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cubic_earth 3 hours ago [-]
It's easiest to visualize in terms of conversion from potential energy.
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.
nlawalker 2 hours ago [-]
> We know intuitively that a ball atop a 20ft ladder has twice the potential energy of a ball atop a 10ft ladder.
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.
hunter2_ 2 hours ago [-]
That's a good question, and I suppose the mgh formula isn't a suitable answer, so my answer would be something like: if you lift an object to some height, and then you repeat that action (lifting it from there to twice the height), you've done twice the work, and doing twice the work requires twice the caloric intake.
SilasX 23 minutes ago [-]
Okay but that depends on the intuitions the question is trying to justify, which makes it circular. We also know, for example, that the body uses more than twice as much energy to do twice as much work (because of fatigue on the muscles or whatever the right term is here). In fact it takes positive energy just told a weight at a fixed height, doing zero mechanical work! So you’re actually appealing to even weaker intuition than the one the question is trying to ground!
hibernator149 3 minutes ago [-]
All intuitions are wrong, but some are usefull. You have to do the experiment, discover the formula, and then adapt your intuitions accordingly.
throwawaytea 1 hours ago [-]
Because things like energy are relative. So if you label the ground 0, and go up 10 feet, you get x energy. Going up another exact same x from your 10 foot ladder spot you could now call 0 again, would mean you gain x energy again.
Since they're both the same height, and you gained the same energy, you could infer double the height has double energy.
nlawalker 1 hours ago [-]
What if you label standing still as 0 mph and start moving 10 mph, gaining x energy, then call that zero and start moving 10 mph from there? It's just as intuitive to say that you would gain x energy in that case, but you don't.
throwawaytea 26 minutes ago [-]
That's clever, and I can't imagine or explain it as easily.
Something to do with a reference point moving away from you so when solving for bringing it back to zero it's different than just adding the two energies back together. You have to add up the energy of catching them all up to the initial starting reference.
I think also because distance is one unit, so moving reference pointe is easier. Moving reference points on distance over time already gets my spidey senses going that it's not something you should do without some real understanding.
hunter2_ 44 minutes ago [-]
When you're already going 10 mph and you're about to add another 10 mph, you can only "call that zero" (i.e., go from 0 mph to 10 mph again) if your point of reference (i.e., the ground) also begins moving with you at that point. Since the ground is stationary, you're definitely about to increase from 10 mph to 20 mph relative to the ground, not from 0 mph to 10 mph, and that's harder to do. But if you're on a treadmill that was stationary for the first change, and then suddenly starts moving at 10 mph right before the second change without affecting your speed relative to the ground, then you can "call that zero" and you'll be able to add another 10 mph (ending up at 10 mph relative to the treadmill and 20 mph relative to the ground) with the same ease as the first go.
cubic_earth 33 minutes ago [-]
I suppose they are both "intuitive", but the example I gave was both intuitive and correct. Probably for anyone who has carried something or themselves up a hill, or climbed a set of stairs can relate to that from firsthand experience. I don't know what the kinetic energy corollary to that would be? "Stand still and I will throw a baseball at you going 15mph, and note how much it hurts. Now I will throw it at you going 30mph. See! It hurts 4x as much" :D
pishpash 33 minutes ago [-]
Not really. Potential energy in a gravitational well obviously has absolute coordinates.
gorgoiler 33 minutes ago [-]
Because physical movement is intuitively transitive. Going from A to B then B to C is the same as going from A to C.
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 :)
raldi 34 minutes ago [-]
Because if the one falling 20ft lands on a seesaw, the other side of it will toss two balls each of the same mass 10ft up.
pishpash 27 minutes ago [-]
Then 20ft should not be used in the explanation. They should just have one ball going at 2x speed hit the seesaw and have 4 of those balls go up at 1x speed.
hunter2_ 2 hours ago [-]
Brilliant. For those wanting more numbers [0], the ball on the 10ft ladder hits the ground at (I'll stick with imperial units) 17.296 MPH, the ball on the 20ft ladder hits the ground at 24.46 MPH or 41.42% faster, and the ball on the 40ft ladder hits the ground at 34.59 MPH or 100% faster.
