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Variants on this question have been asked repeatedly. See for example:

The reason none of these links answers my question is that they all end up deriving the usual expression for kinetic energy from the standard definition of work. My question is whether it is just a historical accident that the factor of two ended up where it did in the work-energy relationship.

In one sense, the question is more about history than physics, but not entirely. To target it more directly on physics, it could be rephrased as follows:

Would physical theory be unaffected if work were defined as 2 x force x distance and all equations of the form "some kind of energy = f(various variables)" were rewritten as "some kind of energy = 2 x f(various variables)" without any change in the many functions of this form?

In elementary physics, I am confident that the answer to my question is "yes." I am posting it here because I do not have a strong enough physics background to be equally confident about post-Newtonian formulations of mechanics, relativity, and quantum mechanics.

Did the arbitrary decision to put a factor of 1/2 in front of $mv^2$ instead of a factor of 2 in front of $mgh$ (the main example of work at the time the work-energy theorem became established) just propagate through all of physics post-1820?

P.S. The choice that Coriolis made is far from obvious. He puzzled himself over whether it was justified. What we call translational kinetic energy was called "vis viva" for 150 years and almost universally defined as $mv^2$, not $\frac{1}2 mv^2$. A blog post on the subject can be found at https://medium.com/@RebelScience/the-controversial-origin-of-the-kinetic-energy-equation-ek-%C2%BDmv%C2%B2-12a6b4e15e7a (accessed 2/14/2023). Not everything in the post should be taken seriously, but it does include a discussion of the topic at hand, including an English translation of Coriolis's comments on it (Coriolis's influential book Du calcul de l'effet des machines has evidently never been translated into English).

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    $\begingroup$ Sure, you can double all values of energy. But the choice we've made is the right one, because (1) it simply relates force and distance simply, and (2) it comes from the deeper foundation of Noether's theorem. Meanwhile, $K = mv^2/2$ is just one formula for one type of energy in a very specific theory (nonrelativistic Newtonian mechanics). It's not worth screwing everything else up just to remove that $1/2$. $\endgroup$
    – knzhou
    Feb 14, 2023 at 23:52
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    $\begingroup$ @knzhou Does Noether's theorem really nail down a preferred normalization for the energy? I thought that the overall normalization of the conserved quantity given by Noether's theorem was arbitrary (since each symmetry generator in the Lie algebra of the symmetry group generates a whole one-parameter group, not a single group element). You could maybe argue that the usual normalization convention is slightly more natural than other conventions, but I don't think it's overwhelmingly so. $\endgroup$
    – tparker
    Feb 15, 2023 at 2:33
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    $\begingroup$ There is a conserved quantity, we call energy, and indeed it is arbitrary whether you measure it in units of joules or (joules/2), call it a 'minijoule'. If you use $mv^2$ for KE, then you must use $2mgh$ for gravitational PE. An overall constant doesn't affect the physics whatsoever. Everything can be reformulated with the new constant. I don't know the historical reason for the choice we have. $\endgroup$
    – Myridium
    Feb 15, 2023 at 3:37
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    $\begingroup$ @tparker Sure, but if you keep the definition of momentum and other conserved quantities the same, then you would need a different normalization for time translations versus other symmetries. $\endgroup$
    – knzhou
    Feb 15, 2023 at 6:18
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    $\begingroup$ Does this answer your question? Why is there a $\frac 1 2$ in $\frac 1 2 mv^2$? $\endgroup$
    – hft
    Feb 15, 2023 at 19:34

6 Answers 6

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There isn't anything arbitrary in the formula, just the formula appears arbitrary to you because you don't know the formal definitions behind it. You'll need knowledge of basic calculus.

I think the easiest way of thinking is by associating with what happens with potential energy. ¿How much potential energy does a particle have? Depends of the reference you choose. There must be a point where you posit that the potential energy is zero, and the potential energy at a point is thus defined as the work you need for bringing the object from a reference point to the point you want to find the energy.

