Is there an intuitive way to grasp why kinetic energy is proportional to the square of velocity instead of just linear?
I know it's a novice thing, but I'd like something without equations.
I've seen the "throwing something upwards reaches the square of the height" analogy, but that just changes one question for another (why is double the height double the potential energy?).
I've also seen the "you have to move faster to keep pushing an object at the same force"
For example, the pistons of a combustion engine need to move ever faster to maintain acceleration, thus consuming more gas.
This last one appeals to me. Is it "correct"?
Kinetic energy
Eli James
Zachary Butler
It's the definiton that comes from work. Do some calculus on w = F * dr
Jayden Brooks
Doe a rocket engine push with the same force constantly? Does that mean it delivers a constant amount of energy to the vehicle per unit of time?
If so, why does the vehicle's energy increase quadratically?
Joseph Davis
shutup pleb brainlet, if its not intuitive now it will never be
Anthony Butler
>Doe a rocket engine push with the same force constantly? Does that mean it delivers a constant amount of energy to the vehicle per unit of time?
No, but the mass of the rocket decreases with time (as rocket fuel is consumed).
James Hill
1. This is only true for closed system (i.e. constant mass).
2. It is derived from momentum and the second law of motion.
Grayson Sanchez
look up your question in physics stack exchange, I kind of remember some Israeli guy gave a very good answer not using work which would be circular and stupid but symmetry
Luke Butler
Can you sculpt an answer out of this starting point?:
An object falls and as it does it accelerates to about 10 m/s in the first second, then to about 20 m/s in the second second.
In the first second though, the object din't fall 10 meters though, it fell a distance equal to the average acceleration, which is exactly half, which is 5m. This is because half the time it was going slower than 5 m/s and half the time it was going faster than 5m/s. You can see how it equals out. We'll call this distance 1 block/rectangle/square/ whatever.
Now, in the second second, the object is already going 10m/s (even though it has traveled only 5m). Therefore, it will cover AT LEAST 10m more. It will accelerate from 10m/s to 20m/s, meaning its average velocity will be right smack dab in the middle: 15m/s. So therefore it will cover 15 more meters in this time.
How far has the bad bitch gone in these two seconds? 20m. thats a total of four times what it traveled the first second. Wow! it has conquered 1 hurdle in the first second but 4 hurdles in the first two seconds! that means its accumulating hurtles a lot faster w/ each passing second, and they continue to build up.
The same thing happens with kinetic energy hurtles that are accumulated by speed as it increases. Just turn the analogy (above) on its side, and you see that speeds that are increasing are not just gaining more steam as they go faster, they are hording and collecting the energy from the speeds they had before, which were naturally not as potent at the speeds they just gained.
Also, doubling a larger speed does not change your speed by the same amount as doubling a smaller speed. Doubling from 5 to 10mph changes your speed by 5. Doubling again from 10 to 20 changes your speed by 10. (This is just to kind of get a feel of the concept)
(cont)
Xavier Myers
(Cont)
So: a car starts out and has conquered 1 hurdle of KE at 5mph. doubling the speed doesn't erase this conquered hurdle, but effectively adds 3 more as the KE's smoothly transition through the averages. Therefore the total is 4 hurdles conquered by the time it reaches 10mph.
Liam Mitchell
The first answer here is the best treatment of this question I've ever seen.
Owen Edwards
[math] \displaystyle
E_k = \frac{1}{2}mv^2 = \frac{m^2v^2}{2m} = \frac{p^2}{2m}
[/math]
Christian Harris
Circular and stupid? It's how the thing is defined lol.
Charles Perez
You drop a ball 1 meter and it lands with a velocity v. If you drop it 2 meters does it land with velocity 2v? No because in the second meter it is already moving so it has less time to pick up speed.
Jacob Gray
What is your point
Ryan Lee
Since the energy that can be gained, potential energy, is proportional to height. Doubling the height doesn't double the velocity it lands at. So kinetic energy must be proportional with velocity squared.
Brandon Perry
I like the symmetry argument a lot better because you don't need things like potential energy, gravity, etc to get the picture. It is a confusing explanation to me because there are so many things going on at once.
Angel James
That's fair. The potential argument relies is essentially just starting with work as already being significant. I find that the argument on the stackexchange doesn't make sense because he mixes up the two reference frames. ie. setting mE(2v) as equal to the two 2mE(v) terms added up from different observers.
Jace Nguyen
Ok OP, here is a straight-up visual.
I set up a simple pulley system to launch a projectile. (Top picture). Note that in the standard setup the hanging box has to consume two separate rope lengths of 1m in order to move 1m down (2m of rope to 1m of fall), while the projectile moves forward 1m every time the rope moves 1 meter. The X marks a tie off to a wall/ solid surface, which holds tension in the loose end of the rope.
I launch a cannon ball and it goes 200mph. Now i want to double that speed. So I add 1 more loop to the pulley so that the box consumes 4 rope sections of 1m for every meter it moves down (4m of rope consumed/ meter). The rope is getting eaten up twice as fast, the projectile still moves 1m/1m of rope...meaning twice the speed right? Wrong. With more ropes, the weight is too light.
So I want to fix my slow launch by doubling the hanging weight in the box. this means there will now be a total of 200g and 4 ropes, equalling 100g for every two ropes, just like in the top pic. Problem solved right? My box will fall with the same speed as before?
Wrong, the projectile now has twice the mechanical advantage on the box, as each rope holding tension represents 1 projectile hanging on the other end if each rope had their own pulley and the system was static.
So I double the weight of the box again. Doing that, along with the doubling of the ropes will produce the same fall speed for the box and therefore twice the velocity.
Nicholas Moore
A rocket is a bad example since it expels mass. You'll just confuse yourself. Use a train or something.
Benjamin Parker
Additional thoughts after sleeping on this:
The point of the picture is not to add confusion w/ gravity or anything. It simply illustrates the connections between an object and it's KE.
Think of the box as the projectile's energy of inertia. The string wrappings are the links between the object and it's inertial energy (imagine all string wrappings are wound directly between the inertial box and the projectile). You can see how the object is more tightly bound to its inertial energy (more links) as it goes faster, but at a price: the energy of inertia doesn't have as much influence per capita over the projectile.
Also, when the # of strings went from 2 to 4 it doubled the mechanical advantage. So now its like that inertial energy is pulling on two projectiles instead of 1. A good way to imagine it is that that extra ball exists in a separate dimension, one that directly overlaps the projectile and pulls in the same direction. I guess this separate dimension value is heat.
Finally, as to not cause confusion the rope in the second pic was not drawn twice as long (as it actually would be). You say, 'But wait, f=ma would seem to imply that pulling w/ twice the force for twice as long would quadruple the speed.' Well f=ma deals more with time than length, and the second projectile would be under tow for far less than double the time. It is still important to acknowledge the string length however, so that Work can be properly calculated and seen in the relationship at 4x.
Aiden Anderson
you mean integrated
Ryan Thompson
I hope you are being sarcastic