Nice. Nitpick: in the middle paragraph you put "speed 10" instead of 100.
cubic_earth 2 hours ago [-]
Fixed. Thanks.
throw0101a 6 hours ago [-]
Fun little anecdote:
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.
> 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
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.
chrisweekly 1 hours ago [-]
1. +1 insightful, thanks for sharing your physics knowledge
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.
tracerbulletx 5 hours ago [-]
But what if the cars are spherical cows?
kibwen 3 hours ago [-]
I'm sorry to inform you that those cows are going to have a hard time braking on that frictionless surface.
PaulDavisThe1st 3 hours ago [-]
Based on a hike in the Carson National Forest 2 days ago, the only reason a cow is on a frictionless surface is that the cow shat all over it.
BigTTYGothGF 5 hours ago [-]
Cows can't roll that fast.
zdragnar 5 hours ago [-]
Not with that attitude
39 minutes ago [-]
elromulous 4 hours ago [-]
Or on shabbos
linzhangrun 4 hours ago [-]
IIHS video shows the relationship between kinetic energy and speed in a very intuitive way:
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.
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.
ThrustVectoring 3 hours ago [-]
OP wasn't explicit about taking the work = force * distance approach to dissipating energy. Two cars with the same mass and braking force (and thus deceleration) will put the same amount of work into the vehicle per unit distance, so will dissipate the same amount of energy in the braking maneuver.
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.
slicktux 5 hours ago [-]
Cool anecdote!
Couldn’t help but notice you misspelled car twice but only when talking about the blue car..
frogulis 42 minutes ago [-]
Perhaps the beginning of a new vowel harmony phenomenon in English
5 hours ago [-]
senectus1 5 hours ago [-]
heh, thats a fun little experiment.
NamlchakKhandro 3 hours ago [-]
In what way is it fun?
zkmon 8 minutes ago [-]
No amount of scientific explanation can exhaustively explain a phenomenon. Feynman puts this nicely with the story of "Why did aunt slipped and fell down" in his talk about magnetism.
For instance we know that the life forms grow via cell division, but no text can address the question of "why". They can only talk about "how".
Infact, science quest is not really about answering "why" all the way down the causal chain. It is about learning how the qualities of things are related and a bit of shallow causal chain inspection.
The causal chain, by nature, does not allow full inspection. It's dependency on temporal constructs means it breaks down where time breaks down. Infact causality might might break down at macro levels as well, leading to loops with no end or beginning (kind of chicken and egg problem).
Tazerenix 2 hours ago [-]
Here's how to appreciate it in terms of the counterfactual:
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!
SyzygyRhythm 4 hours ago [-]
Cheat answer: velocity is a vector, and can be negative, while KE is a scalar and has to be positive. Therefore you have to square v to get rid of the minus sign.
Why not take the absolute value? Nature hates those, probably because the derivative is undefined at 0. So squaring it is.
xeonmc 33 minutes ago [-]
I like to think of it as dot product being the true "natural" space to compare magnitude metrics, whereas absolute value is just a human construct conceived for our mental convenience. A smooth parabolic bowl vs an unnatural sharp conical tip. Also shows up in standard deviation etc.
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) is the most natural universal neural net, 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)
signa11 4 hours ago [-]
why not raise to any other even power ?
meowkit 4 hours ago [-]
One way of thinking about that is higher order even powers just reduce down to two.
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.
qzw 4 hours ago [-]
That doesn’t answer the title question of why it’s quadratic wrt speed.
SyzygyRhythm 4 hours ago [-]
To get speed from velocity, you need a square root, which is also awful (for the same reason that abs is awful).
4 hours ago [-]
aesthesia 4 hours ago [-]
Michael Spivak's Physics for Mathematicians has a lot of arguments like the one in the top answer here, answering questions about why the math of classical mechanics is the way it is.
alok-g 51 minutes ago [-]
Sharing my understanding:
If one starts with Newton's 2nd law (F=ma) assumed, then one can derive kinetic energy to be 0.5mv^2, and this is what most of the answers are explicitly or tacitly doing.
