Force from Potential Energy

For a conservative force, the work done between two positions equals the drop in potential energy: W = −[U(x₂) − U(x₁)]. Shrink the interval to a width dx so small that Fₓ barely changes across it. The work is then Fₓ dx and the energy change is dU, so Fₓ dx = −dU and Fₓ = −dU/dx. The units agree: a joule per metre is a newton-metre per metre, which is a newton. Test the rule on energies you already trust. Near-Earth gravity has U = mgy, so the force along +y is −mg: 19.6 N downward on a 2.0 kg book, whether you put the zero of height at the floor, the desk or the ceiling, because the added constant vanishes when you differentiate. An ideal spring has U = ½kx², so Fₓ = −kx, Hooke's law with its restoring sign already built in.

On a graph of U against x, the slope carries the force and the minus sign turns it round. Where U climbs to the right the force points left; where U falls to the right it points right. Either way the force points downhill, toward lower potential energy. Its magnitude is the steepness: if the tangent at some point rises 0.60 J over 0.15 m, the slope is +4.0 J m⁻¹ and Fₓ = −4.0 N. Notice what does not matter, which is the height of the curve. A point high on a steep flank feels a large force, a point high on a flat plateau feels none, and a point where U is negative can still be pushed hard. When U comes as a table rather than a formula, estimate the slope with a centred difference, [U(x + Δx) − U(x − Δx)]/(2Δx), which straddles the point symmetrically.

Equilibrium means zero net force, so it sits wherever dU/dx = 0, and finding it is ordinary calculus. Take a bead on a straight wire with U(x) = (1.0 J m⁻⁴)x⁴ − (2.0 J m⁻²)x², the same landscape the figure below plots. Then dU/dx = (4.0 J m⁻⁴)x³ − (4.0 J m⁻²)x, which factors as 4.0x(x² − 1) newtons for x in metres and vanishes at x = 0 and at x = ±1.0 m. Sketch it and the three show up: two valleys at x = ±1.0 m, where U = −1.0 J, separated by a hump at the origin, where U = 0. Zero force says nothing about speed. A bead placed at rest on an equilibrium stays there, but a bead that crosses x = 0 while moving simply carries on through: at that instant nothing speeds it up and nothing slows it down.

Displace the object a small distance δ from equilibrium x₀. To first order the force becomes Fₓ(x₀ + δ) ≈ Fₓ(x₀) + F′ₓ(x₀)δ, and since Fₓ(x₀) = 0 and F′ₓ = −U″, that is −U″(x₀)δ. If U″(x₀) > 0 the force opposes the displacement: the point is a minimum and the equilibrium is stable. If U″(x₀) < 0 the force points along the displacement: a maximum, and unstable. For the bead, U″ = (12 J m⁻⁴)x² − 4.0 J m⁻². At x = +1.0 m, U″ = +8.0 N m⁻¹, so a 1.0 cm nudge outward meets about −0.080 N, back toward the valley floor; the exact value is −0.0812 N. At x = 0, U″ = −4.0 N m⁻¹, and a 5.0 cm nudge meets about +0.20 N driving it further out. At a minimum, U″ is the effective spring constant of the well.

The curvature test has a blind spot. At a flat point where U″(x₀) = 0 as well, the linear force vanishes and the next nonzero derivative decides. Take U = cx⁴ with c > 0. Then U″(0) = 0, yet Fₓ = −4cx³ points back toward the origin from both sides, so x = 0 is stable, only with no spring-like region: halve the displacement and the restoring force drops to an eighth. Take U = cx³ instead. Now Fₓ = −3cx² points toward −x on both sides of the origin, so a nudge to the right is pushed back while a nudge to the left runs away, and the point is unstable. Finally, on a stretch where U is constant, as for a puck on a level frictionless table, every position is an equilibrium and a displaced object stays where it is put: neutral equilibrium.

The relation needs a potential energy to exist. Friction, drag and a push from your hand have none, because their work depends on the path, so their forces never come from a U(x) and must be added separately. It also returns only the component along the coordinate you differentiate. A bead on a curved frictionless wire moves along arc length s, so the useful statement is Fₛ = −dU/ds, which for U = mgy is −mg sin θ: about 2.5 N down the wire for a 0.50 kg bead where the wire rises at 30°. The normal force never appears. Differentiate mgy with respect to horizontal x instead and you get −mg tan θ, which is neither the force along the wire nor the horizontal part of the net force. A U(x) with a sharp corner, such as a hard wall, stands for an idealised impulsive push rather than a finite slope. And in three dimensions each force component needs its own derivative of U, taken with the other coordinates held fixed, a step beyond this course.