# Where did the kinetic energy formula come from? [duplicate]

I was looking for a proof for it and all rely on the work-energy theorem. But the work energy theorem relies on the kinetic energy equation. Ergo circular logic.

So where did it come from?

• What formula are you referring to as "the kinetic energy formula"? Commented Jan 2, 2021 at 21:46
• The standard one you learn in high school (v^2m/2) Commented Jan 2, 2021 at 21:49
• If you're worried about "circular reasoning" in physics, see here. Commented Jan 2, 2021 at 22:02
• Another route: physics.stackexchange.com/a/112344/44126
– rob
Commented Jan 2, 2021 at 23:15
• This can be done non-circularly and its hard to say how to fix up whatever version you've seen without knowing what it is. The accepted answer to the question MarkH links to does this nicely. Commented Jan 2, 2021 at 23:19

By "kinetic energy equation", I assume you mean the definition $$KE = \frac{1}{2} mv^2$$

This does indeed arise from the work-energy theorem, which says that the net work done on an object of mass $$m$$ over some interval of time is given by

$$W_{net}=\frac{1}{2}mv_f^2- \frac{1}{2} mv_i^2$$

Looking at that equation, we simply notice that the quantity $$\frac{1}{2} mv^2$$ seems to be useful, so we give it a name - kinetic energy - and then phrase the work-energy theorem as

$$W_{net} = \Delta(KE)= KE_f - KE_i$$

The net work done on an object between times $$t_i$$ and $$t_f$$ is $$W_{net} = \int_{t_i}^{t_f} \mathbf F_{net}(t) \cdot \mathbf v(t) \ dt$$ Newton's second law tells us that $$\mathbf F_{net} = m\mathbf a$$, and so

$$W_{net} = \int_{t_i}^{t_f} \bigg(m \mathbf a(t) \cdot \mathbf v(t)\bigg) dt$$

However, $$\mathbf a(t) = \mathbf v'(t)$$, so $$\mathbf a \cdot \mathbf v = \mathbf v' \cdot \mathbf v = \frac{1}{2} \frac{d}{dt}(\mathbf v \cdot \mathbf v) = \frac{d}{dt}\bigg(\frac{1}{2} v^2\bigg)$$ and so finally

$$W_{net} = \int_{t_i}^{t_f} \frac{d}{dt}\left(\frac{1}{2} m v^2\right) dt = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$

• How did the work-energy theorem arise then? The work theorem REQUIRES that the KE equation holds true. This is circular reasoning. Unless you mean that KE is by definition which doesn't answer my question. How would you arrive at that definition without the work-energy theorem? Commented Jan 2, 2021 at 21:50
• @LucasFrykman No it isn't. I updated my answer to show that $W_{net} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$ is a consequence of Newton's second law. We then define kinetic energy to be equal to $\frac{1}{2} mv^2$. Commented Jan 2, 2021 at 21:58
• @LucasFrykman Do you understand what it means to define something? A definition is not something you prove. I showed that the net work is equal to $\frac{1}{2} mv_f^2 - \frac{1}{2}mv_i^2$ - that is, the change in the quantity $\frac{1}{2} mv^2$ - which motivates us to give $\frac{1}{2}mv^2$ a name. Somebody a long time ago decided that the name for that quantity should be kinetic energy. Commented Jan 2, 2021 at 22:09
• @Neelim I’m not sure what you’re talking about. My answer makes no reference to momentum. Commented Jan 2, 2021 at 22:12
• @Neelim Once you define what work is, the work-energy theorem is an immediate consequence of Newton’s second law. What motivates you to define the quantity we call work is irrelevant. Commented Jan 2, 2021 at 22:18

OK, derivation of the work-energy theorem from F=ma

The qualification 'theorem' is indeed appropriate.
If we accept Newton's second law as axiom, and we accept as axiom that space is Euclidean, then the work-energy theorem follows logically.

First two standard kinematic relations, valid for the case of uniform acceleration. The derivation will capitalize on these relations:

Change of velocity as a function of time:

$$v = v_0 + at \qquad (1)$$

Change of position as a function of time:

$$s = s_0 + v_0t + \tfrac{1}{2}at^2 \qquad (2)$$

With the above we can obtain an expression that is in terms of derivatives of time only.

(1) can be restated in the form of (3), and then you substitute the $$t$$ in (2) with the expression for $$t$$ from (3)

$$t = (v - v_0)/a \qquad (3)$$

It looks hairy, but it turns out a lot of terms drop away against each other.
In the end you arrive at this formula:

$$a(s - s_0) = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \qquad (4)$$

The above expression is also known as Torricelli's formula

The above is not yet physics; it's still only a kinematic relation.

By combining (4) and F=ma we obtain a dynamics statement.

$$F \Delta s = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2$$

Reminder: the unit of force is called the 'Newton'. The dimension are: $${\displaystyle 1\ {\text{N}}=1\ {\frac {{\text{kg}}\cdot {\text{m}}}{{\text{s}}^{2}}}.}$$

General discussion

Other answers to this question proceed according to the following strategy: define a concept called 'work done' and then show that this implies an expression $$\tfrac{1}{2}mv^2$$, that expression can then be defined as 'kinetic energy'.

In Dynamics we are accustomed to thinking in terms of accumulation over time. An equation of motion is a function of time; future position is computed as a function of time

The work-energy theorem doesn't fit that mould. The work-energy theorem describes accumulation over distance.

In the history of physics the work-energy theorem was recognized quite late. I think it was first stated around 1800 or so.

Generalization

Using (4) is of course not a general way to derive the work-energy theorem. The kinematic relations used are for uniform acceleration.

