Conservative & Nonconservative Forces

Carry a 2.0 kg package at steady speed from A to B, where B is 8.0 m to the right of A and 3.0 m higher. The straight route is √(8.0² + 3.0²) = 8.54 m long. A detour climbs 4.5 m, crosses 8.0 m and drops 1.5 m: 14.0 m. Gravity does work only on the vertical legs, −(2.0)(9.81)(4.5) = −88.3 J going up and +29.4 J coming down, a total of −58.9 J. That is exactly −mgΔy, the same as on the straight route. Air drag of constant size 4.0 N does −4.0 × 8.54 = −34.2 J on the straight route but −4.0 × 14.0 = −56.0 J on the detour. A force whose work between two points is the same along every route joining them is conservative. One matching pair is evidence, not proof: the claim covers every route between every pair of points.

For a force set by where the object is, reversing a route reverses every step dr but leaves F alone, so the work changes sign. Out along route 1 and back along route 2 therefore gives ∮ F⋅dr = W₁ − W₂, and zero on every loop is the same claim as agreement on every pair of routes. The loop is usually the easier test, one integral with no endpoints to match, but its verdicts are lopsided: one nonzero loop proves a force nonconservative, while one zero loop proves nothing. Drag breaks the argument at its first step. Turn round and the drag turns round with you, so the return leg costs work instead of repaying it. Out by the detour and back along the straight route, gravity gives −58.9 J + 58.9 J = 0, while drag gives −56.0 J − 34.2 J = −90.2 J.

Potential energy is built from the force's own work. Pick a reference point r₀, give it any value U(r₀), usually zero, and define U(r) = U(r₀) − ∫ F⋅dr from r₀ to r. That definition makes sense only if the integral ignores the route; otherwise a point would receive a different U for every path used to reach it, and U would not be a function of position at all. When it does make sense, work between any two points needs no integral: W_c = U(A) − U(B) = −ΔU. For the package, put U = 0 at A; then U(B) = +58.9 J, and gravity's work from A to B is −58.9 J. Put U = 0 at B instead and U(A) = −58.9 J, with ΔU still +58.9 J: the reference is free because only differences appear. And U belongs to the package–Earth system, since it describes their interaction.

Kinetic friction and drag point against the velocity on every step, so F⋅dr = −f ds whichever way the object moves, with ds the length of the step. Adding the steps gives W = −f s, where s is the distance travelled, not the displacement, and that is negative round every loop, so no potential energy can reproduce it. Drag a 5.0 kg crate at steady speed 6.0 m across a floor with μₖ = 0.30 and back again: f = 0.30 × 5.0 × 9.81 = 14.7 N, and over 12.0 m friction does −177 J on a crate that ends where it began. Those 177 J have not vanished. They are thermal energy in the crate's underside and the floor, random atomic motion that no position of the crate can record, and reversing the motion cannot bring them back: that is dissipation. Not every nonconservative force dissipates, though. The hand dragging the crate does +177 J, and its force has no potential energy either.

Ask a few questions before writing U. Does the force depend on velocity or on the direction of motion, as friction and drag do? Then no. Does it change with time at a fixed place? Then the work along a route depends on when you travel it, and no fixed U(r) can hold it. Is the motion along one line with the force set by x alone? Then yes: any closed trip on a line retraces itself, so U(x) = U(x₀) − ∫ F dx always exists. Constant forces and central forces F(r) r̂ also pass, because F⋅dr only ever sees the change of position along one fixed direction, or the change in r. In two dimensions, depending on position is not enough. Take F = b(−y î + x ĵ) with b = 3.0 N m⁻¹ counter-clockwise round the 2.0 m × 1.0 m rectangle with one corner at the origin: no work on the two sides where F is perpendicular or zero, +6.0 J up the right side and +6.0 J back along the top. The loop gives 12 J = 2bA, so no potential exists.