The same goes with kinetic energy. The work you need for bringing an object from A to B ¿How is defined the work done in some object between two points?:

$$W_{AB} := \int_{A}^{B} \vec{F} d\vec{s} = \int_{A}^{B} \frac{d\vec{p}}{dt} d \vec{s} = \int_{A}^{B} m \frac{d\vec{u}}{dt} \vec{u} dt = \int_{A}^{B} m \vec{u} d\vec{u} = \int_{A}^{B} m u du = \frac{1}{2} m (u_B^2 - u_A^2)$$

Now, if you bring an object from $A$, and, from your reference system (restudy Galileo's relativity), you posit that in $A$ the object is at rest (no speed, no kinectic energy), then the kinectic energy will be the total work you have to do in the particle to make it have a certain speed, hence $K = \frac{1}{2} m u²$. As you see, the formula is just a conclusion that is followed from the 'authentic' definition of work.

The result I got only is valid for classic mechanics. For Einstein theory you can't assert $\vec{F} = m \vec{a}$, so despite the work definition (first equality) is valid, the final formula you'll get will be different. If you see, also the formula $\vec{F} = m \vec{a}$ isn't the formulation of Newton's second Law, is just a particular case when the mass is constant. It's analogous. 99% of the definitions in physics are given in the integro-differential form because they are suitable for the most general situations. The simpler formulas that are explained in introductory courses are just particular cases of the definitions that makes easier the explanation when you don't have the mathematical basis.

I don't know if this kind of answer is what you were waiting for. And:

Would physical theory be unaffected if work were defined as 2 x force x distance and all equations of the form "some kind of energy = f(various variables)" were rewritten as "some kind of energy = 2 x f(various variables)" without any change in the many functions of this form?

You can't change the formula of kinetic energy since it was deduced and not defined. No matter than in elementary physics (kinder physics, i.e. w/o knowledge of calculus) the formula is presented as a definition, in the current corpus that is not standard. To accomplish $K = m u²$, you can do two things:

  1. Change the standard definition of work which is what allow us to define kinetic energy. If you let $W_{AB} := 2 \cdot \int_{A}^{B} \vec{F} d\vec{s}$, the 1/2 factor would vanish as you'd like. But since it must be also $W_{AB} = \Delta U$, you will probably end with predictions that are qualitatively different when dealing with both kinetic and potential energy. In other words, while in the standard world a car with some speed and kinetic energy $K = 0.5 m u²$ that is unable to climb a mountain of a certain height, in your $K = m u²$ world it might manage to do it since its energy is bigger there.

  2. In addition to 1. , you could redefine everything scaled with a factor of two , so you would do $W_{AB} = 2 \cdot \Delta U = 2 \cdot \int_{A}^{B} \vec{F} d\vec{s}$. And now, that would be the same qualitative world, but is just an unnecessary scaling, and when you could do the same qualitatively predictions you wouldn't share the common feeling about what a quantity represents one Newton or one Joule.

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    $\begingroup$ This is also where the factor of half comes from pretty much everywhere you see it next to a squared variable e.g. $s = .5 at^2$, $T = .5 mv^2$, $U_{spring} = 0.5 kx^2$, etc. Integrating a variable with respect to its own differential element gets you half the variable squared, so any time quantities can be defined in proportion to one another (like compression distance and spring force, time and displacement, etc) an equation of this shape is going to pop up. $\endgroup$
    – g s
    Feb 15, 2023 at 2:50
  • $\begingroup$ @gs, yes yes- like work stored in a charged capacitor or a stretched spring. $\endgroup$ Feb 15, 2023 at 7:29
  • $\begingroup$ @tac is certainly right that the question hinges on formal definitions. I do know enough calculus to understand that if one defines work as it is defined in the above derivation and accept the work-energy theorem, the factor of 1/2 goes where it goes. The question is "If we define work as twice force times distance does all the rest of physics work out ok, albeit with the elimination of a lot of factors of 1/2 and introduction of some new factors of two. $E = 2mc^2$ ??? $\endgroup$
    – CaveMan
    Feb 15, 2023 at 17:31
  • $\begingroup$ @CaveMan answer on the last paragraph $\endgroup$
    – tac
    Feb 15, 2023 at 17:35
  • $\begingroup$ @tac thanks for your comment, which makes it clear that your answer to my question is "yes." Coriolis defined work as force x distance and all else followed, including $E=mc^2$. There is not yet a clear consensus that this is the case, although I suspect it is. Many commentators consider the accepted convention as the most "natural" one, for one reason or another, but I see no justification for this view. I think it is just the one we all grew up with. However, I still may be missing something. $\endgroup$
    – CaveMan
    Feb 15, 2023 at 23:35
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Accidents and Probability

So, insofar as all of these conventional choices are “historical accidents” the answer to your question of “is this just a historical accident that everyone settled on this definition” is yes, but it is important to understand that this choice was also significantly more likely to win out.

