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

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

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Work energy theorem Statement and Derivation

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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.

$W=K_{f}-K_{i}$

$W=\Delta K$

Derivation of the Work-energy theorem:

According to the 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 $A$

$W=\Delta K$

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}} $

$W=\frac{1}{2}mv^{2}_{B}-\frac{1}{2}mv^{2}_{A}$

$W=K_{f}-K_{i}$

$W=\Delta {K}$

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

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

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