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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