Hacker News Board Hacker News, browsed like a classic message board
Live data from Hacker News

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

physics.stackexchange.com

Post reply on HN
Print view View original

51–60 of 231 posts

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

#51

Post by ThrustVectoring »

cucumber3732842 wrote: post #24
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…

>same intensity and deceleration. 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. Wh…

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.

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

#52

Post by ajross »

firebot wrote: post #19
ajross wrote: post #16

Earlier quoted context omitted.

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 th…

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 energ…

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.' ...?

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

#53

Post by c1ccccc1 »

drivebyhooting wrote: post #44

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.

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.

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

#54

Post by c1ccccc1 »

ajross wrote: post #16
firebot wrote: post #11

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.

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 th…

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.

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

#55

Post by cubic_earth »

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.

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

#56

Post by c1ccccc1 »

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:

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.

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

#57

Post by jacknews »

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.

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

#58

Post by esalman »

ajross wrote: post #16
firebot wrote: post #11

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.

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 th…

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.

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

#59

Post by hunter2_ »

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…

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.

[0] https://www.omnicalculator.com/physics/free-fall

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

#60

Post by card_zero »

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…

Nice. Nitpick: in the middle paragraph you put "speed 10" instead of 100.
Post reply on HN