Is the classical formula for kinetic energy wrong?

[BruceW]

I have never seen that formula before, is there anywhere you know of I can see how it is derived, as that would probably solve the problem (assuming the proof doesn't just rearrange formulae I know of, like most other proofs I have encountered)?

E(\mathrm{transformed}) = \int_0^t f v(t') dt'
Where $v(t')=at'$. Also, $a$ is a constant, since $f$ and $m$ are constants and $a=f/m$. Therefore,
E(\mathrm{transformed}) = af \int_0^t t' dt'
and doing the integration,
E(\mathrm{transformed}) = \frac{af}{2} t^2
and now use $a=f/m$ gives:
E(\mathrm{transformed}) = \frac{1}{2m} f^2 t^2

Now, what I started with is power = force * velocity. Maybe that is just another formula that you know of, so this will be no use to you. But just maybe, this is a formula that you haven't seen before, and that you are happy with. Sorry for the sarcastic tone. Uh, to be honest, you have to start with a postulate of some kind, which defines what energy is. Or you can start with energy as simply "something which is conserved with time", then you can derive the KE=1/2mv² for Euclidean space.

 

[my2cts]

To me F=dP/dt and dE/dP=P/E.
The latter can be integrated to E^2=P^2+m^2, with m a constant of your choice.
I did not bother with factors c.
When electromagnetism, gravity etc. are present , some more effort is required.
Also you may want to convert this into wave mechanics.

 

[A.T.]

that the velocity at the bottom the fall is proportional to the square of the height the object is dropped from.

The orher way around.

 

[Nugatory]

The other way around.

Or maybe I left the word "root" out :)
Either way, I'm wrong and you're right - thanks.

 

[A.T.]

Or maybe I left the word "root" out

I think the original question is how you justify (non-mathematically) the root, instead of linearity. May be like this:

Doubling the fall height, doubles the kinetic energy on impact. But it cannot double the velocity, because in the lower half of the height the object is faster, and therefore has less time to accelerate further, so it gains less speed than in the upper half.

 

[DrStupid]

If the fuel is burning at a constant rate, then the energy is being spent at a constant rate. The rate of energy is power, so the rocket generates constant power.

But that does not mean that constant force is being generated. Power is force times velocity.

Not in case of a rocket. A rocket engine with constant burning rate actually generates a constant force and constant power. 256bits already explained seanhenley's error: he forgot the kinetic energy of the exhausted reaction mass.

 

[G. Cavazzini]

Brucew wrote: "for example, gravity, a basic intuitive idea of the concept is that massive things are attracted to each other. like there is some invisible arm between the two objects, pulling them towards each other. Using just this intuition, you can already say something about gravity. You already have some physics here".

There is no basic intuitive idea that massive things are attracted to each other. This is only an INTERPRETATION of what we observe, if you want to interpret the phenomenon this way.
It is NOT A FACT. It is an HYPOTHESIS. Using this hypothesis (IT IS NOT AN INTUITION AT ALL!) you cannot say NOTHING about gravity except your hypothesis. You do not have some physics of gravitation: you have only an hypothesis about physics of gravitation!
My God.

 

[Ronie Bayron]

Recently I have been troubled by several things which have been stated in my school physics classes, mainly concerning the classical kinetic energy formula and other formulae derived from it (as well as the 'Work Energy Theorem').

The classical formula for kinetic energy '(mv^2)/2' was first found by two lone french scientists in the early 18th century, with neither proof nor definitive experimental evidence for the key to this formula - the v squared coefficient. If one looks up the proof for this formula, there are several proofs to choose from. The problem with them all is that they utilise the 'Work Energy Theorem', which in itself uses the Kinetic Energy formula in its proof - it doesn't take much to realize something is wrong.

Consider this. In a friction-less environment, away from any gravitational influence, there is a rocket. It has an engine with an unlimited fuel supply (to keep mass the same) that is burning at a constant rate. Its rate of fuel consumption is constant (the chemical energy contained within it is being used at a constant rate), and therefore the engine is providing a constant force. This is where I am most likely to wrong in my deduction (if I am).

