Posted by ProxyTracer 17 hours ago
We know intuitively that a ball atop a 20ft ladder has twice the potential energy of a ball atop a 10ft ladder. And we also know when they fall, by the time they reach the ground and all the potential energy has been converted to kinetic energy, the previously higher ball will have twice the kinetic energy too.
But a twice higher ball won't have even close to twice the speed at impact. So let's look at why not.
The force of gravity is a constant force that causes constant acceleration in free fall regardless of speed. (Ignoring air resistance, inverse sq considerations, etc.)
Suppose it takes 1 second for the ball on the 10ft ladder to hit the ground with kinetic energy of 10 and a speed of 100. Again, gravity as a constant acceleration force is speed increase per time... not speed per distance. In the ladder example, it took 1 full second for gravity to accelerate the object to speed 100.
Now think about the 20ft ladder: the ball is dropped. How much kinetic energy and speed does the ball have after it has fallen 10 feet (but still has 10 left to go)? Well it has the same exact amount as the other ball did after falling 10 feet for a duration of 1 second: kinetic energy of 10 and speed of 100.
Now the crux: thinking about when the final 10 feet of the fall look like. We know for sure the ball still has 10 ft of potential energy to covert into kinetic, and that that will happen as it falls. But what of the impact speed? Since the current velocity of the ball as it enters the last 10 feet is already 100, we know it will spend less time transiting this distance than it did the first half where it started at off at speed 0. Since gravity imparts speed in free fall as a function of time - consequently less speed will be imparted over the second 10 foot interval. That concept is enough to prove the relationship isn't linear.
If you do the actual calculation or tests, you will see one ball needs to be dropped from 4x the hight of another to hit the ground at 2x the speed, but yet with still 4x the kinetic energy.
What makes this intuitive? The foundation of the asker’s question is that it seems intuitive that kinetic energy would increase linearly with speed, but that turns out to be wrong.
I would not say we have the same intuition for kinetics. Increasing walking/running from 0 to 5 km/h doesn’t feel the same as than moving from 5 to 10, which does not feel the same as moving from 10 to 15. I don’t think we have an experience of linear relationship between running speed and effort, or other types of speed/energy types of relationships.
But more to the point the kinetic energy here is being turned into gravitational potential energy. If you move to a place with a weaker gradient in gravitational potential of course the same amount of kinetic energy moves you farther up.
Getting up from a seat als walking a couple steps feels that same at home and in a flying airplane (or does it?). But the base speed is 0 in the former and several hundred mph in the latter case
Acceleration is a real force that we can feel. But once moving at a constant speed, physics dictates that it’s all the same. That’s also why you can throw a tennis ball up on a plane and not have it fly backwards immediately smacking into the person behind you.
In the reference frame of you and the aircraft, you are not moving at all and neither is the plane. In the reference frame of the ground you and the plane are moving.
and also pushing that reference frame down when moving up
You’re introducing two new intuitions, and it’s not intuitively obvious how they are related to each other. Why would work correlate 100% with caloric intake, and caloric intake 100% with kinetic energy?
Certainly, ‘work’ is highly counterintuitive. If I move a concrete block over loose sand on a beach, I’m doing zero work, in the physics definition, so moving it over a kilometer should be as easy as moving it for a millimeter.
Even ignoring the difference between caloric intake and caloric expenditure, it also isn’t intuitive to me that caloric expenditure is independent of the speed at which one lifts an object.
In the end, the answer is “because the math works out that way, and kinetic energy is a useful concept”
Stacking a weight on top of a table holds it at a fixed height and requires zero mechanical work.
The failure in intuition here relates to physiology and the mechanism by which muscles work, not physics. Myosin and actin are constantly cycling through bonding and release during muscle contraction, as this is how the shortening action actually occurs. In fact, muscle contraction is particularly unintuitive because people frequently consider ATP the "energy currency", yet the ATP-consuming steps are actually the release/relaxation and preparation for binding, not the pulling action. This is also why the phenomena of rigor mortis upon death occurs.
How I got banned from some reddit channel. Flip this around ask if a ball were fired out of a gun up into the air what height would it reach? A ball twice as fast goes up 4 times as high. If energy is force times distance it had 4 times the energy.
The journey from Y to Z might feel more tiring than the journey from A to B, but only if you do them all in one day :)
Not really, no. Not all forces are conservative.
But in the end, it's all up to the units/quantities we choose to measure, no? If we, say, decided to measure "Squenergy" in Sqoules, with 1Sq² = 1J, then suddenly, squenergy does increase linearly with speed! The formula for kinetic Squenergy becomes sqrt(m/2)v.
Of course this complicates other stuff, like potential Squenergy becoming sqrt(MgH), it not being additive, etc.
