Physics

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

##### SOLUTION

Since we  know that,

The dimension of energy $[E]=[M^1L^2T^{-2}]$
The dimension of mass   $[M]=[M^1L^0T^0]$
Dimension of velocity$[V]=[M^0L^1T^{-1}]$

Let $[E]=k[M]^x[V]^y$

Where k is the proportionality constant which is a dimensionless quantity.
Therefore,
$[M^1L^2T^{-2}]=k[M^1L^0T^0]^x[M^0L^1T^{-1}]^y$

So, $x=1,y=2$

Thus, We have
$E=kmv^2$

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

Hence, $E=\dfrac{1}{2}mv^2$

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Subjective Medium Published on 18th 08, 2020
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