Lesson one measured what a force costs. Lesson two measured what motion is worth. The work–energy theorem welds them together: $W_{\text{net}} = \Delta E_k = \tfrac{1}{2} m \left(v^{2} - v_0^{2}\right)$. Read aloud: net work equals the change in kinetic energy. The symbols: $W_{\text{net}}$ is the net work in joules — every force's contribution added up, not just the one you like; $\Delta$ (Greek delta) means change in; $m$ is the mass in kilograms; and the subscript convention is the one you have used since kinematics — $v_0$ (v-nought) is the speed BEFORE, $v$ with no subscript is the speed AFTER, both in m/s.

The sign carries the story, and it is final minus initial, always. Speed up, and $v^2 > v_0^2$: the net work is positive, something paid in. Slow down, and the net work is negative — friction and brakes are the usual withdrawers. Cross a level floor at a constant speed and the net work is exactly zero, no matter how tired you are.

One trap deserves naming before you meet it: $\left(v - v_0\right)^{2}$ is not $v^{2} - v_0^{2}$. Difference the squares; never square the difference. Check it once with 10 and 6 — 64 against 16, and the two answers are not even close.