To understand why, you have to fully appreciate that vis viva was standing between two distinct historical subfields of mechanics, and the definition $\vec F\cdot\Delta \vec r$ makes a lot more sense in one of them, while the definition $mv^2$ makes a lot more sense in the other. So the one that's vastly more likely to win is going to be the one that is taught chronologically first in the standard didactic approaches of the time, and that was the first one.

Lagrange as a notable precursor

To illustrate this let's go back to Joseph-Louis Lagrange’s seminal 1788 two volume textbook, Mécanique analytique, 40 years before Coriolis. Pages 17-18 of the first book begin the treatment of mechanics with, mm

  1. Je viens enfin au troisième principe, celui des vitesses virtuelles. On doit entendre par vitesse virtuelle, celle qu'un corps en équilibre est disposé à recevoir, en cas que l'équilibre vienne à ètre rompu, c'est-à-dire la vitesse que ce corps prendrait réellement daus le premier instant de son mouvement; et le principe dont il s'agit consiste en ce que des puissances sont en équilibre quand elles sont en raison inverse de leurs vitesses virtuelles, estimées suivant les directions de ces puissances.

    Pour peu quon examine les conditions de l'équilibre dans le levier et dans les autres machines, il est facile de reconnaitre cette loi, que le poids et la puissance sont toujours en raison inverse des espaces que l'un et l'autre peu vent parcourir en même temps: cependant il ne parait pas que les anciens en aient eu connaissance. Guido Ubaldi est peut-ètre le premier qui l'ait apercue dans le levier et dans les poulies mobiles reconnue ensuite dans les plans inclinés et dans moufles. Galilée l'a machines qui en dépen dent, et il l'a regardée comme une propriété générale de l'équilibre des machines. oyez son Traité de Mécanique et le scolie de la seconde pro position du trois Dialogue, dans l'édition de Bologne de 1655.

Translated by Boissonnade & Vagliente,

  1. Finally, I come to the third principle, the one of virtual velocities. One must understand by the term virtual velocity, the velocity which a body in equilibrium would take if the state of equilibrium ceased to exist, that is, the velocity that the body would have in the first instant of its motion. The principle requires the forces to be in a state of equilibrium if they are to be inversely proportional to their virtual velocities taken in the direction of these forces.

    For one who examines the conditions of equilibrium of the lever and other devices, it is easy to recognize the law that the acting forces and resisting weights are always inversely proportional to the distances which they traverse in the same interval of time. Yet, it appears that the Ancients never knew this law. Guido Ubaldo was perhaps the first to perceive it in the lever and in the block and tackle. Galileo recognized it later in the inclined plane and in machines which depend on the properties of the inclined plane. He considers it as a general property of the equilibrium of machines. Refer to his treatise on Mechanics and to the scholium to the second proposition of the Third Dialogue in the Bologna edition of 1655.

This is a definition of literally $P =\vec F\cdot\vec v$, what we would today call the power exerted by the force, but in response to some infinitesimal velocities that need to be compared across the entire static structure. Thus we are invited to consider actually $\mathrm dW = \vec F\cdot\mathrm d\vec{\ell},$ which is an infinitesimal work of the first definition! Lagrange goes on to also justify it in terms of a pulley system if there is any doubt left.

In our modern riches we can immediately see that for a lever this claim that $F\propto 1/L$ is indeed saying that energy is conserved by the lever. But Lagrange did not yet have this vocabulary evidently, for a very important reason: this book invents that vocabulary on page 268,

  1. Dans le cas où la quantité $P~dp + Q~dg + R~dr + \dots$ est intégrable. lequel a lieu lorsque forces $P, Q, R, $etc., tendent à des centres fixes à des corps du même système, et sont fonctions des distances $p, q, r, $ etc., en faisant $$P~dp+Q~dq+R~dr+\dots= d\Pi,$$ l'équation précédente devient $$S\left(\frac{dxd^2x+dyd^2y+ dzd^2}{dt^2} + d\Pi\right)m= 0,$$ dont l'intégrale est $$S\left(\frac{dx^2+dy^2+ dz^2}{2 dt^2} + \Pi\right)m= H,$$ dans laquelle $H$ désigne une constante arbitraire, égale à la valeur du pre mier membre l'équation dans instant donné.

