Where did the kinetic energy formula come from? 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?
 A: 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
$$W_{net} = \int_{t_i}^{t_f} \frac{d}{dt}\left(\frac{1}{2} m v^2\right) dt = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$
A: 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.
A: 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.
A: 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.
A: 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:
$$\Delta W = \frac{1}{2}mv_ 2^2 - \frac{1}{2}mv_ 1^2$$
A: 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
For reference: https://en.wikipedia.org/wiki/%C3%89milie_du_Ch%C3%A2telet#Advocacy_of_kinetic_energy
