The Work-Energy Theorem stated that the net work on an object equals its change in kinetic energy. With the chain-rule trick from Acceleration That Depends on Position and the work integral from Work by a Variable Force, you can prove it from Newton's second law in three lines.
Why it works
Start from $\Sigma F = m\dfrac{dv}{dt}$ along the line of motion. Rewrite the acceleration with the chain rule: $\dfrac{dv}{dt} = \dfrac{dv}{dx}\dfrac{dx}{dt} = v\dfrac{dv}{dx}$. So $\Sigma F = mv\dfrac{dv}{dx}$. Integrate both sides over the motion: $$\int_{x_1}^{x_2} \Sigma F\,dx = \int_{v_1}^{v_2} mv\,dv = \tfrac{1}{2}mv_2^2 - \tfrac{1}{2}mv_1^2.$$ The left side is the net work; the right side is the change in $K = \tfrac{1}{2}mv^2$. In three dimensions the same steps work with $\vec{F}\cdot d\vec{r}$.
Key idea
$$W_{\text{net}} = \int \Sigma\vec{F}\cdot d\vec{r} = \Delta K = \tfrac{1}{2}mv_2^2 - \tfrac{1}{2}mv_1^2.$$ It holds for any forces, constant or not. "Net" matters: add the work of every force, or integrate the net force.
Worked example: Pushing against friction
A $5\ \text{kg}$ box moving at $1\ \text{m/s}$ is pushed $4\ \text{m}$ across a floor by a horizontal $30\ \text{N}$ force, with $\mu_k = 0.2$. Find its final speed.
- Friction: $f_k = 0.2(5)(9.8) = 9.8\ \text{N}$.
- Net work: $(30 - 9.8)(4) = 80.8\ \text{J}$. Gravity and the normal force are perpendicular to the motion and do no work.
- $\tfrac{1}{2}(5)v^2 = \tfrac{1}{2}(5)(1)^2 + 80.8 = 83.3\ \text{J}$, so $v = \sqrt{33.3} = 5.77\ \text{m/s}$.
Answer: $5.77\ \text{m/s}$.
Worked example: A force that varies with position
A $2\ \text{kg}$ cart starts from rest; the only horizontal force on it is $F(x) = 6x^2$ (N, m). How fast is it moving at $x = 2\ \text{m}$?
- $W = \int_0^2 6x^2\,dx = 2x^3\Big|_0^2 = 16\ \text{J}$.
- $\tfrac{1}{2}(2)v^2 = 16$, so $v = 4\ \text{m/s}$.
- No need to find $x(t)$: the theorem skips time entirely.
Answer: $4\ \text{m/s}$.
Why bother, when $F = ma$ already holds? Because the theorem answers "how fast at this place?" without ever finding $x(t)$, which for a force that depends on position can be hard or impossible to write down. It also splits naturally: conservative forces' work moves into potential energy (Potential Energy from Force), and what's left is the work of everything else.
Common mistake
Using the work of one force, often the applied push, as if it were the net work. In the first example the push alone does $120\ \text{J}$, which would predict $v = 7\ \text{m/s}$, ignoring the $39.2\ \text{J}$ friction takes back. Check: list every force on the object, find each one's work (zero for forces perpendicular to the motion), and add them with signs before setting the total equal to $\Delta K$.