    Cette dernière équation renferme le principe connu sous le nom de Conservation des forces vives. En effet, $dx^2 + dy^2+ dz^2$ étant le carré de l'espace que le corps parcourt dans l'instant $dt$, $\frac{dx^2+dy^2+dz^2}{dt^2}$ serra la carré de sa vitesse, et $\frac{dx^2+dy^2+dz^2}{dt^2}~m$ sa force vive. Done, $S \left(\frac{dx^2+dy^2+dz^2}{dt^2}\right)m$ sera la somme des forces vives de tous les corps, ou la force vive de tout le système; et l'on voit, par l'équation dont il s'agit, que cette force vive est égale à la quantité $2H-2S\Pi m$, laquelle dépend simplement des forces accélératrices qui agissent sur les corps, et nullement de leur liaison mutuelle, de sorte que la force vive du système est à chaque instant la même que les corps auraient acquise si, étant animés par les mêmes puissances, ils s'étaient mus librement chacun sur la ligne qu'il a décrite. C'est ce qui a fait donner le nom de Conservation des forces vives à cette propriété du mouvement.

Above Lagrange keeps the conventional definition of forces vives as $mv^2$, but you will notice that inside of the action function $S$ he places a ½ factor, it is the most natural place to put it! And as a result here at the outset he says that the conserved quantity is twice these other quantities. Throughout, Lagrange defines $T =\frac12 mv^2$ types of terms containing the prefactor, it is just clear that he constrains himself by the definition of vis viva of the times, which has no prefactor.

Now why, why does the first statement appear defining something like work in the very opening pages, while conservation of energy is appearing all the way later on page 250+?

Because Lagrange organized his textbook to start with statics and then move to dynamics. The traditional division of mechanics was between statics and dynamics; when you are dealing with static things, you have a very firm notion of forces, they are all balancing each other out. On the other hand, if you try to push something which can be moved, it “gives”, so there is a sense in which one becomes uneasy talking about a force. There is no good reason for this unease after all: Newton had given us the dispositional definition we needed, a force tries to accelerate a thing but whether it actually accelerates depends on the sum of all of the forces. But the full unification of statics and dynamics which this allows, seems to have taken a while to seep into public consciousness.

In statics, this work energy principle, or more properly the conservation of energy, turns out to be the central theme of leverage, and that is how Lagrange presents it. It is a primary point, he does not derive it from the conservation of energy, that is an advanced topic to come later in the Dynamics part of the course.

And why? Because what vis viva, that's why. You're doing statics, so $v=0,$ so $mv^2=0$. There is literally nothing to viva.

Coriolis calls it an embarrassing annoyance

Coriolis says this outright in his Du calcul de l'effet des machines, saying,

Si l'on avait éprouvé comme moi combien les élèves sont embarrassés par les dénominations mal choisies, je crois qu'on ne blâmerait pas ce léger changement. Il est très gênant d'avoir un nom pour le double d'une quantité qu'on retrouve à chaque instant.

If one has felt, as I have, how the students get so embarrassed by a name poorly chosen, I believe they wouldn’t blame this slight change. It is so very annoying to have a name for the double of a quantity that comes up every instant.

It was an intentional break with tradition, Coriolis states so explicitly, and he states why, and it's because this factor of 2 frustrated his students.

Why would he say that work times distance comes up “every instant”? Because it does, in statics. If you think about what machines rapidly became—the biggest example would be mechanical watches and other clockworks! These all depend on this crucial idea of leverage, ideas like gear ratios and so forth. You find many occasions to multiply force times a distance it moves through, and Lagrange has given you the reassurance that this is the basic correct truth of how to think about leverage.

