Work Done by a Force and the Joule

Work is done when a force moves an object through a displacement, W = Fs cos θ, where θ is the angle between the force and the displacement; the unit of work is the joule (J), and 1 J is the work done by a force of 1 N moving an object 1 m in the direction of the force. In brief: only the component of the force along the direction of motion does work. Work is a transfer of energy, so the work done on an object equals the energy transferred to it. A force that does not cause displacement does no work, however large it is.

In physics, holding a heavy box still does no work, but lifting it does. This guide defines work and the joule, explains the effect of the angle between force and displacement, shows how to find work from a force-distance graph, compares the cases in a table, works through examples and corrects common mistakes.

What is work?

When a constant force F moves an object through a distance s in the direction of the force, the work done is W = Fs. The SI unit is the joule: 1 J = 1 N·m. Work is a scalar. Work is a way of transferring energy: when a force does positive work on an object, energy is transferred to the object (as kinetic energy, potential energy or thermal energy, depending on the situation). The kilojoule (kJ) and megajoule (MJ) are used for larger amounts.

How does the angle matter?

If the force makes an angle θ with the displacement, only the component F cos θ along the displacement does work: W = Fs cos θ. The component perpendicular to the displacement does no work. For θ = 0° (force along the motion), cos θ = 1 and W = Fs; for θ = 90°, cos θ = 0 and W = 0; for θ = 180° (force opposite to the motion), W = −Fs. This is the same as the dot product of the force and displacement vectors.

How is work found from a graph?

On a graph of force (along the displacement) against distance, the area under the line is the work done. For a constant force this is a rectangle, F × s. For a force that varies steadily from F₁ to F₂, the work is the area of a trapezoid, ½(F₁ + F₂)s. For a spring, F = kx, the area is a triangle: W = ½kx².

Work in different situations

Worked example: calculating work

1. Force along the motion. A student pushes a box with a horizontal force of 40 N over a distance of 6.0 m. W = Fs = 40 × 6.0 = 240 J.
2. Force at an angle. A rope pulls a sled with a force of 50 N at 30° above the horizontal, and the sled moves 10 m. W = Fs cos θ = 50 × 10 × cos 30° = 500 × 0.866 = 433 J.
3. Lifting. A 5.0 kg bag is lifted at constant speed through 1.2 m. The lifting force equals the weight: F = 5.0 × 9.8 = 49 N. W = 49 × 1.2 = 58.8 J, equal to the gain in gravitational potential energy.
4. Force-distance graph. A spring with k = 200 N/m is stretched from 0 to 0.10 m. The force rises from 0 to 20 N. W = ½ × 20 × 0.10 = 1.0 J (= ½kx² = ½ × 200 × 0.01).
5. Zero work. A waiter carries a tray horizontally at a constant velocity. The upward force on the tray is perpendicular to the displacement, so the work done by it is zero.

How did the concept of work develop?

The term “work” in its modern sense was introduced by Gaspard-Gustave Coriolis in 1829. The idea that work and energy are related was developed in the early 19th century in the study of steam engines, whose output was measured in “foot-pounds”. James Prescott Joule, who proved the equivalence of work and heat in the 1840s, gave his name to the unit.

Which mistakes are common?

Mistake 1: Thinking that effort means work

Pushing against a wall until you are tired does no work on the wall, because it does not move.

Mistake 2: Forgetting the cosine

With a force at an angle, use only the component along the displacement.

Mistake 3: Using the weight with a horizontal displacement

The weight acts vertically and does no work on a horizontal motion.

Mistake 4: Confusing the joule with the newton

The newton is the unit of force, and the joule (N·m) is the unit of work and energy.