Physics

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

##### SOLUTION
Units of K.E
= $\begin{array}{l}N{m^{ - 1}} = Kgm{s^{ - 2}}m\\ = kg{m^2}{s^2}\end{array}$
Let $\begin{array}{l}{(m)^a}{(v)^b}\alpha K.E\\k{(M)^a}{(LT)^b}\alpha K.E\\K.E = k{(M)^a}{(L)^b}{(T)^{ - b}}\end{array}$
by compairing
$\begin{array}{l}a = 1\\b = 2\\ - b = - 2\\(a = 1\& b = 2)\end{array}$
thus K.E$\begin{array}{l} = k{(m)^a}{(v)^b}\\ = km{v^2}\end{array}$

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Subjective Medium Published on 18th 08, 2020
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