Given that everyone agreed that statics was the natural precursor to dynamics, and statics has a lot of force times displacement ideas but no idea of vis viva because nothing is moving, the overwhelming probability was that we were going to end up with $\vec F\cdot\Delta \vec r$ and not $2 \vec F\cdot\Delta \vec r$ anticipating $mv^2$ potentially a semester later.

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  • $\begingroup$ Thanks. This seems right to me. Work squeezes into the space between statics and dynamics. The method of virtual work makes that clear. I have flagged this post as answering my question. $\endgroup$
    – CaveMan
    Feb 20, 2023 at 19:04
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Already tac's answer is correct. But to give a speedier answer, it is of purely mathematical nature, because the derivation of kinetic energy is the same as from any squared value. Let's assume a value $y$ from mass $m$ and velocity $v$:

$$y = mv^2$$

Then differentiate it by speed $v$:

$$\frac{dy}{dv} = \frac{mvdv}{dv}$$

And solve for momentum $p$:

$$\frac{dy}{dv} = 2mv = 2p$$

This means that the momentum $p = mv$ is only half of the derivation of y. Therefore we must calculate for the kinetic energy:

$$KE = \frac{1}{2}y = \frac{1}{2}mv^2$$

The integral would be the other way round, start with momentum $p = mv$ and integrate it to get $KE = \frac{1}{2}mv^2$. It can be found here.

Tip: Try to calculate with real-world values without the "$\frac{1}{2}$" and see what values you get. You will be constantly off by $\frac{1}{2}$. It took Newton a lot of empiric measurements to find this out. He finally learned: Calculation in physics is calculus, not just algebra, so it's about the nature of change.

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  • $\begingroup$ I should have made it more explicit that I understand this facet of the problem. I am fine with the possibility that if I change the definition of work to 2 x force x distance and plough through all the consequences of doing so, I will just be off by a factor of 1/2 anytime I calculate energies. Then I just redefine the Joule to half its current value, and all is good. In this case, the answer to my question is "yes." $\endgroup$
    – CaveMan
    Feb 15, 2023 at 17:44
  • $\begingroup$ Incidentally, user442 answered an earlier variant of this question (physics.stackexchange.com/questions/35987) as I do in my comment (just redefine the Joule and everything else will work out) more than 10 years ago. His answer, which I still think is the best of all answers to questions of this form, got one vote, while various other answers have accumulated hundreds of votes. $\endgroup$
    – CaveMan
    Feb 15, 2023 at 17:56
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The only factor in the article not approached by the previous answers is centripetal force $F_c=\frac{mv^2}{r}$. Centripetal force is not a force. It is the mass of an object multiplied by the acceleration necessary to hold that object on a circle of the indicated radius when moving at the indicated speed along the circular path. The acceleration itself can be derived by motion alone, actually based on the geometry of two similar triangles, resulting in $\frac{a}{v}=\frac{v}{r}$. This derivation is in all standard calculus-based physics textbooks. Any force or combination of forces can be used to provide this net force, so long as the magnitude of the sum is as stated earlier and the direction of the net force is always toward the center of the circle. There is no physical relationship between kinetic energy and the sum of all force vectors that in combination hold an object moving at constant speed to a circular path of specific radius. A person cannot even fall back to concepts such as work. The net force of uniform circular motion is perpendicular to the motion at all times. That which is called centripetal force does not do any work on the object.

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  • $\begingroup$ my only reason to include the link to the RebelScience post was because it clarifies a point of history. I agree with you that the post's discussion of circular motion makes little sense. $\endgroup$
    – CaveMan
    Feb 15, 2023 at 23:40
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The factor 1/2 is there because the expression $\tfrac{1}{2}mv^2$ arises from an integration: specifically, integration with respect to the position coordinate.

Using the standard letters t, s, v, a:
t time
s position
v velocity
a acceleration

As an example of how the factor 1/2 arises: the case of uniform acceleration.