...no ? dropping something 10 times from 1ft is nowhere near energetic/damaging as once from 10tf
Force = change in momentum with time
Energy = Force x distance
Now consider how much energy can be dissipated by a tiny change in momentum over a small distance dx, when we are at a given velocity v:
dE = Fdx = (dp/dt)dx = m(dv/dt)dx = mdv(dx/dt) = mv*dv
The intuition is that in order to apply a force through some distance, I have to change the velocity of an object by dv. But, the distance I just traveled also depends on the current velocity v. That's why the total energy available isn't just simply proportional to velocity - every time we change v, the amount of force available goes down, too.
Summing all the little bits of energy dE over our velocity changes dv, from the starting velocity down to zero, and we get the formula for kinetic energy.
BTW, the intuition here really starts from the idea that force = momentum change with time. The definition of "force", "momentum", and "energy" can be maddeningly circular, even if we have clear mathematical representations and a common world we experience.
A blue care is travelling along at 70 units, and a red car (exact same make and model) is catching up to it going 100. When they're both right beside each other a bend in the road reveals an obstacle blocking both lanes, so both cars brake at the same intensity and deceleration.
The blue care stops right before the obstacle. Since the red car was going at a faster speed, and braked at the same rate, it doesn't managae to stop: but what speed is it going when it hits the obstacle?
The blue car, using ½mv², shed (~70²=) 4900 units of energy (we'll hand wave away the constants). So the red car, which had (100²=) 10000 units of kinetic energy to start, also shed 4900 units, which means it had 5100 units of energy when it collided, and so was going (√5100~) 71.
* Numberphile: https://www.youtube.com/watch?v=i3D7XYQExt0
But if the cars produce downforce this is no longer true because you brake harder (more friction available) at higher speeds!
This is how F1 cars pull 4G when breaking. Some custom cars (like one of Ken Block’s last monsters or the Valkyre) use active aero braking to even greater effect.
2. I know you know this, but for the sake of others, it's when _braking_ (applying the brakes), not _breaking_ (becoming broken).
I'm not a pedant. But these errors jump out at me and I'm always a bit surprised and dismayed at this dichotomy; in our field, somehow the requisite attention to detail, the precision inherent to communicating scientific concepts, code, algorithms and formulae, is so often just abandoned when it comes to prose.
Honestly that was a typo and I noticed too late to edit. Thanks for catching
(I was suprised to see a cow jumping up on a ~3m rock ledge like it was nothing)
Point is that’s not always true. If they are the same type of car, and the car happens to be the kind with downforce, then their rate of deceleration greatly depends on air speed. A downforce car decelerates faster at higher speeds.
This is why you often see race cars lock their wheels towards the end of the braking zone, never at the beginning. The driver has to release the brakes as the car decelerates because there’s less friction available. You go from pulling 4G at the beginning of the braking zone to pulling the usual 1G once your speed drops enough for downforce to become negligible.
Alos! Many non-race cars actualy produce lift. Meaning the faster car decelerates at a slower rate than the slower car (0.8G vs 1G), making the effect from OP even more pronounced.
That’s not the only reason, and I’m not even sure it’s the majority reason.
Braking in a straight line offers more braking traction than braking while turning. What happens towards the end of a braking zone? The turn in. (Which also shifts weight to the outside tire and away from the inside tire.)
https://www.youtube.com/watch?v=RWwGFDynOHo
For these basic virtual car experiments, BeamNG.drive is a pretty good physics simulator. You can open its built-in tools and run braking tests directly.
5 km/h = 0.13 meter
30 km/h = 4.5 meter
60 km/h = 14 to 18 meter
65 km/h = 21 to 24 meter
You need 150% the distance at 65 vs 60.
It cannot be both. It mathematically cannot be both. They can brake at the same rate (acceleration) or intensity (conversion of kinetic energy into heat) but because they are traveling different speeds those two values cannot be the same for both cars.
The math you did was for intensity, not force/acceleration, which because of the ^2 in the KE equation exaggerates the difference. Whereas if you did the math based on force you'd get a mild, linear, difference.
> and braked at the same rate,
You're being a bit sly with word choice here. You're doing the math for conversion of KE into heat whereas in common parlance "rate" means force/acceleration.
Braking "at the same rate" [of energy conversion] is way less actual braking force for the faster car.
This is basically the same kinetic energy into heat math wherein you can descend a grade at a low speed, apply a force and be fine and descend the same grade at a higher speed and apply the same force and cook the brakes. Or you can apply less force, and get the same amount of energy conversion into heat (i.e. your wording trick in the proposed scenario)
You've taken what's basically the math behind trucks descending a grade (rate of energy conversion is actually limited by ability of brakes to shed heat, not friction) and re-framed it as cars stopping to create a trick question.
You are right that the faster car is converting kinetic energy into heat faster per unit time. It also has less time to do so. The work formulation of the problem makes it obvious that these have to cancel out exactly.
Couldn’t help but notice you misspelled car twice but only when talking about the blue car..
I have been thinking about it and only been able to come up with something that feels intuitive but not at all precise and I don't know how correct.
When you stand still you may use your surroundings to gain some speed, like by pushing against a wall.