One could however start with Lagrangian formulation along with KE = 0.5mv^2 and drive F=ma. This is where one needs an explanation for why KE = 0.5mv^2, and the first answer (@Ron Maimon) is providing an explanation.
Most books I have come across on Lagrangian formulation secretly assume Newton's laws.
In my opinion, Lagrangian formulation can proceed without Newton's and without even defining momentum as mv, however, now needs KE = 0.5mv^2.
jurschreuder 41 minutes ago [-]
If someone walks by and you want to push him in the back to go a bit faster.
Or someone runs by and you want to push him in the back to go faster.
You will have to push with great vigor, unless you first get up to speed yourself (also takes energy).
hyperhello 4 hours ago [-]
I didn’t think this was that weird. When you double your speed you are also going to be going twice as far in the same time, not just twice as fast, and they both have the effect of work.
faustlast 27 minutes ago [-]
Has anyone here read Lanczos 1952 book on variational mechanics? It is beautifully written.
c1ccccc1 2 hours ago [-]
A stationary but hot object has kinetic energy due the the motion of the individual atoms that make it up, even though its overall momentum is 0. I.e.
∑ⱼ mⱼ v⃗ⱼ = 0⃗
where the mⱼ are the masses of the parts of the object and the v⃗ⱼ are the velocities of those parts.
If the object initially has 0 velocity, its kinetic energy is:
T = ½∑ⱼ mⱼ v⃗ⱼ²
Now we give the object a kick (or just switch reference frames) to change its velocity by Δv⃗. The new kinetic energy is:
If M is the total mass of the object, then we can substitute this into the sum in the last term. And we already saw that the sum in the middle term was 0. So:
T' = ½(∑ⱼ mⱼ v⃗ⱼ²) + Δv⃗⋅0⃗ + ½Δv⃗² M
T' = ½∑ⱼ mⱼ v⃗ⱼ² + ½MΔv⃗²
So in terms of the original kinetic energy T, which was purely thermal energy, we get:
T' = T + ½MΔv⃗²
In other words, because of the quadratic kinetic energy formula, we can see that the total kinetic energy T' of a hot object is just its thermal kinetic energy T plus the usual mechanical kinetic energy ½MΔv⃗².
acchow 28 minutes ago [-]
Looks like in your 2nd equation you've already assumed kinetic energy is quadratic with speed
T = ½∑ⱼ mⱼ v⃗ⱼ²
sixo 1 hours ago [-]
Kinetic energy is, strangely, quite a bit like a least squares cost function in an optimization problem. The "dt"s in "dx/dt" hardly matter; it basically represents "dx^2" between the current state and the next.
snarfy 2 hours ago [-]
When you push something you don't change its velocity - you change its acceleration.
2 hours ago [-]
drivebyhooting 4 hours ago [-]
I don’t find the answer convincing. It assumes one can measure heat at a distance and it is a conserved quantity between reference frames.
Energy is actually not a conserved quantity in Galilean relativity.
c1ccccc1 3 hours ago [-]
Energy is conserved in Galilean relativity. The thing you're trying to say is that it's not invariant across reference frames.
The answer linked above actually takes advantage of the fact that energy is not the same in different reference frames in order to make the argument work.
I think you are overthinking the heat thing. If you have a train car full of hot water and you slow the train down (extracting kinetic energy from it) until it stops, the water in the train car does not change temperature at all, other than a bit of sloshing around and loss of heat to the surroundings.
itemize123 1 hours ago [-]
thinking aloud here - so it seems like 2 things are taken as intuitive here:
a) energy is conserved in any frame of reference.
b) energy can vary in 2 frame of references.
but then what it feels like is that when you reference the energy as mE(v), the v is actually not the only variable, and it will be more like mE(v, v_moving_reference)?
so we also must take intuitive that c) E(v, v_moving_reference) == E(v - v_moving_reference)
casey2 5 hours ago [-]
Mikes' answer is the most intuitive, but he rephrases the question in a possibly non intuitive way.
