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JEE questions often hand you a graph of potential energy U against position x and ask where the force is zero, which equilibrium is stable, or how far a particle can go. You do not need the particle's equation of motion for any of this. Four rules about slopes and horizontal lines are enough. This post states them, works one example in full and lists the traps.
Where does this come from?
The NCERT Class 11 chapter on work, energy and power defines the potential energy V(x) of a conservative force through F(x) = −dV/dx, and shows that kinetic plus potential energy stays constant when only conservative forces do work. The JEE (Main) 2026 syllabus lists kinetic and potential energies and conservative and non-conservative forces under Unit 4. Check the syllabus for your exam year. Below, U(x) means the same as NCERT's V(x).
What are the four rules?
Force is minus the slope. F = −dU/dx. Where the graph slopes up to the right, the force points left, and the reverse. The force always pushes the particle downhill on the graph.
Equilibrium is where the slope is zero. dU/dx = 0 means F = 0.
Minimum means stable, maximum means unstable. OpenStax University Physics explains why: near a minimum, the force on either side points back towards the equilibrium point, a restoring force; near a maximum, it points away. A flat stretch where U is constant gives neutral equilibrium.
Draw the energy line. Total energy E is a horizontal line. Since kinetic energy K = E − U cannot be negative, the particle can only be where U ≤ E. Where the line meets the curve, K = 0 and the particle turns round. OpenStax calls these turning points.
Worked example: a cubic potential
A 0.5 kg particle moves along x in a potential U(x) = x³ − 3x, with U in joules and x in metres. These are sample values for practice. No other forces act.
Step 1: the force. F = −dU/dx = −(3x² − 3) = 3 − 3x². At x = 0, F = +3 N (pointing in +x). At x = 2, F = 3 − 12 = −9 N.
Step 2: equilibrium points. 3x² − 3 = 0 gives x = +1 and x = −1.
Step 3: stability. Use the second derivative, d²U/dx² = 6x.
At x = +1, it is +6, so U has a minimum: stable equilibrium, with U(1) = 1 − 3 = −2 J.
At x = −1, it is −6, so U has a maximum: unstable equilibrium, with U(−1) = −1 + 3 = 2 J.
Step 4: a particle released from rest at x = 0. Its total energy is E = U(0) = 0. Turning points are where U = E: x³ − 3x = 0 gives x = 0 and x = ±√3. The force at x = 0 points in +x, so it moves right, through the valley at x = 1, and turns round at x = √3 ≈ 1.73 m. It then swings back to x = 0, and repeats.
Step 5: maximum speed. Speed is greatest where U is lowest along its path, at x = 1. There K = E − U = 0 − (−2) = 2 J, so v = √(2K/m) = √(2 × 2/0.5) = √8 ≈ 2.83 m/s.
Step 6: how much energy before it escapes? The valley's left wall is the peak at x = −1, with U = 2 J. If E = 2 J exactly, U = E gives x³ − 3x − 2 = 0, which factors as (x + 1)²(x − 2) = 0. So the particle is trapped between x = −1 and x = 2. With E above 2 J, a particle moving left gets over the peak and, since U falls forever as x becomes very negative, never comes back.
How do you read these from a graph with no formula?
Most exam graphs come without an equation. Use the rules visually:
Steepest part of the curve: largest force.
Flat points: equilibrium. Valley bottoms are stable, hilltops unstable.
Speed at any point: the vertical gap between the energy line and the curve gives K.
Allowed region: everywhere the curve lies below the energy line.
OpenStax works a similar example with U(x) = 2(x⁴ − x²), which has two stable valleys and an unstable hump between them. Sketch that one yourself as a check.
Common mistakes and how to avoid them
Getting the force direction backwards. Force points downhill on the U graph, so F = −slope.
Treating U = 0 as equilibrium. The zero of potential energy is arbitrary; equilibrium depends on the slope, not the value.
Calling every flat point stable. Check whether it is a valley, a hill or a level stretch.
Saying the particle stops at a hilltop. It stops at turning points, where U = E. It passes a hilltop if E is larger than U there.
Forgetting that K = E − U must not be negative.
A practice routine
Day 1: Sketch U(x) = x³ − 3x by hand, mark both equilibria, and draw energy lines at E = −1, 0 and 3 J. Describe the motion for each.
Day 2: Repeat with U(x) = 2(x⁴ − x²). Find all three equilibria and their stability.
Day 3: Three graph-only questions: identify force direction and turning points without any formula.
Day 4: Link this to springs. Our posts on spring-block energy and energy in SHM graphs show the same rules for U = ½kx².
The rolling-ball picture is easier to trust once you have seen it move. The Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read. If a step in your working goes wrong, the Fermi Problem Library has a step-by-step solver that explains the reasoning.
Tonight, find the speed of the particle in the worked example when it passes x = 0.5 m, and check it is about 2.35 m/s.