Therefore the energy transformed from chemical energy to kinetic energy can be said to be proportional to the force provided by the engine multiplied by the time it is running. Most importantly, neither the rate of fuel nor the force provided by the engine will change depending on how fast the rocket is moving. From this I get the formula:

E (transformed)=f*t

Two other formulae which are well known and I assume are accurate:

F=ma
v=at

Simple algebraic manipulation will find that E=mat, and therefore E=mv

Mass*velocity, as we all know, is the formula for momentum, NOT kinetic energy. But what exactly is the difference? Why should kinetic energy be proportional to velocity squared and not just velocity. It just doesn't make any sense to me. Isaac Newton thought the same, that the 'quantity of motion ' should be mv and not (mv^2)/2, and after all, momentum is conserved in all collisions, not kinetic energy, so surely the definitions should be reversed at least.

Anyway, I would like someone to tell me exactly where I have gone wrong in these deductions (if I have that is). If I have gone wrong, then I will at least understand why, and if I haven't, then why exactly has no one else noticed this?

You start with momentum =mv and d(mv)/dt-the change of momentum with time is force. You end up like F=d(mv)/dt so F= (mdv+vdm)/dt this is the complete equation for force.
Theoretically, work energy is W=F*ds, likewise ds = vdt, so W = [(mdv+vdm)/dt]*vdt= mvdv+v²dm.

The part m∫vdv = 1/2mv² this became the Kinetic Energy of a particle with constant mass, m.
The part v²∫dm is a term something like the energy of diminishing or gaining mass(this is the one, you should refer your problem with-at space, you may use). Fortunately, Einstein found this one to be E=mc², taking c=v, if only Newton take note of the importance of the other term in force, Einstein might have ended up without the credit.
There's nothing wrong.
Further, the sum of the derivative of momentum actually describes the dynamics stable and radioactive elements.
Stable elements tend to follow (statics and dynamics) the classical, Newtonian physics.
Radioactive materials on the other hand, behaves v²∫dm equation of energy.
Have you noticed that? Or am I the only one who recognize it?

 

[DrStupid]

The part v²∫dm is a term something like the energy of diminishing or gaining mass

No, that would dE=dm·v²/2. The work-energy-theorem doesn't hold for open systems.

 

[mfig]

Perhaps the derivation of the kinetic energy through the definition of work will satisfy you. Let's use an external force acting on an object of mass m, through some distance in free space. The work done on the object by the external force is given by:

$W_{ab} = \int_{a}^{b} \vec{F}\cdot d\vec{x}$

But in this case

$\vec{F}\cdot d\vec{x} = m\vec{a}\cdot d\vec{x} = m\frac{d\vec{v}}{dt}\cdot d\vec{x} \frac{dt}{dt} = m\frac{d\vec{v}}{dt}\cdot \vec{v}dt$

This can further be simplified into an exact differential as follows

$m\frac{d\vec{v}}{dt}\cdot \vec{v}dt = \frac{m}{2} \frac{d}{dt}(\vec{v}\cdot\vec{v})dt = d(\frac{mv^2}{2})$

Now substituting in for the work expression gives

$W_{ab} = \int_{a}^{b} \vec{F}\cdot d\vec{x} = \int_{a}^{b} d(\frac{mv^2}{2}) = \frac{{mv_b}^2}{2} - \frac{{mv_a}^2}{2}$

So the work done by an external force on an object is equal to the change in some quantity given by $\frac{mv^2}{2}$. As this quantity is zero when the object does not move, and as this quantity has the units of energy, we call it the kinetic (from the Greek meaning 'to move') energy.

Hope that helps.

 

[Ronie Bayron]

No, that would dE=dm·v²/2. The work-energy-theorem doesn't hold for open systems.

Hi, how do you suppose to solve or model propulsion in space for example?

 

[Philip Wood]

Since PF's automated email service has told me that this thread has reopened, I take the liberty of referring anyone interested back to my post 10, in which I've used the simplest math(s) possible to establish the equation. I've also put the derivation in the context of a silly, but easily visualised, scenario, but it's obvious that it can be made perfectly general (within Newtonian Physics).

