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Why does kinetic energy increase quadratically, not linearly, with speed? (2011)

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Re: Why does kinetic energy increase quadratically, not linearly, with speed? (2011)

#71

Post by chrisweekly »

Swizec wrote: post #12
throw0101a wrote: post #2

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 ra…

> 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…

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.

Re: Why does kinetic energy increase quadratically, not linearly, with speed? (2011)

#72

Post by nlawalker »

throwawaytea wrote: post #69
nlawalker wrote: post #61

Earlier quoted context omitted.

> 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.

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.

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.

Re: Why does kinetic energy increase quadratically, not linearly, with speed? (2011)

#73

Post by lazide »

davidivadavid wrote: post #17
lazide wrote: post #13

Earlier quoted context omitted.

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.

Also, physics (the discipline) is also a construction by and for humans.

To find tools applicable to reality. Not to construct reality.

Re: Why does kinetic energy increase quadratically, not linearly, with speed? (2011)

#74

Post by alok-g »

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.

Re: Why does kinetic energy increase quadratically, not linearly, with speed? (2011)

#75

Post by hunter2_ »

nlawalker wrote: post #72
throwawaytea wrote: post #69

Earlier quoted context omitted.

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.

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.

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.

Re: Why does kinetic energy increase quadratically, not linearly, with speed? (2011)

#76

Post by frogulis »

slicktux wrote: post #5
throw0101a wrote: post #2

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 ra…

Cool anecdote! Couldn’t help but notice you misspelled car twice but only when talking about the blue car..

Perhaps the beginning of a new vowel harmony phenomenon in English

Re: Why does kinetic energy increase quadratically, not linearly, with speed? (2011)

#77

Post by jurschreuder »

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).

Re: Why does kinetic energy increase quadratically, not linearly, with speed? (2011)

#79

Post by raldi »

nlawalker wrote: post #61
cubic_earth wrote: post #55

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…

> 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.

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.

Re: Why does kinetic energy increase quadratically, not linearly, with speed? (2011)

#80

Post by xeonmc »

SyzygyRhythm wrote: post #36

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.

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) 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)

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