What happens to the kinetic energy of an object if its velocity is doubled? Explain.

Step 1:Formula and given data

Kinetic energy can be expressed as

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

where, m is the mass and ‘v’ is the velocity

Given new velocity, $\mathit{v}\mathbf{\text{'}}\mathbf{=}\mathbf{2}\mathit{v}$

Step 2:Finding new KE

New Kinetic Energy:

$\mathit{K}\mathit{E}\mathbf{\text{'}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}\mathit{v}{\mathbf{\text{'}}}^{\mathbf{2}}$

Substituting $v=2v$in the formula

$$\begin{gathered} \mathit{K}\mathit{E}\mathbf{\text{'}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathbf{\left(}\mathbf{2}\mathbf{v}\mathbf{\right)}}^{\mathbf{2}}\mathbf{} \\ \mathit{K}\mathit{E}\mathbf{\text{'}}\mathbf{=}\mathbf{4}\mathbf{\times }\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}} \\ \mathit{K}{\mathit{E}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{4}\mathbf{}\mathit{K}\mathit{E} \end{gathered}$$

Hence, when velocity is doubled, kinetic energy becomes $4$ times.