Work in physics is not work in life
In everyday language, holding a heavy suitcase while you wait for a train is exhausting 'work'. In physics it is exactly zero work. Physics reserves the word work for something specific: energy transferred to an object by a force acting through a displacement. No displacement, no work.
The formula and the angle
When a constant force F pushes an object through a displacement d, only the part of the force pointing along the motion counts. If the force makes an angle \theta with the displacement, we keep the component F\cos\theta.
Work done by a constant force: force times distance times the cosine of the angle between them.
Compactly, work is the dot (inner) product of the force and displacement vectors.
Three cases make the angle intuitive. At \theta = 0^\circ (force along the motion) \cos\theta = 1 and work is maximal and positive. At \theta = 90^\circ (force sideways) \cos 90^\circ = 0 and the force does no work. At \theta = 180^\circ (force opposing the motion) \cos 180^\circ = -1 and the work is negative — the force removes energy, as kinetic friction does.
Positive, negative, and zero work
The sign of work tells you which way energy flows. Positive work pours energy into the object (it speeds up); negative work drains energy out (it slows down). Friction and air resistance almost always do negative work.
When several forces act, the total (net) work is the sum of the work done by each one — which equals the work done by the single net force. That total is the quantity that changes an object's speed, as the next guide proves.
When the force changes: area under the curve
Most real forces are not constant. Stretch a spring and it fights back harder the farther you pull. For a varying force we chop the path into tiny steps, compute F\,dx on each, and add them up. That sum is an integral — the work done by a variable force is the area under the force-versus-position graph.
For a variable force, work is the integral of force over the displacement — the area under the F–x curve.
For a spring obeying Hooke's law the force grows linearly, F = kx, so the area is a triangle: \tfrac{1}{2}kx^2. We will meet exactly that expression again as elastic potential energy in Guide 4.
Worked example: pushing a crate
A worker pushes a 20 kg crate 5.0 m across a level floor with a steady 60 N horizontal force. Kinetic friction pushes back with 15 N. Find the work done by each force and the net work. Set-up: the applied force is along the motion (\theta = 0^\circ), friction opposes it (\theta = 180^\circ), and gravity and the normal force are both vertical (\theta = 90^\circ).
The push adds 300 J; friction removes 75 J.
Gravity and the normal force do zero work; the net work is 225 J, all of which goes into the crate's kinetic energy.