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Check whether the given equation \(F.s=\frac{1}{2}mv^2-\frac{1}{2}mu^2\) is dimensionally correct, or not where, m is mass of the body, v is its final velocity, u is its initial velocity, F is applied force and s is distance covered.

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Best answer

The dimension of L.H.S.

F.s = [MLT-2 ].[L]

=[ML2T-2 ]

The dimensions of R.H.S.

\(\frac{1}{2}mv^2 or \frac{1}{2}mu^2\) = [M][LT-1]2

= [ML2T-2]

So, dimensions of L.H.S.= R.H.S.

So, given equation is dimensionally correct.

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