Equilibrium, turning points and allowed motion
You will be able to: Use U, its slope and a total-energy line to identify equilibrium and accessible positions.
What can a potential-energy graph reveal before solving for motion?
A small object near the bottom of an energy valley feels a force back toward the bottom when displaced. At the top of an energy hill it is pushed farther away. The curve’s slope explains both behaviors.
A useful starting point: Read force from a potential-energy slope →
Words and symbols before equations
- Equilibrium
- A position with zero net force; in this model dU/dx = 0.
- Stable / unstable
- A small displacement gives a restoring force / a force farther from equilibrium.
- Total mechanical energy E
- K + U, constant for the isolated conservative model.
- Turning point
- An accessible boundary with K = E − U = 0 where motion can reverse.
What this picture assumes
One-dimensional particle of mass 1 kg in a conservative potential. Valley U = (2 J/m²)x²; hill U = 8 J − (2 J/m²)x². An energy line identifies allowed positions, not a complete trajectory. Outside the shown region the formula still defines the ideal model.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- Stable valley at x = 0. At x = 1 m: U = 2 J; Fₓ = -4 N. K = 6 J; speed 3.464 m/s, direction unspecified. E = U at x = ±2 m.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
At a smooth local minimum, force on either side points toward the minimum: stable equilibrium. At a smooth local maximum, force points away: unstable equilibrium. A flat interval gives neutral equilibrium. A zero slope alone is insufficient; inspect the nearby shape.
Draw a horizontal E line. The vertical gap E − U is kinetic energy, so classical motion is allowed only where U ≤ E. Speed is √[2(E − U)/m]. The graph does not choose whether the object moves left or right.
A simple intersection of E with U at a nonzero slope is a turning point. Equilibrium and turning point are different ideas: equilibrium concerns force, turning concerns speed. At a potential maximum with E exactly equal to that maximum, a particle initially at rest can remain at the unstable equilibrium; this is not an ordinary finite-time crossing.
A worked example, step by step
For U = (2 J/m²)x², a 1 kg particle has E = 8 J. Identify stable equilibrium, the accessible interval and speed at x = 1 m.
- Fₓ = −(4 N/m)x, so the zero-force position x = 0 is a stable minimum.
- Solve E = U: 8 = 2x² in SI units, giving endpoints x = −2 m and +2 m.
- At x = 1 m, U = 2 J and K = 8 − 2 = 6 J.
- Speed is √(2×6/1) = √12 = 3.46 m/s. Direction is unspecified; at ±2 m speed vanishes and nonzero restoring force reverses it.
Zero speed is not zero force. A turning point usually has nonzero force; an equilibrium point need not have zero speed as an object passes it.
At the bottom of a valley, must a moving particle stop?
Compare with an explanation
No. Its force is zero at that instant, but its speed can be greatest there because U is smallest.
Predict. Change one thing. Explain.
Choose the energy valley, then change E and position. Predict turning positions from the intersections. Switch to the hill and compare force directions around x = 0.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Stable valley at x = 0. At x = 1 m: U = 2 J; Fₓ = -4 N. K = 6 J; speed 3.464 m/s, direction unspecified. E = U at x = ±2 m.
One-dimensional particle of mass 1 kg in a conservative potential. Valley U = (2 J/m²)x²; hill U = 8 J − (2 J/m²)x². An energy line identifies allowed positions, not a complete trajectory. Outside the shown region the formula still defines the ideal model.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant work, system boundary, energy or calculus relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA 2 kg particle moves in U = (1 J/m²)x² with E = 4 J. Find turning positions, speed at the center, and explain stability at the center.
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Compare with the answer and four-point rubric
- 1 point: Set U = E: x = ±2 m are turning positions.
- 1 point: At the center U = 0 and K = 4 J.
- 1 point: Speed = √(2×4/2) = 2 m/s.
- 1 point: Fₓ = −(2 N/m)x points toward the center from either side, so equilibrium there is stable.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What does the gap between E and U measure?
Kinetic energy, which cannot be negative.
RECALL 2How do you distinguish stable and unstable equilibrium?
Displace slightly and check whether force restores or pushes farther away.
RECALL 3Can U > E describe an accessible position in this model?
No. It would require negative kinetic energy.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Equilibrium, turning points and allowed motion
- Allowed positions satisfy E ≥ U(x).
- v = √[2(E − U)/m].
- Equilibrium: U′ = 0; a minimum is stable and a maximum unstable.
Remember: Zero speed is not zero force. A turning point usually has nonzero force; an equilibrium point need not have zero speed as an object passes it.
Conditions: One-dimensional particle of mass 1 kg in a conservative potential. Valley U = (2 J/m²)x²; hill U = 8 J − (2 J/m²)x². An energy line identifies allowed positions, not a complete trajectory. Outside the shown region the formula still defines the ideal model.
Refresh Kid · AP Physics C: Mechanics Unit 3 (official Unit 3) · Objectives 3.3.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 3.3, objectives 3.3.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026 alongside the Fall 2026 clarifications. This is Mechanics Unit 3: Work, Energy, and Power. The unit covers Topics 3.1–3.5. Calculus connects work to force integrals, force to potential-energy derivatives, and power to the rate of energy transfer. Models distinguish object-only and multi-object systems; translational models exclude rotational energy unless explicitly noted. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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