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 (\

(The Advance Learning Institute of Physics and Technology)