Work as Dot Product
In everyday language, 'work' means any effort you put in. In physics, the definition is far more precise — and more interesting. Work is done on an object only when a force causes a displacement, and crucially, only the component of the force that acts along the direction of displacement contributes. Push a wall all day and you do zero work on it, because the wall doesn't move. Carry a heavy bag horizontally across a room and, surprisingly, gravity does zero work — it acts downward while your displacement is horizontal. This precision is captured beautifully by the dot product (also called the scalar product) of two vectors. If a force F acts on an object that undergoes displacement d, the work done is W = F · d = Fd cosθ, where θ is the angle between the two vectors. The dot product extracts exactly the component of force that is parallel to the displacement, multiplies it by the magnitude of displacement, and returns a single scalar number — positive, negative, or zero — telling us the energy transferred. Understanding work as a dot product connects vectors to energy, tying together two major themes of physics. It also explains why the angle between force and displacement is so critical: at 0° the force is fully effective; at 90° it does nothing; at angles between 90° and 180° the work is negative, meaning energy is removed from the object. This framework underpins the work–energy theorem and the entire concept of mechanical energy.
- Vectors and their directions
- The dot product defined
- Applying the dot product to work
- Why the result is a scalar
- Variable forces and integration
Vectors and their directions
Both force and displacement are vector quantities — they have magnitude and direction. Force F might point at an angle to the path of motion. To find what fraction of the force actually drives the object along its path, we need to project F onto the direction of d. That projection is F cosθ, where θ is the angle between the two vectors when placed tail-to-tail.
The dot product defined
For any two vectors A and B, their dot product is A · B = |A||B|cosθ. It multiplies the magnitudes of both vectors by the cosine of the angle between them, yielding a scalar. Alternatively, in component form, A · B = AₓBₓ + AᵧBᵧ + A_zB_z. Both forms are equivalent and give the same number.
Applying the dot product to work
Work W = F · d = Fd cosθ. When θ = 0° (force and displacement in the same direction), cosθ = 1 and all the force contributes: W = Fd (maximum positive work). When θ = 90° (force perpendicular to displacement), cosθ = 0 and W = 0. When θ = 180° (force opposing motion, like friction or braking), cosθ = −1 and W = −Fd (negative work — energy is taken from the object).
Why the result is a scalar
Energy has no direction — it is a scalar quantity. The dot product is designed precisely to combine two vectors and produce a scalar, making it the natural mathematical operation for defining work. This distinguishes it from the cross product, which produces a vector (used in torque and magnetic force). The scalar nature of work means energies from different sources simply add arithmetically.
Variable forces and integration
When the force changes direction or magnitude along the path (as with a spring or curved motion), the work is found by summing tiny contributions: W = ∫ F · dr. Each infinitesimal displacement dr is dotted with the force at that point. This integral generalises the dot-product definition and is essential for problems beyond constant forces.
The Key Equation
- W — Work (Joule (J)): The scalar energy transferred to or from an object by the force; positive if the force aids motion, negative if it opposes motion.
- \mathbf{F} — Force vector (Newton (N)): The force applied to the object, with both magnitude and direction.
- \mathbf{d} — Displacement vector (Metre (m)): The displacement of the object from its initial to final position, with both magnitude and direction.
- \theta — Angle between F and d (Degrees or radians (dimensionless)): The angle between the force and displacement vectors when placed tail-to-tail; determines how much of the force contributes to work.
Worked Example
Problem. A person drags a suitcase along a flat floor with a force of 50 N applied via a strap that makes an angle of 35° above the horizontal. The suitcase moves 8.0 m horizontally. Calculate the work done by the applied force.
Answer. The work done by the applied force is approximately 328 J.
Origins · 1 minGaspard-Gustave de Coriolis and how this was discovered
Coriolis formally introduced the modern concept of 'work' (travail) as a scientific term in his 1829 treatise on the calculation of the effect of machines. He defined work as the force acting over a distance in the direction of the force — the same scalar dot-product relationship used today. He also introduced the factor of one-half in kinetic energy (½mv²), placing energy mechanics on a rigorous mathematical footing. Although Coriolis is most famous for the 'Coriolis effect' — the apparent deflection of moving objects in rotating reference frames — he never studied atmospheric science. The meteorological application of his rotational dynamics work came only after his death.
Related Concepts
- Work Done by a Constant Force — Directly applies the dot-product definition to calculate work when force magnitude and direction are constant throughout the displacement.
- Kinetic Energy and the Work–Energy Theorem — The work–energy theorem (W_net = ΔKE) is the principal consequence of the work definition, linking the dot product calculation directly to changes in an object's kinetic energy.
- Power and the Rate of Doing Work — Power is defined as the rate at which work is done (P = W/t = F·v), extending the dot-product definition of work into a time-dependent context.
- Scalar and Vector Quantities — Understanding why force and displacement are vectors, and why their dot product produces a scalar, is essential mathematical background for the definition of work.