Potential Energy Curves & the Stability of Equilibrium
- Recover force from a potential energy function using F = −dU/dx
- Identify equilibrium points and classify them as stable, unstable or neutral
- Use a potential energy diagram with total energy to find turning points
Force is the negative slope of potential energy
For any conservative force, F = −dU/dx. The minus sign says that the force points downhill on the potential energy curve — toward lower U — which is why a ball released on a slope rolls down rather than up. This relation is the inverse of the definition of potential energy as the negative of the work done, and it means a graph of U against x contains the complete force law: the force at any point is just the negative slope there.
Why minima are stable
At any equilibrium the slope of U is zero, so the force is zero. What distinguishes stable from unstable is what happens after a small displacement. In a valley, moving right puts you on a rising slope, so the force −dU/dx points left — back toward equilibrium. On a hilltop, moving right puts you on a falling slope, so the force points right, away from equilibrium. A restoring response defines stability, and for small displacements about a minimum the curve is approximately parabolic, so the motion is approximately simple harmonic. That is why SHM appears everywhere in physics.
Reading turning points off the diagram
Draw a horizontal line at the total energy E on a U-versus-x plot. Since E = K + U and kinetic energy cannot be negative, the particle can only exist where U ≤ E. The points where the curve crosses the line are turning points: there K = 0, the particle momentarily stops and reverses. If the line intersects the curve on both sides of a minimum, the particle is trapped in that well and oscillates between the two turning points. If the total energy exceeds the height of a barrier, the particle escapes over it.
A particle moves along x with potential energy U(x) = 2x³ − 9x² + 12x joules, x in meters. Find the equilibrium positions and classify each.
- 1.Differentiate: dU/dx = 6x² − 18x + 12 = 6(x² − 3x + 2) = 6(x − 1)(x − 2).
- 2.Setting dU/dx = 0 gives equilibria at x = 1.0 m and x = 2.0 m.
- 3.Second derivative: d²U/dx² = 12x − 18.
- 4.At x = 1.0: d²U/dx² = −6 < 0, a maximum, so unstable. At x = 2.0: d²U/dx² = +6 > 0, a minimum, so stable.
A particle sits at a point where the potential energy curve has a local maximum. This equilibrium is —
Picture the U(x) curve as a physical landscape and the particle as a marble on it. The marble rolls downhill, oscillates in valleys, balances precariously on peaks, and can only reach places no higher than the height at which you released it. Nearly every question on this topic answers itself under that picture.
On a potential energy diagram, a turning point of the motion occurs where —
Answer the 2 checkpoints as you read.
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