In this answer I will derive the work-energy theorem twice: Section 1 presents a derivation that straightaway adresses the case of an arbitrary acceleration profile. This approach requires integration. The derivation provides the means to understand why the work-energy theorem relates specifically to kinetic energy.
The purpose of section 2 is to show that it is also possible to derive the work-energy theorem using means that were already available before integral calculus was developed. First the case of uniform acceleration is treated, and then it is demonstrated that the result generalizes to an arbitrary acceleration profile.
Section 1
In preparation: derivation of a kinematic relation between position $s$, velocity $v$, and acceleration $a$:
$$ v = \frac{ds}{dt} \ \Leftrightarrow \ ds = v \ dt \tag{1.1} $$
$$ a = \frac{dv}{dt} \ \Leftrightarrow \ dv = a \ dt \tag{1.2} $$
In the next four steps the differential is changed twice, using (1.1), and then (1.2): first the differential is changed from $ds$ to $dt$, with corresponding change of limits, and next the differential is changed from $dt$ to $dv$, again with corresponding change of limits.
$$ \int_{s_0}^s a \ ds \tag{1.3} $$
$$ \int_{t_0}^t a \ v \ dt \tag{1.4} $$
$$ \int_{t_0}^t v \ a \ dt \tag{1.5} $$
$$ \int_{v_0}^v v \ dv \tag{1.6} $$
The reason that (1.3) can be restated in the form of (1.6): acceleration and position are not independent; they are connected by differentiation.
Evaluating (1.6):
$$ \int_{v_0}^v v \ dv = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \tag{1.7} $$
The end result:
$$ \int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \tag{1.8} $$
Note that up to this point the treatment does not refer to any physics; (1.8) is purely a mathematical property that follows from the definitions (1.1) and (1.2), and from the properties of integration and differentiation.
With the above in place we are ready to derive the work-energy theorem:
The starting point: Newton's second law:
$$ F=ma \tag{1.9} $$
Next: integration of both sides of (1.9); integration with respect to position coordinate:
$$ \int_{s_0}^s F \ ds = \int_{s_0}^s ma \ ds \tag{1.10} $$
Use (1.8) to express the right hand side in terms of velocity. The result is the work-energy theorem:
$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \tag{1.11} $$
If $F=ma$ is granted as an axiom then the work-energy theorem follows as a theorem.
The definition of potential energy slots in with the work-energy theorem. Definition: change of potential energy is the negative of the change of work done:
$$ \Delta E_p = \Delta W \tag{1.12} $$
That is: change of potential energy is defined as in terms of evaluating an integral: (It's the negative of an integral, but the more important property is that it is an integral.)
$$ \Delta E_p = - \int_{s_0}^s F \ ds \tag{1.13} $$
The table below is set up to emphasize there is a parallel relation. The transition from the first row to the second row is integration with respect to position coordinate.
| Force |
acceleration |
| $F$ |
$ma$ |
| work |
kinetic energy |
I think the best way of understanding the work-energy theorem is in terms of how it relates to $F=ma$: integration with respect to position coordinate.
Section 2
Before the introduction of differential calculus scholars already had the following two relations for the case of uniform acceleration:
$ s - s_0 = v_0 t + \tfrac{1}{2} a t^2 \tag{2.1} $
$ v - v_0 = a t \tag{2.2} $
The factor $\tfrac{1}{2} a t^2$ was arrived at through geometric reasoning:
Create a diagram with time along the horizontal axis, and speed along the vertical axis.
In the case where the velocity is constant: the distance covered during a time interval $t$ is given by the product of the velocity and that time interval: $s - s_0 = v_0 t$. Geometrically that product corresponds to an area.
In the case of uniform acceleration: in the diagram the line that represents the velocity as a function of time is a diagonal line. It was noticed that the covered distance is still in proportion to the area between the diagonal line and the horizontal axis. That area is the area of a right angled triangle.
This area rule is also referred to as the Merton rule of uniform acceleration
(In retrospect we see that this area consideration for a specific case anticipated the later concept of integration.)
Take (2.1), and multiply both sides with acceleration $a$:
$$ a(s - s_0) = a v_0 t + \tfrac{1}{2} a^2 t^2 \tag{2.3} $$
The following rearrangement of the right hand side makes it easier to recognize the coming substitution:
$$ a(s - s_0) = v_0 (a t) + \tfrac{1}{2} (a t)^2 \tag{2.4} $$
Use (2.2) to substitute the product $(at)$:
$$ a(s - s_0) = v_0 (v - v_0) + \tfrac{1}{2}(v - v_0)^2 \tag{2.5} $$
Work out the multiplications:
$$ a(s - s_0) = v_0 v - v_0^2 + \tfrac{1}{2} v_0^2 - v_0 v + \tfrac{1}{2} v^2 \tag{2.6} $$
The crossterms $v_0 v$ drop away against each other, leaving only squared terms:
$$ a(s - s_0) = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \tag{2.7} $$
(2.7) has a remarkable property: it generalizes to an arbitrary acceleration profile.
First consider the case with two consecutive stages of different accelerations, each uniform, first from $s_0$ to $s_1$, and then from $s_1$ to $s_2$
Expressions for each of the two stages:
$ a_1(s_1-s_0) = \tfrac{1}{2}v_1^2 - \tfrac{1}{2}v_0^2 $
$ a_2(s_2-s_1) = \tfrac{1}{2}v_2^2 - \tfrac{1}{2}v_1^2 $
Since these expressions are for consecutive intervals they can be added; the result of the addition is a valid expression:
Upon addition the intermediate term $\tfrac{1}{2}v_1^2$ drops out; only the outer terms $\tfrac{1}{2}v_2^2$ and $\tfrac{1}{2}v_0^2$ remain:
$$ a_1(s_1-s_0) + a_2(s_2-s_1) = \tfrac{1}{2}v_2^2 - \tfrac{1}{2}v_0^2 \tag{2.8} $$
The above result generalizes: the total distance can be subdivided into any number of subdivisions; after adding everything together only the outer terms remain; all of the intermediate terms drop out.
The generalization of (2.8) to an arbitrary number of subdivisions of the total distance covered can be expressed as a summation:
$$ \sum_{i=1}^n a_i(s_i - s_{i-1}) = \tfrac{1}{2}v_n^2 - \tfrac{1}{2}v_{i-1}^2 \tag{2.9} $$
In the limit of $n \rightarrow \infty$:
$$ \int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 $$
