They splatter in the air instead of on the wall, coming to a complete stop. They were both still moving with speed **v**, but in opposite directions. When we do this experiment, we find that the amount of deformation each ball undergoes in this collision is the same as when they hit a wall. This makes sense – each ball has energy **E**, so the entire collision has energy **2E**, but it’s spread out through twice as much mass. Each ball undergoes an **E**’s worth of explosion, the same as if they each hit a wall.
Our second step is to change this experiment once again. Throw the balls a each other, but now observe them from a car moving forward at speed **v**. The ball we’re following looks like it’s hanging in the air, while the other ball is moving toward it at speed **2v**. Now to see what happens when they collide, you’ll need to keep track carefully. To a person standing on the ground, the two balls stopped moving completely, and fell straight down. But since we’re in a car moving forward at speed **v**, the collided balls are now moving backward at speed **v**. From the car, it looks like the **2v** ball came in, collided, and knocked the combined system backwards.
Now the crucial step – other than this change in relative speeds, the collision looks identical. Us being in the car doesn’t change the events. The splatter of each ball is the same size. So the moving ball went from **2v** to **v**, and yet it delivered **2E**’s worth of energy. Going from **0** to **v** grants an object **E** worth of energy, and yet going from **v** to **2v** grants *at least* **2E** (in reality more, since the two-ball system is still moving, so it didn’t lose all its kinetic energy to the splatter). We got here not by assuming any kind of mechanics, but by assuming that the laws of physics work the same when we’re moving as when we’re stationary – if we do all our math from a moving car, we should still observe the same things happening, even if we observe them happening at different speeds.
To get the exact quadratic relationship, we have to account for the fact that the two balls are now moving backwards with velocity **v**, which means there is an extra **2E**’s worth of energy stored in their motions. That’s a total of **4E**. If velocity **v** has energy **E**, while velocity **2v** has energy **4E**, that means doubling the velocity quadruples the energy, which is only possible if **E(v) ~ v^(2)**. Then **E(2v) ~ (2v)^2 ~ 4 v^2**.
This argument suggests that this relationship does not come from the inherent properties of objects, the way electromagnetic energy does. That energy is somehow stored in the bonds between nuclei and electrons. Kinetic energy is some kind of property of the geometry of space and time, so just thinking about the symmetries of motion can tell us what the relationship between motion and energy must be.
Other have explained the frame of reference part but let’s look at the square part.
The reason is the square of the velocity is at high speed you need to apply the force for a longer distance.
(1) W = F *s
where W =Work (transferred enerergy), F = force and s= applied distance
(2) s = v *t
Where v is the speed and t is the time.
(3) a= F/m => F =m *a
where a is the acceleration and m is the mass.
(4) v= a * t => t= v/a
Consider if you an object with a mass of 1 kg that you accelerate with a force of 1N that means the acceleration is 1m/S
If you start at a speed of 0 the speed after 1 second is 1m/s so the average speed for that second is(0+1)/2= 0.5 m/s and you only travel 0.5m That is the distance you apply the force. So the work is 1 N * 0.5m = 0.5 Joules.
If you start at 10m/s the speed after 1 second is 11m/s so the average speed is (10+11)/2=10.5m/s . So you traveled 11.5m. That means the work is 1 N * 11.5m = 11.5 joule
If we just have constant acceleration 0 to v the average speed is v/2.
Combine 1 and 3 and we get (5) W= m *a *s
If we use 2 with the average speed you get (6) s= v/2 *t
Combine 5 and 6 and we get (7) W= m *a * v/2 *t
Combine 7 with 4 and we get W= m *a * v/2 * v/a = m * v/2 *v * a/A = m * v/2 *v = (m * v^2)/2
So the square is a result of that you need to apply the force over a longer and longer distance the higher the speed is and the work gets larger and larger.
It can also explain what we see in a different frame of reference. When a car slowed down from 120 to 90 by applying breaks the force is applied from the road. So from your point of view, the car change is speed but pushing on something that moved backward at a speed of 120 relatives to you. You need to include that the ground is not stationary compared to you.
For it to do less work whilst slowing down it needs to apply force to something that moves at the same speed as you.
You can compare this to you walkin’ in a train where you start waking forward in the direction of the moment. Your feet apply the force relative to the train so the distance is short. But at the same time, the train needs to apply the same force relative to the ground, if it did not it would slow down.