Odd that nobody mentioned power, which scales linearly with speed. Of course if you add linear increasing amounts of power to the system the energy will increase quadratically.
Power scaling linearly is more intuitive because doubling your speed requires twice the power to maintain the same force, why does it require twice the power? because you have half the time to power it.
jacknews 2 hours ago [-]
The first example only tells me that the energy is dependent on your frame of reference, since the collision seen from the train appears to have more energy than the head-on collision, simply due to the moving viewpoint, whereas they must be the same.
Agingcoder 5 hours ago [-]
Physics is an endless source of frustration to me. It feels like a mix of random tricks, most of which I don’t understand.
I find math and compsci reasonably understandable, can read research papers in both fields ( and have published papers) etc. There’s something specific about physics I don’t get but I’ve never been able to figure out what. The main symptom is that most cause -> consequence in such demonstrations , which are seemingly obvious to everyone, make no sense to me.
Am I the only one ? Are there good resources to learn it?
digdugdirk 5 hours ago [-]
It seems that we're exact opposites! But if mathematics is your thing, it might be interesting for you to explore trying to learn things from a lagrangian perspective first?
Not sure if it'll help you with gaining an intuitive understanding, but at least it'll be interesting!
Lagrangian / Hamiltonian mechanics, the principle of least action, always seemed neat, in L&L and other places I encountered it, until I tried doing exactly what you're saying: gaining an intuitive understanding. At that point it just never made sense to me and seemed like a gratuitous deus ex machina that happens to work beautifully but for no apparent reason. You won't be surprised to learn I dropped out of my STEM program shortly after, though I keep a keen interest in the topic.
5 hours ago [-]
davidivadavid 5 hours ago [-]
More than twenty years ago, I quit a program that taught math/cs/physics (the notorious French "classes préparatoires") ~almost precisely over this: I felt like I was being taught physics like it was an axiomatic system where the tricks should not be questioned, they just work so "shut up and calculate" (and you don't even need to be doing quantum mechanics for that).
I just felt like we never got to the heart of the matter of why the models work and how to approach developing them, it was all about learning a bag of tricks.
Meanwhile, math and CS being a lot more axiomatic by nature, they also made a lot more sense to me.
That being said, that specificity of physics, the unbridgeable gap between reality and the models we build to describe it, in retrospect, is what makes it more interesting to me today (it's not just a "closed" system in the sense that math is — of course the relationship between math and physics is itself fascinating but that's yet another topic), but I still feel like I haven't found the right pedagogical approach to make it fit my mindset.
joshAg 5 hours ago [-]
Your issue with physics but not with math reminds me a little of Hume's law. The difference that has always made that difference "make sense" to me is that math rules, even the axiom we use, are entirely chosen by the people using them, but the rules of physics are only useful if they match/predict what happens in the real world. For math we get to pick the ones that happen to be useful at a given time for a given problem (my go-to example of "it's all made up and the points don't matter" is why 1 isn't considered prime). For physics we're constrained to pick what best describes the real world. It probably helped that nearly all the physics course I had in high school/university had lab components focused on experimentally validating those rules/using those rules to predict results.
davidivadavid 4 hours ago [-]
I think what it boils down to is that in my experience physics education lacks a clear historical component about how the current state of the art is a gradual process of proposing new models and rejecting old ones and figuring out the gaps between reality and the model. Instead, it feels like a God-given set of equations (that lots of people consider "the truth" for some reason), that you apply to cookie-cutter problems you must learn by rote. Though I understand the practical concerns (but then let's call it "physics for engineering"), as far as I'm concerned, you couldn't treat physics in a worse way.
joshAg 4 hours ago [-]
that (edit: the way you were taught) sounds like an altogether awful way to learn any hard science
lazide 5 hours ago [-]
The world just is, regardless of what we think about it. Physics is our best attempt so far to understand and predict it at a low level, but it will always be incomplete.
Maths (and especially compsci!) are constructions by and for humans.
