Work and the Work–Energy Theorem

Lean into a stalled car and shove it down the road; draw a bow until the string bites into your fingers; haul a crate up a loading ramp. In each case you push, the thing moves, and afterwards something has changed — the car is rolling, the bow is cocked and dangerous, the crate is higher up. Physics gives that "something you spent" a precise name: work. Work is the currency in which forces pay for changes in motion and stored energy, and this page is about computing it exactly — even when the force keeps changing as you go.

The everyday word is slippery, so we pin it down. Work is not "effort" and not "how tired you feel": stand holding a heavy suitcase perfectly still and you do zero physical work on it, no matter how much your arm aches. Work is done only when a force acts through a displacement — and only the part of the force that lies along the motion counts. That single idea, taken seriously and pushed through calculus, gives us two things at once: a way to add up work along any winding path, and a beautiful bookkeeping law — the work–energy theorem — that ties the total work done on a body to the change in its kinetic energy.

Work as a line integral

Start with a single instant. A force \vec{F} acts on a particle while the particle moves through a tiny displacement d\vec{r}. Only the component of \vec{F} along that step contributes, and the clean way to extract "the part along" is the dot product. So the little bit of work done over that step is

dW = \vec{F}\cdot d\vec{r} = |\vec{F}|\,|d\vec{r}|\cos\theta,

where \theta is the angle between the force and the direction of motion. To get the total work as the particle travels along a whole path C — a curve through space that may bend and climb — we add up all those infinitesimal contributions. That sum is a line integral of the force field along the curve:

Read those three lines as one idea in three costumes. The first is fully general — any force, any curly path. The second says: if you know the force and velocity moment by moment, integrate their dot product over time. The third is the workhorse for a first course — push something in a straight line and you are just finding the area under a graph of force against position. We spend the rest of the page living in that third form and then cashing it out into energy.

Constant force: W = F d \cos\theta

The simplest case: a constant force pushing a body through a straight displacement of length d. Nothing changes along the way, so the integral is trivial — the force factors straight out:

W = \int_C \vec{F}\cdot d\vec{r} = \vec{F}\cdot\vec{d} = F\,d\cos\theta.

Worked example — hauling a sledge. You drag a sledge d = 20\ \text{m} across flat snow by a rope pulling with force F = 50\ \text{N}, held at \theta = 30^\circ above the horizontal. Only the horizontal part of the pull moves the sledge along its path, so

W = F d\cos\theta = (50)(20)\cos 30^\circ = 1000 \times 0.866 \approx 866\ \text{J}.

The vertical part of the rope's pull, F\sin\theta, is perpendicular to the motion and does no work at all — it only lightens the load on the snow. Notice the unit: a newton times a metre is a joule (1\ \text{J} = 1\ \text{N·m}), the SI unit of work and of energy. And the sign is carried entirely by \cos\theta: a force pulling along the motion (\theta < 90^\circ) does positive work; one pushing against it (\theta > 90^\circ, e.g. friction, or you lowering a weight gently) does negative work; and a force exactly perpendicular to the motion (\theta = 90^\circ) does precisely zero.

Exactly zero, over a full loop and at every instant in between. For a body in a circular orbit the gravitational pull points to the centre while the velocity points along the circle — the two are always at right angles, so \vec{F}\cdot\vec{v} = 0 the whole way round. Gravity bends the path without ever speeding the satellite up or slowing it down, which is why a circular orbit has a constant speed. The same logic explains why the tension in a whirling string, or the normal force from a track, so often does no work: it is there to steer, not to energise. Whenever a force turns a motion without changing its speed, you can bet it is sitting perpendicular to the velocity and quietly doing nothing, work-wise.

Variable force: work is the area under the F–x graph

Most forces are not constant. Stretch a spring and it fights back harder the further you pull; a spring obeying Hooke's law exerts a restoring force F_{\text{spring}} = -kx, where k is the spring constant (in \text{N/m}) and x the stretch from its natural length. The minus sign says the spring always pulls back toward x = 0. To stretch it you must supply the opposite, F_{\text{you}} = +kx — small at first, growing steadily as x grows.

Because the force changes with position we cannot just multiply; we must integrate. The work you do stretching the spring from 0 to x is

W_{\text{you}} = \int_0^{x} kx'\,dx' = \tfrac{1}{2}k x^2,

and the work done by the spring over the same stretch is the negative of that, W_{\text{spring}} = \int_0^{x}(-kx')\,dx' = -\tfrac{1}{2}kx^2 — it resists, so it does negative work as it is stretched, and stores that \tfrac{1}{2}kx^2 as elastic potential energy ready to fling back. Look at the graph below: your applied force F = kx is a straight line from the origin, and the work is the triangular area underneath it — base x, height kx, so area \tfrac{1}{2}\cdot x\cdot kx = \tfrac{1}{2}kx^2. Drag the slider to change the stiffness and watch that area — and the work — grow.

