Potential Energy & the Force-Energy Relationship

You’ll be able to

Define potential energy for a conservative force as U = −∫F dx
Recover the force from a potential energy function using F = −dU/dx
Read a potential energy curve to locate equilibria and judge their stability

Conservative forces store energy

A conservative force (gravity, an ideal spring) does work that depends only on the endpoints, not the path, so the energy it "spends" can be fully recovered. We bookkeep that stored energy as potential energy U, defined so that the work done by the force lowers U: ΔU = −W = −∫F dx. Near Earth this gives U = mgh; for a spring it gives U = ½kx².

Potential energy functions

U_grav = mgh U_spring = ½kx²

Each is minus the work done by the conservative force. Potential energy is always measured relative to a chosen reference where U = 0.

Force is minus the slope of the energy curve

Because ΔU = −∫F dx, differentiating undoes the integral: the conservative force is the negative derivative of the potential energy, F = −dU/dx. The force points "downhill" on the U-versus-x graph, toward lower potential energy. Where the curve is flat (dU/dx = 0) the force is zero — an equilibrium. A valley (local minimum) is a stable equilibrium because a small push produces a restoring force; a hilltop (local maximum) is unstable.

Force from potential energy

F = −dU/dx

The force is the negative slope of the potential energy curve. Equilibrium occurs where dU/dx = 0; it is stable at a minimum of U, unstable at a maximum.

Worked example

A particle moves along the x-axis with potential energy U(x) = 2x² − 4x (joules, x in meters). Find the force as a function of x and locate the equilibrium position.

1.Apply F = −dU/dx. First differentiate: dU/dx = 4x − 4.
2.Negate to get the force: F(x) = −(4x − 4) = 4 − 4x newtons.
3.Equilibrium is where the force vanishes: 4 − 4x = 0 → x = 1 m.
4.Since U is an upward parabola (a minimum at x = 1), the restoring force makes this a stable equilibrium.

Answer: F(x) = −dU/dx = 4 − 4x N; equilibrium at x = 1 m, and it is stable (U has a minimum there)

Checkpoint

A particle has potential energy U(x) = 3x² (joules, x in meters). What is the force on the particle at x = 2 m?

Checkpoint

On a potential energy curve U(x), a point where dU/dx = 0 and U is a local minimum represents:

On the exam

Move fluently in both directions: integrate a conservative force to get U (U = −∫F dx), and differentiate U to get the force (F = −dU/dx). On a U-versus-x graph, remember the force points downhill and equilibria sit where the slope is zero.

Answer the 2 checkpoints as you read.