Work energy theorem Statement and Derivation

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.

Motion of particle between two position

$W=\Delta K$

Derivation of the Work-energy theorem:

According to the equation of motion:

$v^{2}_{B}=v^{2}_{A}-2as $



$mas=\frac{m}{2} (v^{2}_{B}-v^{2}_{A})$

$Fs=\frac{1}{2}mv^{2}_{B}-\frac{1}{2}mv^{2}_{A} \qquad (\because F=ma)$

$W=\frac{1}{2}mv^{2}_{B}-\frac{1}{2}mv^{2}_{A} \qquad (\because W=Fs)$


$K_{f}$= Final Kinetic Energy at position $B$
$K_{i}$= Initial Kinetic Energy at position $A$

$W=\Delta K$

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}} $



$W=\Delta {K}$