Where does ½mv² come from?

We know two things from earlier topics:

• doing work on an object transfers energy to it ($W = Fs$);
• a resultant force makes it accelerate ($F = ma$).

Put them together and the work must show up as kinetic energy. The derivation is just bookkeeping: write the work in terms of speed, and see what comes out.

A mass $m$ is pushed by a constant resultant force $F$ through a distance $s$.

It speeds up from an initial speed $u$ to a final speed $v$ along a straight line. The force points the same way as the motion, so all of the work goes into speed.

The work done by the force is

$W = Fs$

Newton's second law says the resultant force and the acceleration are linked by $F = ma$.

Substituting into the work equation:

$W = Fs = mas$

Now everything is in terms of $m$, $a$ and $s$ — but we want speed. That is where the equations of motion come in.

The equation of motion that links $u$, $v$, $a$ and $s$ — with no time in it — is

$v^2 = u^2 + 2as$

Rearrange it to make $as$ the subject, because $as$ is exactly what appears in $W = mas$:

$as = \frac{v^2 - u^2}{2}$

Put that into $W = mas$:

$W = m\left(\frac{v^2 - u^2}{2}\right)$

$W = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

The work done equals the increase in the quantity $\frac{1}{2}mv^2$. So that quantity is the energy of motion — the kinetic energy:

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

and the general statement is $W = \Delta E_K$.

Here is the same result without algebra, and it makes the $\frac{1}{2}$ obvious.

Start from rest and accelerate steadily. The speed grows from $0$ to $v$, so the average speed is $\frac{v}{2}$, not $v$.

The distance covered is therefore $\frac{v}{2} \times t$, and the force is $F = ma = \frac{mv}{t}$. Multiply:

$W = Fs = \frac{mv}{t} \times \frac{v}{2} \times t = \frac{1}{2}mv^2$

The $t$ cancels, and the $\frac{1}{2}$ is simply the average-speed factor. Nothing mysterious — you do not travel at the top speed for the whole trip.