Using E_K = ½mv²: the squared bit that bites

There is not much to memorise here: $E_K = \frac{1}{2}mv^2$, with mass in $\text{kg}$ and speed in $\text{m s}^{-1}$, giving energy in $\text{J}$.

What matters is the shape of the formula. $v$ is squared, so a small change in speed makes a big change in energy.

Square a few speeds to feel it:

• $v = 10 \Rightarrow v^2 = 100$
• $v = 20 \Rightarrow v^2 = 400$ (four times bigger)
• $v = 30 \Rightarrow v^2 = 900$ (nine times bigger)

Very often the question gives you the energy and wants the speed, for example "how fast is it going after falling this far?" or "what speed does this engine give it?".

Make $v$ the subject in two steps:

$\frac{1}{2}mv^2 = E_K$

$v^2 = \frac{2E_K}{m}$

$v = \sqrt{\frac{2E_K}{m}}$

The square root is not optional. Forgetting it gives an answer that is too big — and the wrong unit pattern is a clue, since $v^2$ has units $\text{m}^2\text{s}^{-2}$.