How kinetic energy is proportional to velocity squared

[exponent137]

I hear somewhere explanation
why kinetic energy is proportional to $v^2$ and not to $v^3$, for instance.
https://www.physicsforums.com/threads/quadratic-forms-and-kinetic-energy.826884/
I wish to find reference for this.

Now I find your thread with the same question and very similar answers:

https://www.physicsforums.com/threa...gy-is-proportional-to-velocity-squared.78484/

Someone suggested a book Landau-Lifshitz Mechanics, but I do not find this derivation in it.
The final explanation to why kinetic energy is proportional to velocity squared

My question is, from where this derivation is.

 

[sophiecentaur]

I hear somewhere explanation
why kinetic energy is proportional to $v^2$ and not to $v^3$, for instance.
https://www.physicsforums.com/threads/quadratic-forms-and-kinetic-energy.826884/
I wish to find reference for this.

Now I find your thread with the same question and very similar answers:

https://www.physicsforums.com/threa...gy-is-proportional-to-velocity-squared.78484/

Someone suggested a book Landau-Lifshitz Mechanics, but I do not find this derivation in it.
The final explanation to why kinetic energy is proportional to velocity squared

My question is, from where this derivation is.

I think the derivation is pretty straightforward if you consider the work done by bring a moving object to rest under a constant force (SUVAT equations is all that's needed). The distance (as in Force times Distance) needed is proportional to the initial velocity squared. (Energy will be conserved.)

 

[exponent137]

But, I wish derivation which omits $Fdt=mvdv$. The above derivation is based on a collision of two bodies. We should respect that momentum is conserved, that energy is conserved and this gives that $E_{kin}\propto v^2$. As it is already derived in two above links.

 

[sophiecentaur]

The above derivation is based on a collision of two bodies.

OK Use the same derivation and accelerate the body from rest. The work done is Fd and the energy it acquires if mv²/2
Why are you being 'picky' about the derivation? Is there something invalid about it?

 

[anorlunda]

https://en.wikipedia.org/wiki/Lagrangian_mechanics#From_Newtonian_to_Lagrangian_mechanics

That article shows a completely separate way to derive it using the Euler Lagrange equation and the calculus of variations.

 

[OmCheeto]

My interpretation of the question:

Why is kinetic energy proportional to v² and not to v³, and who decided that?​

My guess is that it was derived experimentally.
I'm also guessing that someone in antiquity thought that a rock sitting on a shelf should have a "linear" value of something called "energy" in relation to its height above the ground. Someone had to start somewhere.
I'm also pretty sure you can't get anything other than ke ∝ v² when you do the experiments.
At least, near the surface of the Earth that is.

I didn't factor in the reduction of gravity with altitude when I analyzed the Blue Origin flight yesterday, and the conservation of energy went all cattywampus.

ps. My apologies for dragging this down to the "B" level, but this seems like a really simple problem.

 

[anorlunda]

why kinetic energy is proportional to $v^2$ and not to $v^3$, for instance.

Another thing to consider is that kinetic energy must be an even power of v because it is unsigned.

Consider car moving at 60mph colliding with a second car at 60mph. It makes an enormous difference if they are moving in the same direction or opposite direction. That is momentum, it is signed and it is proportional to $v^1$ which is an odd power.

But it takes the same fuel to accelerate from 0 to 60mph east or 0 to 60mph west. That is kinetic energy; it is unsigned and proportional to $v^2$ which is an even power.

 

[robphy]

In Special Relativity, the kinetic energy is $K_{sr}=mc^2\left( \frac{1}{\sqrt{1-(v/c)^2}} -1 \right)$, also an even function of $v$, which reduces to $\frac{1}{2}mv^2$ in the $v\ll c$-limit.
This follows from the relativistic version of the Work-Energy Theorem.

Thus, as others have suggested, look at the work-energy theorem.