Is it any wonder it is as you describe? It would be odd if it was any other way.
davidivadavid 5 hours ago [-]
My point is precisely that I was often taught physics as if it was mathematics, where there is in fact a profound ontological difference between the two.
davidivadavid 5 hours ago [-]
Also, physics (the discipline) is also a construction by and for humans.
lazide 56 minutes ago [-]
To find tools applicable to reality. Not to construct reality.
esikich 5 hours ago [-]
Weird, I always loved physics because I felt like I didn't have to straight up memorize everything. In a pinch (years ago) I felt like I was able to pretty much derive everything I needed if I couldn't remember the exact formulas. It's all just forces and vectors.
symian 4 hours ago [-]
Same for me. I wanted to major in physics and I quickly realized that I have no intuition for physics. Math made sense to me and I went to graduate school in math and still don’t understand anything in physics. Differential geometry, no problem. Electromagnetism makes no sense to me.
JoshMandel 5 hours ago [-]
I identify with this perfectly. (I mean, was able to get by in physics but it never crystallized into intuition for me the way math and CS do.)
casey2 5 hours ago [-]
Physics? Yes. Feynman Lectures On Physics and Computation. Landau & Lifshitz. If you like SICP you might like SICM. Nielsen & Chuang's Quantum Computation and Quantum Information then Faulkner's Modern Quantum Mechanics and Quantum Information
General advice take a look at the references in works you've recently read and look for lower level topics that interest you, after repeating a few times you'll find your way to physics or chemistry and you can use the above as reference works. The best resource is the one you actually use. If https://www.youtube.com/learning works better for you then use it.
rustyhancock 5 hours ago [-]
What's the problem exactly? Could you not follow the example in the text?
The standard text to build understanding in physics is University Physics by Sears & Zemansky.
It's worth remembering you're quite far from the ground in physics, and it's mostly taught with "neat" cases that give insight into physics. I.e. the thought experiment to show why kinetic energy must scale quadratically with velocity is carefully designed to show that conclusion. You shouldn't expect to have come up with it off the cuff.
AngryData 5 hours ago [-]
This is also why splitting wood with a maul is way more work than using an axe. You can swing an axe at incredibly speeds which gives incredibly transfers of energy, but a maul is going to always have "meh" levels of speed because it is too much mass to accelerate over such a short distance as a swing. Also why you don't see framers using 3 lb hammers. You can put in more effort and get your lighter hammer swing to twice the normal speed, no way in hell you are doubling the speed of a 3 lb hammer though.
gorfian_robot 25 minutes ago [-]
have you ever had to split wood?
5 hours ago [-]
11101010010001 5 hours ago [-]
read Ron Maimon.
symian 4 hours ago [-]
He has interesting perspectives in math which is an area I know about. I assume the same for physics. People should read his answers.
lngnmn2 4 hours ago [-]
[dead]
firebot 5 hours ago [-]
Because it's not momentum. ;p
F=ma (Force equals mass times acceleration)
W=Fd (work equals force multiplied by distance)
V^2=2ad (velocity squared equals two times acceleration times distance)
So W = Fd = ma(v^2/2a)
Finally: W=1/2mv^2 (work equals 1/2 mass times velocity squared)
So this explains why car crashes can be so dramatic, as a doubling of speed results in 4x the kinetic energy.
ajross 5 hours ago [-]
Actually, it is momentum, sorta. Galilean 3D momentum isn't conserved under special relativity. The energy-momentum four-vector, however, is, under all lorentz-transformed frames.
So in some sense energy is momentum in the time direction (though it's not a Euclidean 4D space, so beware of assumptions). For an object at rest, this becomes its E=mc² equivalence. Kinetic energy is just a straightforward "rotation" of the frame.
esalman 2 hours ago [-]
Original comment is correct, it's not momentum. Work (hence, energy) is integral of force over distance, momentum is integral over time. There's not "sorta" about high school physics.
c1ccccc1 3 hours ago [-]
If you use the right formula for calculating it (which approximates p=mv at low speeds), momentum is actually conserved in special relativity, and so is energy.
