The formula is $K = \frac{1}{2}mv^2$.
Which part is the square supposed to influence? Does it affect just $v$, $mv$ or $\frac{1}{2}mv$?
The problem is $\frac{1}{2} (1.2\cdot 10^5 [kg]) \cdot 40.8^2 \left[\frac{m}{s}\right]$. How would this resolve?
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I think this question is better suited for physics.SE. Also: The question does not exhibit any attempt at a solution! Here I leave a hint: Try to check whether the units of the quantities on both sides of the equality are the same. – Nox Mar 18 '17 at 09:50
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One has $$ \frac 12 mv^2=\frac 12 m\cdot v\cdot v\ne\frac 12 (mv)^2=\frac 12 mv \cdot mv=\frac 12 m^2 \cdot v^2 $$ giving here $$ K=\frac 12 \cdot 1.2 \cdot 10^5 \cdot (40.8)^2=\cdots, $$ hope you can finish it.
Olivier Oloa
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But if I do the calculations, it's 1997.5 X 10^5, yet the book says 2.0 X 10^8. It doesn't make any sense. Math is too inconsistent. – user416503 Mar 18 '17 at 10:20
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