 

[russ_watters]

Since this thread was two years old when revived and most original posters aren't coming back, I think it is better to let it die. Locked.

 

[Othin]

First of all, you can't prove anything in physics (or any natural science by the way). What you can do is either build models which are in agreement with a certain theory or with a new one you're willing to create. If the predictions arriving from you model agree with the experiments, it's a good model, otherwise, it's a bad one and won't be used. Nobody proved a formula for kinetic energy, people simply recognized that a certain expression showed up so often it was worth giving it a name. That said, there are some things to consider regarding its form.
You can find a very good explanation regarding the form of the kinetic energy expression in Landau's classical mechanics book, which you should check if you can. When we talk about classical mechanics, some things are tacitly assumed , uniformity of time and space included. We must assume that there is a function which contains all mechanical information of the system, the Lagrangian. Let's think about the simple case of ONE free particle. It can be described in generalized coordinates, but let's do it in Cartesian terms, so L is, in general, a function of the spatial coordinates, time, and velocity (which are here independent). But time and space are postulated to be uniform, so, if our function is to be consistent with the said uniformity of space and time, it can't contain time and position explicitly (why, if there's no force acting on my particle I can't expect it to change simply because time passed, nor should I expect that one particular point in space is at all different of the others). Also, space is considered isotropic, so I must be able to rotate my entire system without changing it's physical properties. Some requirements are already clear: my function can't, for instance, be a function of v^k , with k odd, 'cause otherwise a rotation of my system, which changes the sign of velocity, would change my Lagrangian, and that's not allowed. My function must depend only on the absolute value of velocity. Now comes the point when me must choose which function, among the infinite that respect those requirements. Let's take L= L(v^2) for now. Yes, we coude have choosen differently, but I'll come to it later.
Now, given this assumption, let's change our reference frame, and choose one that moves with a infinitely small velocity \delta in relation to our previous fame. By Galileo's principle, this change can't alter the physical properties of my system, so the Lagrangian in this new frame must obey: L'=L(v'^2)=L([v+\delta]²). A Taylor expansion around \delta of the rightmost expression ( remember that \delta can be made as small as desired) , renders : L(v'^2)=L(v^2) + \frac{\partial L}{\partial v^2}2V\delta , where V\equiv \delta - V' (I assumed up there that our new reference frame moved with a small velocity in relation to the previous one, so here I'm just repeating myself in mathematical terms!)
We have two different Lagrangians representing the same system . Since the laws of physics are still the same, the equations of movement must still be equal, dispate this difference. The only way this could happen is if they differ only by a total time derivative (if you didn't know it, examine the Euler-Lagrange and you will see). Hence:
\frac{dL(v^2)}{dt} \equiv \frac{\partial L}{\partial v^2} \frac{\partial L}{\partial t} = \frac{\partial L}{\partial v^2}2V\delta \to \frac{\partial L}{\partial t} = 2V\delta \implies L= \alpha v^2 , where \alpha is the constant which arrives from integration. Giving it the value (m/2), we arrived at the desired expression.
Now, could I choose a different value for the constant? Mathematically, yes. But since our question was simply whether the usual formula was wrong, I've proved that no, it isn't. It's consistent with the axioms (Notice that I used nothing but the very basic axioms of mechanics, I didn't even specify the form of the Lagrangian at first.), so there's zero possibility that it's wrong. Any formula that can be proved right by the same method would be mathematically right. If you want to call \alpha by a different name, you wil arrive in a consistent, but completely different (and probably less convenient) formalism for mechanics. Any different choices would imply a new form for Newton's second law, a new definition of what we call mass, and so on. If, instead of v², you wished to use v⁴, for example, you could be able to come up with a self consistent formalism in which the kinetic energy isn't given by the usual expression, but if you want any of the expressions to agree with Newton's formulation and to our basic, intuitive concepts, you would need to make exact the choices made above.