A closer examination:
(1) and (2) are closely related: when you differentiate (2) you get (1). As we know, diffentiation and integration are essentially inverse operations of each other. (4) should be seen as the result of integration.

Generalization to the more general case (acceleration a function of something else) is straightforward.

The derivation presented in this answer is not as general as it can be. I chose to present this derivation to emphasize: the work-energy theorem follows directly from F=ma.

• Ok then my question changes in to how you do you know newtons multiplied with distance compute joules/energy? Those are two completely different units. The work energy theorem states that those units are the same. Commented Jan 3, 2021 at 10:43
• @LucasFrykman Break each set of units into fundamental meters, kilograms, and seconds. You should be able to do that easily. Commented Jan 3, 2021 at 13:24

For classical mechanics the kinetic energy $$T$$ is defined as $${1 \over 2} m v^2$$.

$${d \over {dt}} ({1 \over 2} m v^2) = \vec F \cdot \vec v$$. So $$T_2 - T_1 = {1 \over 2}mv_2^2 - {1 \over 2}mv_1^2 = \int_{t_1}^{t_2} \vec F \cdot \vec v \enspace dt$$. Since $$\vec v \enspace dt = d \vec r$$, $$T_2 - T_1 = \int_{r_1}^{r_2} \vec F \cdot d \vec r$$, which is the work done by force $$\vec F$$ between $$r_1$$ and $$r_2$$.

See a physics classical mechanics text, such as Mechanics by Symon.

The 1/2 is a matter of definition. If we changed it to some other unitless number, then other equations would also have to change, e.g., Newton's second law.

The proportionality to $$m$$ is not an arbitrary definition. We want a conserved quantity, and conservation laws are additive. If we had used $$m^2$$ or something, the we wouldn't have had an additive quantity.

The dependence on $$v^2$$ is not an arbitrary definition, and in fact is not even correct. It's just the lowest nonvanishing term in the Taylor series of the relativistic expression.

Newton's laws are logically equivalent to conservation of energy and momentum. If you start from either one, you can derive the other. Any experiment that establishes one is also an experiment that establishes the other. Any experiment that disproves one, such as experiments that show relativistic effects, disproves the other.

The work done by a force is $$\Delta W = \int_{x_1}^{x_2} \mathbf{F.dx}$$. When $$\mathbf F$$ is the resultant of the forces in a body, the second law applies: $$\mathbf F = m\mathbf a$$.

So, $$\Delta W = \int_{x_1}^{x_2} m\mathbf {a.dx} = m\int_{x_1}^{x_2} \frac{\mathbf {dv}}{dt}\mathbf {.dx}$$

As $$\mathbf x$$ is a function of t, $$\mathbf {dx} = \frac{\mathbf {dx}}{dt}dt$$

$$\Delta W = m\int_{t_1}^{t_2} \frac{\mathbf {dv}}{dt}\frac{\mathbf {dx}}{dt}dt$$

Integrating by parts, we get 2 identical integrals:

$$\Delta W = m\int_{t_1}^{t_2} \frac{\mathbf {dv}}{dt}\frac{\mathbf {dx}}{dt}dt = m\left [\frac{\mathbf {dx}}{dt}\frac{\mathbf {dx}}{dt}\right]_{t_1}^{t_2} - m\int_{t_1}^{t_2} \frac{\mathbf {dv}}{dt}\frac{\mathbf {dx}}{dt}dt$$

And finally: $$\Delta W = \frac{1}{2}mv_ 2^2 - \frac{1}{2}mv_ 1^2$$

The formula for kinetic energy is derived from the formula for work done, but the formula for work done is not derived further from any more fundamental underlying formulas. It comes from the empirical results of an experiment done in the 18th century. The experiment was basically dropping balls on soft clay and measuring the distance from which it was dropped, and the impact. What the experiment found is that the impact was proportional to the distance. So they came up with the formula, $$W=Fd$$. Then if you want to get the formula for kinetic energy, you have to combine the formula for $$W$$ with Newton's second law i.e. $$F=ma$$ and kinematics. The exact derivation is as follows:

$$W = Fd$$ $$F=ma=m\cdot\frac{v-u}{t}$$

If you replace the F in the first equation with the value for the second equation and you consider the K.E.(kinetic energy) to be the change in energy i.e. work done on an object to achieve its velocity from an initial velocity $$u$$ of 0, you get:

$$K.E. = m\cdot\frac{v-u}{t} \cdot d = m\cdot\frac{v-u}{t} \cdot \frac{v+u}{2} \cdot t = \frac{1}{2}mv^2$$

And that is how you get the formula for kinetic energy

• Omg did no one read my post? That is clearly circular logic since work-energy theorem RELIES on the KE equation. This is like A implies B because B implies A. Commented Jan 2, 2021 at 21:58
• no I did read your question, and I also addressed it. The formula for kinetic energy comes from the formula for work. But the formula for work comes from an experimental result
– user283752
Commented Jan 2, 2021 at 21:59
• Aha thanks. Yeah I wished more physics textbooks clarified which formulas are theorems and which are experimental. Commented Jan 2, 2021 at 22:07
• I get your issue, I had the same question. I even had doubts about whether the formulas for work done and kinetic energy were correct or not. I still have doubts whether they are universally correct
– user283752
Commented Jan 2, 2021 at 22:09
• Work done can be related to total energy. Total energy is the conserved quantity that arises as a result of time-invariance of the Lagrangian. So whilst initial motivation for energy was probably empirical today it is derived from very fundamental principles. Therefore your introduction is misleading
– Cryo
Commented Jan 3, 2021 at 0:07