Develop the notations of work and kinetic energy and show that it leads to work-energy theorem.

Work : The dot product of force and displacement is called as work.
$\text{ Work }=\mathrm{w}=\overline{\mathrm{F}}\cdot \overline{\mathrm{S}}$
W = F S Cos$\mathrm{\theta }$
$$\begin{gathered} \mathrm{Case} \mathrm{I} : \mathrm{\theta }=0 \\ \mathrm{W}=\mathrm{FScos}0=\mathrm{FS} \\ \mathrm{Case} \mathrm{I} : \mathrm{\theta }=90° \\ \mathrm{W}=\mathrm{FScos}90=0 \end{gathered}$$
Case III : $\mathrm{\theta }={180}^{\circ }$
W = – FS cos180 = -FS
Kinetic Energy :
The energy possessed by a body by virtue of its motion is called kinetic energy.
$\mathrm{KE}=\frac{1}{2}{\mathrm{mv}}^2$
Work-Energy Theorem :
The work done is equal to change in its KE
$\text{ i.e. }\mathrm{W}=\frac{1}{2}{\mathrm{mv}}^2-\frac{1}{2}{\mathrm{mu}}^2$
Proof : Let mass of a body is m moving with initial velocity u be acted on by a force F and its final velocity is V.

Initial Velocity = u
Final Velocity = V
Time Taken by the body = t
Distance travelled by the body = S
From ${\mathrm{v}}^2-{\mathrm{u}}^2=2\mathrm{as}$
$\mathrm{a}=\frac{{\mathrm{v}}^2-{\mathrm{u}}^2}{2\mathrm{s}} ...\left(1\right)$
Workdone = W = FS
But F = ma
$\therefore \mathrm{W}=\mathrm{mas} ...\left(2\right)$
Eqn (1) is substituting in eqn (2)
$\begin{array}{l}\mathrm{W}=\mathrm{m}\times \left(\frac{{\mathrm{v}}^2-{\mathrm{u}}^2}{2\mathrm{S}}\right)\times \mathrm{S}\\ \mathrm{W}=\frac{1}{2}\mathrm{m}\left({\mathrm{v}}^2-{\mathrm{u}}^2\right)\\ \mathrm{W}=\frac{1}{2}{\mathrm{mv}}^2-\frac{1}{2}{\mathrm{mu}}^2\\ \mathrm{W}=\text{ final KE - Initial }\mathrm{KE}\end{array}$
i.e. workdone W = Change in KE
Hence work-energy theorem is proved.