To find an expression for covered distance as a function of time. First velocity as a function of time:

$$ v = at \tag 1 $$

To find distance covered as a function of time: integrate (1) with respect to the time coordinate:

$$ s = \int_0^t at \ dt = \tfrac{1}{2}at^2 \tag 2 $$

With the intermediate step omitted:

$$ s = \tfrac{1}{2}at^2 \tag 3 $$

Incidentally, (1) and (3) are already sufficient to obtain an expression known as Torricelli's formula:

Multiply both sides of (3) with acceleration $a$:

$$ as = a\tfrac{1}{2}at^2 = \tfrac{1}{2}a^2t^2 = \tfrac{1}{2}(at)^2 = \tfrac{1}{2}v^2 \tag 4 $$

With the intermediate steps omitted:

$$ as = \tfrac{1}{2}v^2 \tag 5 $$

So on the left hand side we have $as$, the product of acceleration and position coordinate, and on the right hand side we have $v^2$ The position coordinate $s$ has been rescaled to velocity by dividing it with the time factor $t$ and the acceleration value has been rescaled to velocity by multiplying it with the time factor $t$.

We generalize (5) to accommodate an initial velocity $v_0$, and an initial position $s_0$:

$$ a(s -s_0) = \tfrac{1}{2}v^2 - \tfrac{1}{2}{v_0}^2 \tag 6 $$

Anyway, so far (6) has been demonstrated for uniform acceleration, but we need generalize to non-uniform acceleration

For non-uniform acceleration: perform the integration.

To develop the integral the following two relations will be used.

$$ v = \frac{ds}{dt} \qquad \Longleftrightarrow \qquad ds = v \ dt \tag 7 $$

$$ a = \frac{dv}{dt} \qquad \Longleftrightarrow \qquad dv = a \ dt \tag 8 $$

The integral:

$$ \int_{s_0}^s a \ ds = \int_{t_0}^t a \ v \ dt = \int_{t_0}^t v \ a \ dt = \int_{v_0}^v v \ dv = \tfrac{1}{2}v^2 - \tfrac{1}{2}{v_0}^2 \tag 9 $$

With the intermediate steps omitted:

$$ \int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}{v_0}^2 \tag {10} $$

As we know: integration is addition of rectangular strips.

The integral $\int_{s_0}^s a \ ds$ specifies rectangular strips with width $ds$ and height $a$.

The integral $\int_{v_0}^v v \ dv$ specifies rectangular strips with width $dv$ and height $v$.

The dimensions of the rectangular strips are rescaled; the height is multiplied with the factor $t$ and the width is divided by the factor $t$, that is why the areas are the same.



We obtain the work-energy theorem by combining $F=ma$ with (10):

$$ \int_{s_0}^s F \ ds = \int_{s_0}^s ma \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}m{v_0}^2 \tag {11} $$

With the intermediate step omitted:

$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}m{v_0}^2 \tag {12} $$



In hindsight:
As you mention: the concept of 'vis viva' was in circulation for 150 years, quantatively expressed as $mv^2$ In hindsight the physics community should have moved to the $\tfrac{1}{2}mv^2$ form at the first opportunity.

Here is why I think it could remain in that $mv^2$ form for so long: I think for most of that time it was customary to express problems and their solutions in terms of ratios. It would be in the form: C is to D as A is to B:

$$ \frac{C}{D} \qquad \Longleftrightarrow \qquad \frac{A}{B} \tag {13} $$

When everything is expressed as ratios then you can use any multiplication factor in front of $\tfrac{1}{2}mv^2$; that does not change the outcome. So in hindsight we can say that in effect a multiplication factor of '2' was used: $2 \ * \tfrac{1}{2}mv^2 = mv^2$


The main message:
There is nothing arbitrary about the factor $\tfrac{1}{2}$: it is there because the quantity $\tfrac{1}{2}mv^2$ is obtained through integration.

What you want is that your set of laws of motion is interconnected. In order for the concept of kinetic energy to be connected to $F=ma$ the factor $\tfrac{1}{2}$ has to be there.

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By definition, work is $\int_{1}^{2} \vec F \cdot d \vec s$. $\int_{1}^{2} \vec F \cdot d \vec s = \int_{1}^{2} m {d\vec v \over dt} \cdot d \vec s = m\int_{1}^{2} \vec v \cdot d\vec v$.

$\vec v \cdot d \vec v = {1 \over 2} d(v^2)$.

Therefore, work is ${1 \over 2}m\int_{1}^{2} d(v^2) = {1 \over 2}m(v_2^2 - v_1^2). $

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