University Physics I · Work & Kinetic Energy · 6.7
Work Along a Path
The one-axis integral assumes force and motion share a line. On a curve they do not. Cut the route into steps, keep only the force along each step, and add.
Build the model
Connect the measurement to the mechanism.
Work is not a property of two endpoints. It is a running total accumulated along the route the object actually travels. Cut the path into steps short enough to count as straight, take F·Δr on each — which keeps only the force component along that step's direction of travel — and add.
In the limit that sum is the line integral W = ∫ F·dr, and C names the route. Sometimes the route cancels out: a constant force comes through the integral untouched, so gravity near the ground gives −mg Δy on every path. Friction never does that; it charges f for every metre travelled.
Deciding which case you are in is the physics.
- Simple definition
- Work along a path is the sum of F·Δr over every small step of the route, written W = ∫ F·dr along the curve C.
- Example
- A 5.0 N force pointing east acts while a cart goes 4.0 m east and then 3.0 m north. Only the east leg counts: W = 20 J.
Add F·Δr over every tiny step of the path; the integral is that sum made exact.
C names the actual route taken; W in joules, J.
Only the force component along the direction of travel counts, and θ changes step by step.
θ is the local angle between F and the tangent; ds is arc length in m.
Written out, the dot product is nothing but these per-axis terms added together.
One term per axis, each in N·m = J; a third axis adds one more term.
Write F and v as functions of t, dot them, and integrate over time — one ordinary integral.
dr = v dt, so this is W = ∫ P dt with P = F·v in watts.
A force that never changes comes out of the integral, so only the net displacement is left.
s = |Δr| in m; F must be constant in magnitude and direction.
Friction opposes motion at every step, so every metre travelled subtracts and detours cost extra.
L is path length in m, not displacement; f = μₖN in N.
Horizontal parts of any path contribute nothing, so only the height change survives.
Δy in m with upward positive; the route never appears.
The one-axis integral runs out of road
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.
Chop the path, keep the tangential part
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.
Parametrise, and it becomes an ordinary integral
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.
Same endpoints, three different answers
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.
Friction charges by the metre, gravity by the height
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.
When the shortcut is still legal
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.
Change one variable at a time
Make the relationship visible.
Set n = 1 and route C lands on the dashed line; raise n and the route hugs the floor, spending its x-travel where the force is weak, so the work falls below the straight-line value.
WORK ALONG C2.67 J
WORK, STRAIGHT LINE4.00 J
W(C) − W(LINE)-1.33 J
W(C) ÷ W(LINE)0.67
Live interpretationWORK ALONG C: 2.67 J. WORK, STRAIGHT LINE: 4.00 J. W(C) − W(LINE): −1.33 J. W(C) ÷ W(LINE): 0.67
Catch the common trap
Explain before calculating.
A 2.0 kg block is pulled horizontally at steady speed between points 3.0 m apart (μₖ = 0.30): straight, then round a semicircle on that line. Friction's work?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA constant force F = (12 N) î + (5.0 N) ĵ acts on a crate that is dragged 4.0 m east, then 3.0 m north. Find the work done, and check the answer against W = F s cos θ.
- The force never changes, so it comes straight out of the integral: W = ∫F·dr = F·∫dr = F·Δr. Only the net displacement counts, and the corner in the route is irrelevant.
- Add the legs: Δr = (4.0 m) î + (3.0 m) ĵ.
- Take the dot product one axis at a time: W = FₓΔx + FᵧΔy = (12 N)(4.0 m) + (5.0 N)(3.0 m) = 48 J + 15 J = 63 J.
- Check with the magnitude form, taking the angle from the directions themselves rather than from the answer: |F| = √(12² + 5²) = 13 N pointing arctan(5/12) = 22.6° north of east, while s = √(4.0² + 3.0²) = 5.0 m at arctan(3/4) = 36.9°. So θ = 36.9° − 22.6° = 14.2°, and W = (13 N)(5.0 m)cos 14.2° = (13)(5.0)(0.969) = 63 J.
AnswerW = 63 J
MediumA 4.0 kg box is pushed at steady speed across a level floor (μₖ = 0.25) along a quarter-circle arc of radius 2.0 m. Find the work done by friction, and compare it with the straight route between the same two points. Take g = 9.81 m s⁻².
- Kinetic friction points opposite the velocity at every step, so cos θ = −1 and F·dr = −f ds. The line integral collapses to W = −∫f ds = −f L, with L the distance actually travelled.
- The floor is level and the speed steady, so N = mg and f = μₖmg = 0.25 × 4.0 kg × 9.81 m s⁻² = 9.81 N — constant, so it comes out of the integral.
- Arc: a quarter circle of radius 2.0 m has L = ¼ × 2π × 2.0 m = 3.14 m, so W = −(9.81 N)(3.14 m) = −30.8 J.
- Chord: the straight route joining the same two points is √(2.0² + 2.0²) = 2.83 m, so W = −(9.81 N)(2.83 m) = −27.7 J.
- The detour costs 3.1 J more. Friction charges by the metre, so naming the endpoints is not enough — you have to name the route.
AnswerArc: W = −30.8 J. Chord: W = −27.7 J — the arc costs 3.1 J more.
HardA force F = (βy) î with β = 4.0 N m⁻¹ acts on a bead carried from (0, 0) to (2.0 m, 2.0 m). Find the work along (a) the straight line y = x and (b) the parabola y = x²/(2.0 m).
- F has no y-component, so W = ∫(Fₓ dx + Fᵧ dy) = ∫βy dx. The path equation is what supplies y in terms of x — and it is the only place the route enters.
- Straight line, y = x: Fₓ = βx, so W = ∫₀² (4.0 N m⁻¹)x dx = (4.0 N m⁻¹)(2.0 m)²/2 = 8.0 J.
- Parabola, y = x²/(2.0 m): Fₓ = (4.0 N m⁻¹)(x²/2.0 m) = (2.0 N m⁻²)x², so W = ∫₀² (2.0 N m⁻²)x² dx = (2.0 N m⁻²)(2.0 m)³/3 = 5.3 J.
- The parabola runs below the line, so it spends its x-travel where the force is weaker and earns 2.7 J less between the same endpoints. This F is not conservative, and "the work from (0, 0) to (2.0 m, 2.0 m)" is not a question with an answer until the route is named.
AnswerStraight line: W = 8.0 J. Parabola: W = 5.3 J (16/3 J).