Work-Energy Theorem — Work Equals Change in Kinetic Energy

What 'work' means in physics

In everyday speech, 'work' means effort. In physics it has a precise meaning: work is done when a force acts through a distance.

$$W = Fd\cos\theta$$

• $F$: the force's magnitude
• $d$: the displacement's magnitude
• $\theta$: the angle between them — only the force component along the motion counts.

Push a crate forward while it moves forward ($\theta = 0°$): full credit, $W = Fd$. Carry a bag horizontally at constant height ($\theta = 90°$): zero work on the bag. Push against a wall that doesn't move ($d = 0$): zero work, no matter the strain.

The work-energy theorem

Add up the work of every force — the net work — and you get one of mechanics' great shortcuts:

$$W_{\text{net}} = \Delta KE = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mu^2$$

The work-energy theorem: net work done on an object equals its change in kinetic energy. Instead of tracking forces moment by moment, you compare two snapshots — before and after — and the theorem bridges them.

Derivation in three steps

The theorem falls out of $F = ma$ in three steps:

1. Start with $F_{\text{net}} = ma$ and multiply by a small displacement: $F\,dx = ma\,dx$.
2. Rewrite acceleration using the chain rule: $a = v\,dv/dx$.
3. Integrate from $u$ to $v$: $\int F\,dx = m\int v\,dv = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mu^2$.

The left side is the definition of work; the right side is the change in $\tfrac{1}{2}mv^2$. The theorem is Newton's second law, accumulated over distance instead of time.

Potential energy enters the picture

Work done against gravity doesn't vanish — it's stored as gravitational potential energy, $PE = mgh$. Lift a book, and your work becomes potential energy; drop it, and gravity converts it back to kinetic energy.

When only conservative forces (gravity, springs) do work, the total $KE + PE$ stays constant — conservation of mechanical energy. Friction breaks the conservation by diverting energy into heat, which is why real roller coasters need a first big hill: every lap costs energy to friction.

Worked mini-example

You push a 25 kg crate with a steady 120 N horizontal force over $6\ \mathrm{m}$ of frictionless floor. It starts at rest. How fast is it moving at the end?

Work done: $W = Fd = 120 \times 6 = 720\ \mathrm{J}$.

By the work-energy theorem, $720 = \tfrac{1}{2}(25)v^2$:

$$v = \sqrt{\frac{2 \times 720}{25}} = \sqrt{57.6} \approx 7.6\ \mathrm{m/s}$$

No kinematics needed — no time, no acceleration computed. Energy methods shine exactly when the details of the motion are messy or unknown.

Variables and units

Common mistakes

Real-world applications

• Roller-coaster designers trade height for speed and back — mgh becomes ½mv² and back — sizing every hill from energy, not forces.

• Hydroelectric dams convert gravitational potential energy of stored water into electrical energy via the work-energy chain.

• Braking systems are sized by kinetic energy: stopping distance grows with v², which is why speed limits matter so nonlinearly.