Check whether the given equation \(F.s=\frac{1}{2}mv^2-\frac{1}{2}mu^2

1.	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.
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.

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.

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.