Question

Derive the expression for kinetic energy.

Answer 1

Step 1: Assumptions

Let, the mass of a body be ‘m’ moving with an initial velocity be ‘u’

Constant force be ‘F’ acts on it and the velocity changes to ‘v’

Distance covered be ‘s’ and constant acceleration be ‘a’

Step 2: Applying equations of motion

We know that,

${\mathbf{v}}^{\mathbf{2}}\mathbf{-}{\mathbf{u}}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathit{a}\mathit{s}$

$\mathbf{s}\mathbf{}\mathbf{=}\mathbf{}\frac{{\mathbf{v}}^{\mathbf{2}}\mathbf{-}{\mathbf{u}}^{\mathbf{2}}}{\mathbf{2}\mathbf{a}}$

$\mathrm{Work}\mathrm{done}=\mathrm{Force}\left(\mathrm{F}\right)\times \mathrm{displacement}\left(\mathrm{s}\right)$

$\mathbf{F}\mathbf{}\mathbf{=}\mathbf{}\mathit{m}\mathit{a}$

Step 3: Calculate the work done

$\mathbf{W}\mathbf{}\mathbf{=}\mathbf{}\mathit{m}\mathit{a}\mathbf{}\mathbf{\times }\mathbf{}\frac{{\mathbf{v}}^{\mathbf{2}}\mathbf{}\mathbf{-}{\mathbf{u}}^{\mathbf{2}}}{\mathbf{2}\mathbf{a}}$

$\mathbf{=}\mathbf{}\mathbf{m}\mathbf{}\mathbf{\times }\mathbf{}\frac{{\mathbf{v}}^{\mathbf{2}}\mathbf{}\mathbf{-}\mathbf{}{\mathbf{u}}^{\mathbf{2}}}{\mathbf{2}}$

$\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{m}\left[v^2-u^2\right]$

When initial velocity, u=0

Then,

$\mathit{W}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}}$

As, Kinetic energy = Work done

Therefore,

Kinetic energy $\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}}$