Work energy theorem statement:
The work between the two positions is always equal to the change in kinetic energy between these positions. This is known as the work energy Theorem.
$W=K_{f}-K_{i}$
Derivation of work-energy theorem:
According to equation of motion:
$v^{2}_{B}=v^{2}_{A}-2as $
$2as=v^{2}_{B}-v^{2}_{A}$
$2mas=m(v^{2}_{B}-v^{2}_{A})$
$mas=\frac{m}{2} (v^{2}_{B}-v^{2}_{A})$
$Fs=\frac{1}{2}mv^{2}_{B}-\frac{1}{2}mv^{2}_{A} \qquad (\because F=ma)$
$W=\frac{1}{2}mv^{2}_{B}-\frac{1}{2}mv^{2}_{A} \qquad (\because W=Fs)$
$W=K_{f}-K_{i}$
Where
$K_{f}$= Final Kinetic Energy at position $B$
$K_{i}$= Initial Kinetic Energy at position $A$
Alternative Method (Integration Method):
We know that the work done by force on a particle from position $A$ to position $B$ is-
$W=\int F ds$
$W=\int (ma)ds \qquad (\because F=ma)$
$W=m \int \frac{dv}{dt}ds$
$W=m \int dv \frac{ds}{dt}$
$W=m \int v dv \qquad (\because v=\frac{ds}{dt})$
If position $A$ is the initial point where velocity is $v_{A}$ and position $B$ is the final point where velocity is $v_{B}$ then work is done by force under the limit-
$W=m \int_{v_{A}}^{v_{B}} v dv$
$W=m[\frac{v^{2}}{2}]_{v_{A}}^{v_{B}} $
$W=\frac{1}{2}mv^{2}_{B}-\frac{1}{2}mv^{2}_{A}$
$W=K_{f}-K_{i}$