Kinetic Energy is $K = \tfrac12 mv^2$. Work by a Constant Force says how much a force adds or removes. Put them together and you get one of the most useful shortcuts in mechanics.
Key idea
The net work done on an object equals its change in kinetic energy: $W_{net} = \Delta K = \tfrac12 mv_f^2 - \tfrac12 mv_i^2$. $W_{net}$ is the sum of the work done by every force, which equals the work done by the net force.
Why it works
For a constant net force along the motion, $W_{net} = F_{net}\,d = ma\,d$. From motion without the time, $a\,d = \tfrac12(v_f^2 - v_i^2)$. Multiply by $m$: $W_{net} = \tfrac12 mv_f^2 - \tfrac12 mv_i^2$. So kinetic energy is exactly the quantity that the net work changes, and the theorem holds for changing forces too (add up the work in small steps).
Worked examples
Worked example: Speed from work
A net work of $400\ \text{J}$ is done on a $50\ \text{kg}$ cart that starts from rest. How fast does it end up?
- $400 = \tfrac12(50)v^2 - 0$.
- $v = \sqrt{\dfrac{800}{50}} = 4\ \text{m/s}$.
Answer: $4\ \text{m/s}$
Worked example: Braking force
A $1{,}200\ \text{kg}$ car moving at $20\ \text{m/s}$ stops in $50\ \text{m}$. Find the average braking force.
- $$\Delta K = 0 - \tfrac12(1200)(20)^2 = -240{,}000\ \text{J}.$$
- Braking does that work: $-F(50) = -240{,}000$, so $F = 4{,}800\ \text{N}$.
Answer: $4{,}800\ \text{N}$
Worked example: Several forces
A $10\ \text{kg}$ box at rest is pushed $5\ \text{m}$ with $40\ \text{N}$ while friction of $15\ \text{N}$ acts. Find its final speed.
- $W_{push} = 200\ \text{J}$, $W_{friction} = -75\ \text{J}$; gravity and the normal force do none.
- $W_{net} = 125\ \text{J} = \tfrac12(10)v^2$, so $v = 5\ \text{m/s}$.
Answer: $5\ \text{m/s}$
The theorem is often the fastest route when a problem gives forces and distances but asks for a speed, with no time involved.
Common mistake
Using the work of one force (usually the push) instead of the net work. The push adds energy but friction takes some away, so the push alone overstates the gain in kinetic energy. Check: list every force, find each one's work with its sign, and add them before setting the total equal to $\Delta K$.