Question

Derive an expression of kinetic energy of a body of mass 'm' and moving with velocity 'v' , using dimensional analysis.

Answer 1

Since we know that,

The dimension of energy $\left[E\right]=\left[M^1{L}^2{T}^{-2}\right]$

The dimension of mass $\left[M\right]=\left[M^1{L}^0{T}^0\right]$

Dimension of velocity$\left[V\right]=\left[M^0{L}^1{T}^{-1}\right]$

Let $\left[E\right]=k\left[M{]}^x\right[V{]}^y$

Where k is the proportionality constant which is a dimensionless quantity.

Therefore,

$\left[M^1{L}^2{T}^{-2}\right]=k\left[M^1{L}^0{T}^0{]}^x\right[M^0{L}^1{T}^{-1}{]}^y$

So, $x=1,y=2$

Thus, We have

$E=km{v}^2$

It is found that $k=\frac{1}{2}$

Hence, $E=\frac{1}{2}m{v}^2$