You already know the everyday version of work: a force moving something through a distance, counted in joules as W = F\,s. That rule is true — but it quietly assumed two things: that the force pointed along the motion, and that it stayed the same size the whole way. Real life is rarely so tidy. You pull a sledge on a rope tilted up at an angle; a spring pushes back harder the more you squash it; a car's brakes bite while the road drags underneath. To handle all of these we need a sharper, more powerful idea of work — and a theorem that ties every joule of it directly to motion.

This page builds that bridge. First we make W = F\,s honest for a force at an angle, then for a force that changes, and finally we prove the crown jewel of the whole topic — the work–energy theorem: the total work done on an object is exactly the change in its kinetic energy. Push on something and the joules you spend do not vanish or scatter at random; they show up, to the last one, as a change in how fast it is going.

Work when the force is at an angle

Drag a sledge across the snow with a rope. Your pull F slopes upwards at an angle \theta to the ground, but the sledge slides horizontally. Only the part of your pull that lies along the direction of travel actually helps drag the sledge forward; the upward part merely tries to lift it and does no dragging at all. Splitting the force into components, the useful horizontal part is F\cos\theta, so the work done is that component times the distance moved:

Read the \cos\theta as a "dial" that measures how well the force is lined up with the motion:

• Force along the motion (\theta = 0^\circ): \cos 0^\circ = 1, so W = F\,s — the old formula, the special case of full alignment.
• Force at right angles (\theta = 90^\circ): \cos 90^\circ = 0, so W = 0. A force perpendicular to the motion does no work whatsoever.
• Force opposing the motion (\theta = 180^\circ): \cos 180^\circ = -1, so W = -F\,s — the work is negative, meaning this force is taking energy away (friction and braking do exactly this).

This last point is deeper than it looks. A ball whirled on a string, a planet circling the Sun, a car rounding a bend — all are held on their curved path by a centripetal force pulling straight towards the centre, always at right angles to the velocity. That force changes the direction of motion but never does any work, which is why an object in a steady circular orbit keeps exactly the same speed (and the same kinetic energy) forever.

Work is the area under a force–displacement graph

Now drop the other assumption — that the force stays constant. Stretch a spring and it fights back harder the further you pull; a rocket's thrust changes as its fuel burns. When F varies, you cannot just multiply one value of the force by the distance, because there is no single value to use. Instead, plot the force against the displacement and read off a beautifully general rule:

Why? Chop the journey into slivers so thin that the force barely changes across each one. Over a sliver of width \Delta s the work is F\,\Delta s — a thin rectangle. Add up all the rectangles and you have the whole area under the curve. (For those who have met it, this sum is the integral W = \int F \,\mathrm{d}s.)

For a constant force the graph is a flat horizontal line, and the area is a rectangle F \times s — recovering W = F\,s yet again. For a spring obeying Hooke's law, the force rises in a straight line from zero, so the area is a triangle:

Try it: shade the area, count the joules

Below is a force–displacement graph. Set the peak force and the displacement with the sliders, and switch the graph between a constant force (a flat line — the shaded region is a rectangle) and a spring (a rising line — the shaded region is a triangle). The shaded area is the work done, worked out live at the top; because of the work–energy theorem we are about to prove, that same number is the change in the object's kinetic energy. Notice that the spring stores exactly half the work of a constant force of the same peak size — the triangle is half the rectangle.

The work–energy theorem

Here is the central result. Add up the work done by every force on an object — that is, the work done by the resultant (net) force — and you get precisely the change in its kinetic energy. Nothing is lost, nothing is invented; the books always balance.

Where it comes from

The proof is short, and it stitches together two laws you already own: Newton's second law and the suvat equations of motion. Take a constant net force F acting along the motion over a distance s.

Step 1 — Newton's second law. The net force produces an acceleration a given by

Step 2 — the net work. Because this net force is along the motion, the net work is

Step 3 — bring in suvat. The equation of motion v^2 = u^2 + 2as rearranges to

Step 4 — substitute. Put that straight into the net-work expression:

And there it is. The \tfrac12 m v^2 that defined kinetic energy on the previous page was not plucked from the air — it falls out inevitably the moment you push F = m\,a through a distance. Work and kinetic energy are two views of the same thing.

Conservative and non-conservative forces

If the work–energy theorem always balances, why does a pushed trolley eventually roll to a halt, its kinetic energy apparently "gone"? Because forces come in two very different kinds.

A conservative force stores the work you do against it and gives every joule back later. Lift a ball and you do work against gravity; that work is banked in the gravitational store and returned in full as kinetic energy when the ball falls. Compress a spring and the elastic store hands the energy straight back on release. For a conservative force the work done depends only on start and end points, never on the path taken — and a round trip does zero net work.

A non-conservative force does not give the energy back. Friction and air resistance dissipate the work done against them into the thermal store — the surfaces simply warm up — and you can never recover it as motion. The trolley stops because friction has quietly turned its kinetic energy into a tiny amount of heat in the wheels and floor. The books still balance: friction did negative net work equal to the kinetic energy lost. Nothing broke the theorem; the energy just left as heat instead of speed.

Worked examples

Example 1 — net work gives the final speed. A 2\ \text{kg} trolley starts from rest (u = 0). A net work of 36\ \text{J} is done on it. How fast is it going? By the theorem W_{\text{net}} = \tfrac12 m v^2 - \tfrac12 m u^2, and with u = 0:

Example 2 — work by an angled force. A traveller pulls a suitcase with a force of 50\ \text{N} along a handle tilted at 60^\circ to the horizontal, wheeling it 20\ \text{m} along a flat concourse. The work done by the pull is

The vertical part of the pull (F\sin\theta) does no work — the case never moves upward — which is why only the \cos 60^\circ = 0.5 slice counts.

Example 3 — spring work as a graph area. A spring needs a force that rises straight to 40\ \text{N} when stretched by 0.10\ \text{m}. The work stored is the triangular area under the force–extension line:

Using 40 \times 0.10 = 4\ \text{J} — the full rectangle — would double-count; the spring only ever pulled with the full 40 N at the very end.

Example 4 — work against friction and braking distance. A 1000\ \text{kg} car travels at 20\ \text{m/s} and brakes to a stop. Its kinetic energy is

The brakes provide a friction force of 8000\ \text{N} opposing the motion. This force must do negative net work equal to all the kinetic energy, so f \times d = E_k, and the braking distance is

Because E_k \propto v^2, doubling the speed quadruples the energy to shed and so roughly quadruples the braking distance — the work–energy theorem is the reason speed limits and stopping distances matter so much.