Work Along a Path

W = ∫ Fₓ dx earns its simplicity from two assumptions: the object moves along one straight axis, and the force lies along that axis too. A cart on a track satisfies both. A bead threaded on a bent wire, a ball on a curved ramp, or a charge crossing a field region satisfies neither — the direction of travel swings around as the object moves, and the force can swing independently of it. Neither one has a single fixed direction to project onto. Until you say which route was taken, the question "how much work?" has no answer to give.

Cut the route into steps small enough that each counts as straight. Over a step Δr the force is effectively constant, so the constant-force result holds: ΔW = F·Δr = F Δs cos θ, with Δs the step length and θ the angle between F and that step's direction. Add the steps, shrink them, and the sum becomes W = ∫ F·dr along the curve. The dot product does its old job — keep the component along the direction of travel, discard the perpendicular part — but now afresh at every point, because θ varies along the path. A force perpendicular to the motion everywhere therefore does no work at all, however large.

To evaluate the integral, describe the path with one parameter. Time is the usual choice: write r(t) = (x(t), y(t), z(t)), note that dr = v dt, and the line integral collapses to W = ∫ F·v dt between two times — a single-variable integral of the kind you already do. Any parameter serves: use x itself when the path is a graph y(x), or an angle when the path is an arc of radius R, where ds = R dφ. In components the same statement reads W = ∫ (Fₓ dx + Fᵧ dy), with the path equation used to express one coordinate in terms of the other. The route enters through that substitution and nowhere else.

Take a force that grows with height but always points along x: F = (βy) î with β = 1.0 N m⁻¹. Move a particle from (0, 0) to (1.0 m, 1.0 m) three ways. Across then up: on the first leg y = 0, so F = 0; on the second, F points along x while dr points along y, so F·dr = 0 — total W = 0. Up then across: the first leg again gives nothing, but on the second y is fixed at 1.0 m, so Fₓ = 1.0 N acts over 1.0 m — total W = 1.0 J. Straight diagonal, where y = x: F·dr = βx dx, so W = ∫₀¹ βx dx = ½β(1.0 m)² = 0.50 J. Same force law, same endpoints, three answers.

Two everyday forces sit on opposite sides of this. Kinetic friction of magnitude f always points opposite the velocity, so cos θ = −1 at every step and W = −∫ f ds = −f L for constant f, with L the length actually travelled. Take the scenic route and you pay more. Gravity near the ground is the other case: with F = −mg ĵ the dot product gives F·dr = −mg dy, every horizontal contribution vanishes, and W = −mg Δy on any path — stairs, ramp, spiral, closed loop. That path-independence defines a conservative force, and it is what lets you replace the integral with a potential energy such as U = mgy.

So when may you still write W = F s cos θ? Whenever the force is constant in magnitude and direction. Then F comes out of the integral, W = F·∫dr = F·Δr, and only the net displacement survives — the path may bend as it likes. That is why uniform gravity gives −mg Δy on any route. A straight path is not the condition; a constant force is. What kills the shortcut is a force that changes: friction turns with the velocity, so s cos θ has nothing fixed to multiply. Before you integrate, ask whether the force is constant, and if not, whether another route between the same endpoints would change the answer.