This "area under the curve" reading is completely general: for any one-dimensional force, W = \int F(x)\,dx is the signed area between the F(x) curve and the x-axis. A constant force gives a rectangle (W = F d); a spring gives a triangle (W = \tfrac{1}{2}kx^2); a lumpy, wiggling force gives a lumpy area you find by integrating. The graph is the calculation.

The work–energy theorem

Now the payoff. Take the net force on a particle of mass m and feed Newton's second law, F_{\text{net}} = ma, into the one-dimensional work integral. The magic is a chain-rule trick that turns a time-derivative into a velocity-derivative:

a = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\,\frac{dv}{dx}\quad\Longrightarrow\quad F_{\text{net}}\,dx = ma\,dx = m\,v\,\frac{dv}{dx}\,dx = m\,v\,dv.

That little identity F\,dx = m\,v\,dv is the whole engine. Integrate both sides from the start of the motion to the end — the left side is the net work, the right side is an elementary integral in v:

W_{\text{net}} = \int_{x_i}^{x_f} F_{\text{net}}\,dx = \int_{v_i}^{v_f} m\,v\,dv = \left[\tfrac{1}{2}mv^2\right]_{v_i}^{v_f} = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2.

The quantity \tfrac{1}{2}mv^2 that pops out on the right is important enough to name: it is the kinetic energy KE = \tfrac{1}{2}mv^2, the energy a body has purely by virtue of moving. So the net work done on a body equals the change in its kinetic energy — nothing more, nothing less.

Worked examples with energy

Example 1 — a spring launches a block. A spring of stiffness k = 200\ \text{N/m} is compressed by x = 0.10\ \text{m} and released, pushing a m = 0.50\ \text{kg} block along a frictionless table. How fast is the block going when it leaves the spring? The spring does work equal to the energy it had stored:

W = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(200)(0.10)^2 = 1.0\ \text{J}.

By the work–energy theorem, starting from rest (v_i = 0), all of that work becomes kinetic energy:

W = \tfrac{1}{2}mv_f^2 \;\Longrightarrow\; v_f = \sqrt{\frac{2W}{m}} = \sqrt{\frac{2(1.0)}{0.50}} = \sqrt{4} = 2.0\ \text{m/s}.

Example 2 — braking a car. A car of mass m = 1200\ \text{kg} travelling at v_i = 20\ \text{m/s} brakes to a stop. How much work do the brakes (plus friction) do on it? The final speed is zero, so

W_{\text{net}} = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2 = 0 - \tfrac{1}{2}(1200)(20)^2 = -2.4\times 10^5\ \text{J}.

The work is negative — the braking force points opposite the motion, draining 240\ \text{kJ} of kinetic energy away as heat. Double the speed and, because of the v^2, the energy to remove quadruples: this is exactly why stopping distances balloon at highway speeds.

Example 3 — a variable push, straight from the graph. A force along the x-axis rises linearly from 0 to 30\ \text{N} over the first 4\ \text{m}, then holds steady at 30\ \text{N} for the next 3\ \text{m}. The total work is just the area under that F–x graph — a triangle plus a rectangle:

W = \underbrace{\tfrac{1}{2}(4)(30)}_{\text{triangle}} + \underbrace{(3)(30)}_{\text{rectangle}} = 60 + 90 = 150\ \text{J}.

No integration formula needed — when the graph is made of straight lines, geometry is the integral.

No — and this is the classic trap. It is tempting to think "the box moved 10 metres, so the same work was done no matter what." But work is \vec{F}\cdot d\vec{r}, a dot product: only the component of the force along the motion counts, through the \cos\theta. Push a trolley straight ahead and every newton pays off; push it at 60^\circ and only half your force (\cos 60^\circ = 0.5) does useful work; push exactly sideways and, for all your straining, you do zero work on its forward motion. So two people applying the same force over the same distance can do completely different amounts of work — direction matters as much as magnitude.

The mirror-image mistake is to assume any force touching a moving object must be doing work. It need not: a perpendicular force (a string tension steering a ball in a circle, the normal force from the floor as you walk forward, gravity on a satellite) does exactly none, because \cos 90^\circ = 0. Always ask which way the force points relative to the motion before you trust a number.