Compare it to walking in a canoe on the water next to the dock. It has very little friction from the water and the mass close to you. If you stand up and walk forward you push the canoe backward. When you reach the front of the canoe you have walked a short distance relative to the dock. So if you use the canoe as the frame of reference you need to consider how it moves relative to the dock. You moving in a train change it a moment too but it will be a lot less compared to you because its mass is many times higher than you
So if you jump off the train and hit the ground, the difference in energy from you walking forward versus stationary when you hit the ground from the work the train needed to do to keep its speed constant.
So it all works out regardless of the frame of reference. You just need to consider everything like what you apply a force relative to and what your action has on the frame of reference
> let’s say I was traveling in a car going 120 mph and I wanted to decelerate to 90 mph. This would take four times as much energy than going from 30 mph to 0.
> But let’s say there were two cars traveling at 120 mph. The car next to me decelerates to 90 mph, but I’m still going 120. From my point of view, the car next to me just started going 30 mph in the opposite direction. Why would this require 4 times as much energy than if both cars were just stationary, and the car next to me actually started going 30 mph in the opposite direction?
This is a really fascinating question. And the answer takes a bit of thinking about, but you’ve already got the basics of it.
In classical mechanics (pre-20th century) energy is a mathematical tool that is useful in figuring out how things interact. It corresponds to how much something has been forced. The more it has been forced, the more energy it has gained. Because forces are symmetric (Newton’s Third Law) the more one thing has been forced, the more other things have forced things; so we can factor this into energy. If one thing has been forced and so gained “been forced stuff” (energy) we can say that the things that forced it have lost “been forced stuff.” We get conservation of energy!
If energy is conserved that can be really useful for understanding situations. Rather than having to worry about all the forces and interactions going on at each moment in some system, we can look at the energy at the start, look at the energy at the end, and you can get a great idea of what is happening.
Anyway. The important thing to note about this is that it is a maths thing; helpful, but has limitations. One of the big ones is that it depends on our frame of reference (which is what you’ve noticed). It isn’t absolute, but relative. But that’s Ok as we’re only ever looking at how it changes, and it just means we have to be careful about switching between reference frames when doing energy calculations (but we should be doing that anyway).
So let’s look at your situation.
Your car is going at 120mph. It want to decelerate to 90mph. That would involve losing more energy than you’d have to lose to decelerate from 90mph to 60mph, and 4 times as much energy as it would take for you to decelerate from 30mph to 0mph. But this is all from a reference frame fixed with the ground.
From *your* point of view, though, you’re at rest. You’re not changing speed, so no need for any change in energy at all! Of course, you’re in a non-inertial reference frame, so the maths gets a bit weird, but this is a good indication that something weird is going on.
So let’s look at another point of view. Some car that was travelling at the same speed as you before you started slowing down.
From their point of view you were going at 0mph, then *sped up* to 30mph (backwards), then sped up further to 60mph and eventually up to 120mph. From their point of view *you have gained energy*! This is getting really weird…
The big thing we’re missing is that energy is being transferred; if something is gaining or losing energy, something else must be losing or gaining it. We’re only looking at half the problem.
Let’s massively oversimplify the problem and assume that as you slow down you are dumping energy into the Earth. Conservation of momentum tells us that as you slow down the Earth must speed up. So from our “with the ground” frame of reference, the Earth will be gaining energy as you lose energy.
But from the other car’s point of view, as you “speed up” the Earth will “slow down,” so as you gain energy, the Earth loses energy. Which is what we expect.
The maths gets a bit messy (and we have to worry about whole systems for conservation of momentum and conservation of energy) but it all works out in the end. Provided we do *everything* (including measuring distances and forces) from the just one reference frame. The calculations will be different and give us different changes in energy, depending on which reference frame we use, but we will get the same overall result for how things move.
Oddly enough, this is kind of how E = mc^2 was first derived; by looking at how energy changes work when viewed from different reference frames. Einstein’s paper on it is only a couple of pages and is a simple thought experiment.
Simply because that is how it remains conserved.
Lets look at the conservation of momentum first. Its a vector quantity. We define momentum p=m×v that way because that is how it remains conserved. Now not only the direction but the “lenght” of that vector is conserved. We can create a scalar quantity based of of that fact. So just a number and call it kinetic energy.
You have correctly identified the fact that kinetic energy is frame dependent and not conserved between frames of reference.
You aren’t really missing anything other than that you should do all comparisons from within the same frame of reference, but what frame you choose can be arbitrary and the math will ultimately all work out.
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