However: Energy and momentum are not invariant under changes of reference frame, though the magnitude of the energy-momentum 4-vector is invariant between frames.
firebot 5 hours ago [-]
P=mv (momentum equals mass times velocity)
This is linear.
One small nuance... saying "kinetic energy is just a straightforward rotation of the frame" is close, but it's the total energy that is the time component of the four-momentum and mixes with the spatial momentum under Lorentz transformations. Kinetic energy is the difference between that transformed total energy and the invariant rest energy. So kinetic energy isn't itself a four-vector component, but it arises from how the time component changes when viewed from a different inertial frame.
ajross 3 hours ago [-]
To nitpick your nitpick: I know. But precision isn't the point here, it's to point out that there's an interesting and deeper symmetry at work. Energy and Momentum are not actually different quantities that vary in different ways but are still conserved via different laws. They're actually expressible as a single conserved vector quantity.
Details about the specifics were hidden behind the scare quotes on "rotation". But sure, my phrasing was loose, how about 'What we ses as "kinetic energy" pops out of the Lorentz "rotations" of that energy in different reference frames.' ...?
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.
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 :)
[0] https://www.omnicalculator.com/physics/free-fall
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.
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.
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..
For instance we know that the life forms grow via cell division, but no text can address the question of "why". They can only talk about "how".
Infact, science quest is not really about answering "why" all the way down the causal chain. It is about learning how the qualities of things are related and a bit of shallow causal chain inspection.
The causal chain, by nature, does not allow full inspection. It's dependency on temporal constructs means it breaks down where time breaks down. Infact causality might might break down at macro levels as well, leading to loops with no end or beginning (kind of chicken and egg problem).
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.
[1] https://www.youtube.com/watch?v=36GT2zI8lVA
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!
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) is the most natural universal neural net, 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.
If one starts with Newton's 2nd law (F=ma) assumed, then one can derive kinetic energy to be 0.5mv^2, and this is what most of the answers are explicitly or tacitly doing.
One could however start with Lagrangian formulation along with KE = 0.5mv^2 and drive F=ma. This is where one needs an explanation for why KE = 0.5mv^2, and the first answer (@Ron Maimon) is providing an explanation.
Most books I have come across on Lagrangian formulation secretly assume Newton's laws.
In my opinion, Lagrangian formulation can proceed without Newton's and without even defining momentum as mv, however, now needs KE = 0.5mv^2.
Or someone runs by and you want to push him in the back to go faster.
You will have to push with great vigor, unless you first get up to speed yourself (also takes energy).
∑ⱼ mⱼ v⃗ⱼ = 0⃗
where the mⱼ are the masses of the parts of the object and the v⃗ⱼ are the velocities of those parts.
If the object initially has 0 velocity, its kinetic energy is:
T = ½∑ⱼ mⱼ v⃗ⱼ²
Now we give the object a kick (or just switch reference frames) to change its velocity by Δv⃗. The new kinetic energy is:
T' = ½∑ⱼ mⱼ (v⃗ⱼ + Δv⃗)²
T' = ½∑ⱼ mⱼ (v⃗ⱼ² + 2v⃗ⱼ⋅Δv⃗ + Δv⃗²)
T' = ½(∑ⱼ mⱼ v⃗ⱼ²) + Δv⃗⋅(∑ⱼ mⱼ v⃗ⱼ) + ½Δv⃗²(∑ⱼ mⱼ)
If M is the total mass of the object, then we can substitute this into the sum in the last term. And we already saw that the sum in the middle term was 0. So:
T' = ½(∑ⱼ mⱼ v⃗ⱼ²) + Δv⃗⋅0⃗ + ½Δv⃗² M
T' = ½∑ⱼ mⱼ v⃗ⱼ² + ½MΔv⃗²
So in terms of the original kinetic energy T, which was purely thermal energy, we get:
T' = T + ½MΔv⃗²
In other words, because of the quadratic kinetic energy formula, we can see that the total kinetic energy T' of a hot object is just its thermal kinetic energy T plus the usual mechanical kinetic energy ½MΔv⃗².