When you have speed it gets harder to gain more speed because the surroundings are (relative to you) moving in the wrong direction, so for every additional unit of speed, it takes more effort to get there.
Back-of-napkin calculation says that if you managed to perfectly match exhaust speed with current speed, leaving all the expelled propellant stationary, it would only take quadratic amounts of fuels to reach higher speeds. Like the kinetic energy equation predicts.
An object which has a constant force applied will have it's distance increase quadratically with respect to time.
Energy is force times distance. Intuition: the energy it takes to lift an object up is proportional to the height you lift it to.
So if you apply a constant force, you get a constant acceleration which leads to a quadratically increasing distance.
If you accept that energy is force times distance, the energy required to move the object in this scenario increases quadratically.
This means that if you apply a force F for 1 second, the amount of energy that is imparted by that force depends on how fast the object is already going. The energy required to apply a force to an already fast moving object is much higher. Intuition: you have to expend all the energy required to get up to the moving object's speed before you can start applying a force. So there's a cost to even get in the game
As an aside, I believe Ron Maimon's account was suspended after he challenged the character of someone who was soliciting votes for a moderator position. Ron Maimon's stance was that if someone was running for an elected position, discussing their character was valid. The SO site had/has a strict challenge-the-question-not-the-person policy, which the moderators used to ban him permanently.
At the time, I remember seeing some posts by Ron talking about how the SO sites were corrupted by their policies and that it was a matter of time before they ceased to provide value. I think this was late 2000s or early 2010s. Looking back it's hard not to feel like his stance was prescient.
Today they are additionally weighed down by increasingly erratic management decisions desperately trying to extract as much monetary value as possible before AI completely obsoletes SE, but the amount of aggression and hostility on the network was unbearable from the start.
I remember dozens of occasions where I looked up something on StackOverflow, intending to be in and out in 10 seconds, and ending up spending several minutes just staring in disbelief at the comments showing how people treat each other on that site.
Suppose kinetic energy was E = m|v| instead, linearly dependent on speed |v|. What does that mean for the universe?
The traditional Lagrangian is L = 1/2 mv^2 - V(x). This kinetic energy gives a different formula:
L = m|v|ln|v|-V(x).
Deriving the corresponding equations of motion, you get:
p = m(1+ln|v|)sgn(v)
ma = |v|F
A few things we can note from these formulas:
1. They are not boost invariant: Galilean relativity is violated. That means there is necessarily a privileged reference frame (i.e. an aether) in which the universe is at rest, and all dynamics must be understood relative to this reference frame.
2. Newton's first law has a pathological interpretation in regards to the above reference frame: If ma = |v|F and |v| = 0 (i.e. you are at rest relative to the aether), then a = 0 no matter what F is. That is, for objects which are stationary with respect to the aether, no motion is possible regardless of what force is applied.
It is still true that objects in motion (relative to the aether) remain in motion unless acted upon by an outside force, and Newton's third law is still true, but such a universe basically makes no sense.
You could essentially argue from the anthropic principle that such a universe would have such pathological dynamics that it could not permit life, and therefore we cannot observe it.
This is the contrapositive of the argument presented on stackexchange. There they say "given Galilean relativity, you get the quadratic scaling law". This argument says "if you don't have the quadratic scaling law, you don't have relativity".
The point of the counterfactual is a bit like Richard Feynman's "why" argument [1]. There is no fundamental reason why this kind of dynamics couldn't exist. We can only ever reduce our explanation to a more fundamental intuition we have about the same universe we live in (i.e. from kinetic energy scaling laws to Galilean relativity). But without a mathematical proof of the incoherence even in principle of the alternative, its perfectly valid to imagine an alternative universe with different dynamics. It's just not our universe.
I've done plenty of this in pure math and stats, but this is the first time I've seen it applied to physics, and I love it! Thank you!
If I saw your derivation when I was 18 years old, who knows, maybe I would have caught the physics bug and went that way, this is super cool!
It's essentially the same argument: the Lagrangian can't have a bare a) position or b) velocity vector or it would violate homogeneity or isotropy of space, respectively.
Why not take the absolute value? Nature hates those, probably because the derivative is undefined at 0. So squaring it is.
Aside: I wonder if complex values neural networks with activation function just being sum(inputs)*conj(sum(inputs)) with threshold normalized by sqrt(num_inputs) could be the most universal, where incoherent inputs will average an absolute value of sqrt(N) and coherent inputs are N like lasers? (square amplitude would be N vs N^2 between uncorrected and correlated population)
For the purpose of inverting a negative vector, you can think of squaring as rotating the vector around the unit circle, 180 degrees, to make it positive. Higher order powers just keep rotating that vector back and forth- from this perspective the other even powers are the same transformation. Obviously with the magnitude being different.
And yet inverse distance laws for potential energy for gravity and electric fields use the absolute value because they require an unsigned distance and how you treat the singularity at zero is extremely important to the structure of those interactions