 

[sophiecentaur]

Why is kinetic energy proportional to v2 and not to v3, and who decided that?
My guess is that it was derived experimentally.

Early on, there was confusion about whether the 'push' of a moving object was mv, mv²/2 or some other power. Both Momentum and KE could be used to predict results but which one worked, depended on the situation. Colliding body calculations seemed to agree with the mv momentum model and throwing object type calculations agreed with the KE approach. I guess there could have been a lot of angry shouting about it at the time because neither proponent was actually wrong.

 

[kith]

There's a nice argument by Ron Maimon which is based on symmetry.

 

[exponent137]

In Special Relativity, the kinetic energy is $K_{sr}=mc^2\left( \frac{1}{\sqrt{1-(v/c)^2}} -1 \right)$, also an even function of $v$, which reduces to $\frac{1}{2}mv^2$ in the $v\ll c$-limit.
This follows from the relativistic version of the Work-Energy Theorem.

Thus, as others have suggested, look at the work-energy theorem.

I know for this derivation. I know also Maimon's derivation.

But, I ask only for reference for this derivation:
The final explanation to why kinetic energy is proportional to velocity squared

I thought that it is Landau Lifshitz's Mechanics book, as one wrote in the above thread, but it is not.

 

[exponent137]

OK Use the same derivation and accelerate the body from rest. The work done is Fd and the energy it acquires if mv²/2
Why are you being 'picky' about the derivation? Is there something invalid about it?

Equation dE=Fdx is not from the first principles. From the first principles is as robphy wrote.
Thus dE=Fdx is a consequence of conservation of momentum and

In Special Relativity, the kinetic energy is $K_{sr}=mc^2\left( \frac{1}{\sqrt{1-(v/c)^2}} -1 \right)$, also an even function of $v$, which reduces to $\frac{1}{2}mv^2$ in the $v\ll c$-limit.
This follows from the relativistic version of the Work-Energy Theorem.

Thus, as others have suggested, look at the work-energy theorem.

Now, my question is more cleary formulated.

regards

 

[Sergio Rodriguez]

I don't know if you are looking after this, where the kinetic energy appears by calculation of the work done by a force:

$$w= \int_A^B F dx
=\int_A^B m \frac {dv} {dt} dx
\, = m\int_A^B dv \frac {dx} {dt}
=m\int_A^B vdv$$
$$w= m\left[ \frac {v^2} 2 \right] _A^B
= m \frac 1 2 v_B^2 - m\frac 1 2 v_A^2$$
$$=KE_B- KE_A$$
$$w=Δ KE$$

 

[exponent137]

I don't know if you are looking after this, where the kinetic energy appears by calculation of the work done by a force:

$$w= \int_A^B F dx
=\int_A^B m \frac {dv} {dt} dx
\, = m\int_A^B dv \frac {dx} {dt}
=m\int_A^B vdv$$
$$w= m\left[ \frac {v^2} 2 \right] _A^B
= m \frac 1 2 v_B^2 - m\frac 1 2 v_A^2$$
$$=KE_B- KE_A$$
$$w=Δ KE$$

No, I wish only reference for this or such one derivation:
The final explanation to why kinetic energy is proportional to velocity squared

 

[anorlunda]

No, I wish only reference for this or such one derivation:
The final explanation to why kinetic energy is proportional to velocity squared

Hehe, I give this thread the zombie "night of the living thread" award. It first died in 2005, came back to life for a day in 2008 and promptly died again, and then came back to life again in 2010 where it has been terrorizing the villagers for a couple of weeks now!

and now it is trying for rebirth in 2018. It seems that the zombie award is well deserved.

 

[exponent137]

and now it is trying for rebirth in 2018. It seems that the zombie award is well deserved.

Because this is an unbelievable thing.

School system has not taught us that equation $dE=Fdx=mvdv$ is only an empirical thing, but more fundamental source of $E_{kin}=mv^2/2$ is hidden in special relativity.