T = ½∑ⱼ mⱼ v⃗ⱼ²
Energy is actually not a conserved quantity in Galilean relativity.
The answer linked above actually takes advantage of the fact that energy is not the same in different reference frames in order to make the argument work.
I think you are overthinking the heat thing. If you have a train car full of hot water and you slow the train down (extracting kinetic energy from it) until it stops, the water in the train car does not change temperature at all, other than a bit of sloshing around and loss of heat to the surroundings.
a) energy is conserved in any frame of reference. b) energy can vary in 2 frame of references.
but then what it feels like is that when you reference the energy as mE(v), the v is actually not the only variable, and it will be more like mE(v, v_moving_reference)?
so we also must take intuitive that c) E(v, v_moving_reference) == E(v - v_moving_reference)
Odd that nobody mentioned power, which scales linearly with speed. Of course if you add linear increasing amounts of power to the system the energy will increase quadratically.
Power scaling linearly is more intuitive because doubling your speed requires twice the power to maintain the same force, why does it require twice the power? because you have half the time to power it.
I find math and compsci reasonably understandable, can read research papers in both fields ( and have published papers) etc. There’s something specific about physics I don’t get but I’ve never been able to figure out what. The main symptom is that most cause -> consequence in such demonstrations , which are seemingly obvious to everyone, make no sense to me.
Am I the only one ? Are there good resources to learn it?
Not sure if it'll help you with gaining an intuitive understanding, but at least it'll be interesting!
https://en.wikipedia.org/wiki/Lagrangian_mechanics
I just felt like we never got to the heart of the matter of why the models work and how to approach developing them, it was all about learning a bag of tricks.
Meanwhile, math and CS being a lot more axiomatic by nature, they also made a lot more sense to me.
That being said, that specificity of physics, the unbridgeable gap between reality and the models we build to describe it, in retrospect, is what makes it more interesting to me today (it's not just a "closed" system in the sense that math is — of course the relationship between math and physics is itself fascinating but that's yet another topic), but I still feel like I haven't found the right pedagogical approach to make it fit my mindset.
Maths (and especially compsci!) are constructions by and for humans.
Is it any wonder it is as you describe? It would be odd if it was any other way.
General advice take a look at the references in works you've recently read and look for lower level topics that interest you, after repeating a few times you'll find your way to physics or chemistry and you can use the above as reference works. The best resource is the one you actually use. If https://www.youtube.com/learning works better for you then use it.
The standard text to build understanding in physics is University Physics by Sears & Zemansky.
It's worth remembering you're quite far from the ground in physics, and it's mostly taught with "neat" cases that give insight into physics. I.e. the thought experiment to show why kinetic energy must scale quadratically with velocity is carefully designed to show that conclusion. You shouldn't expect to have come up with it off the cuff.
F=ma (Force equals mass times acceleration)
W=Fd (work equals force multiplied by distance)
V^2=2ad (velocity squared equals two times acceleration times distance)
So W = Fd = ma(v^2/2a)
Finally: W=1/2mv^2 (work equals 1/2 mass times velocity squared)
So this explains why car crashes can be so dramatic, as a doubling of speed results in 4x the kinetic energy.
So in some sense energy is momentum in the time direction (though it's not a Euclidean 4D space, so beware of assumptions). For an object at rest, this becomes its E=mc² equivalence. Kinetic energy is just a straightforward "rotation" of the frame.
However: Energy and momentum are not invariant under changes of reference frame, though the magnitude of the energy-momentum 4-vector is invariant between frames.
This is linear.
One small nuance... saying "kinetic energy is just a straightforward rotation of the frame" is close, but it's the total energy that is the time component of the four-momentum and mixes with the spatial momentum under Lorentz transformations. Kinetic energy is the difference between that transformed total energy and the invariant rest energy. So kinetic energy isn't itself a four-vector component, but it arises from how the time component changes when viewed from a different inertial frame.
Details about the specifics were hidden behind the scare quotes on "rotation". But sure, my phrasing was loose, how about 'What we ses as "kinetic energy" pops out of the Lorentz "rotations" of that energy in different reference frames.' ...?