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United StatesPhysics C: MechanicsSyllabus dot point

How is potential energy defined for conservative forces, and how do we move between a force and its potential energy by integration and differentiation?

Topic 3.3 Potential Energy: define potential energy for conservative forces, relate force and potential energy by F=βˆ’dU/dxF = -dU/dx, and use gravitational and elastic potential energy, including the general gravitational form.

A focused answer to AP Physics C: Mechanics Topic 3.3, covering conservative forces and potential energy, the relation F=βˆ’dU/dxF = -dU/dx between force and potential energy, gravitational potential energy near a surface and the general βˆ’GMm/r-GMm/r form, elastic potential energy, and reading equilibrium from a potential-energy curve, with calculus-based worked examples.

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  1. What this topic is asking
  2. Conservative forces and potential energy
  3. Force from potential energy
  4. The standard potential energies
  5. Reading a potential-energy curve
  6. Try this

What this topic is asking

The College Board (Topic 3.3) wants you to define potential energy for conservative forces, to move between a force and its potential energy using F=βˆ’dUdxF = -\dfrac{dU}{dx}, and to apply gravitational and elastic potential energy, including the general gravitational form βˆ’GMm/r-GMm/r. The calculus relationship between force and potential energy, and reading stability from a potential-energy curve, are distinctive AP Physics C expectations.

Conservative forces and potential energy

Gravity and the spring force are conservative, so each has a potential energy; friction and drag are non-conservative (path-dependent, dissipating energy as heat), so no potential energy can be defined for them. The minus sign in Ξ”U=βˆ’W\Delta U = -W encodes that when a conservative force does positive work (gravity pulling an object down), the potential energy decreases, and that energy reappears as kinetic energy. This packaging of conservative work into stored energy is what makes energy conservation so convenient.

Force from potential energy

The relationship between a one-dimensional conservative force and its potential energy is a calculus statement:

F(x)=βˆ’dUdx,U(x)=βˆ’βˆ«F(x) dx.F(x) = -\frac{dU}{dx}, \qquad U(x) = -\int F(x)\,dx.

The force is the negative slope of the potential-energy curve. Where UU decreases with xx (negative slope), the force is positive (pushes toward increasing xx); the force always points toward lower potential energy, "downhill" on the UU curve. This lets you extract the force from a given U(x)U(x) by differentiating, or build UU from a known F(x)F(x) by integrating (choosing a reference point that sets the constant). It is a two-way calculus tool the exam tests directly.

The standard potential energies

Three forms recur. Gravitational PE near a surface, where gg is roughly constant, is U=mghU = mgh (with hh measured from a chosen reference). The general gravitational PE of two masses, with zero at infinite separation, is

U=βˆ’GMmr,U = -\frac{GMm}{r},

which is negative because gravity is attractive; you must do positive work to pull the masses apart. Elastic PE of a spring is U=12kx2U = \tfrac{1}{2}kx^2, the integral of the spring force. Each follows from U=βˆ’βˆ«F dxU = -\int F\,dx applied to the corresponding force law, and each has a freely chosen zero point (only differences in UU are physical).

Reading a potential-energy curve

A U(x)U(x) graph is a compact summary of the motion. The force at any point is the negative slope; the kinetic energy at any point is the difference between the total energy (a horizontal line) and U(x)U(x); turning points are where the energy line meets the curve (where K=0K = 0). Wells in the curve trap the particle in oscillation; barriers it cannot surmount limit its range. This graphical analysis, combining force, energy and stability, is a hallmark AP Physics C skill.

Try this

Q1. A conservative force gives U(x)=5x2U(x) = 5x^2 (SI units). Calculate the force at x=2.0x = 2.0 m. [2 points]

  • Cue. F=βˆ’dU/dx=βˆ’10x=βˆ’10(2.0)=βˆ’20F = -dU/dx = -10x = -10(2.0) = -20 N (toward the origin).

Q2. State the gravitational potential energy of a satellite-Earth system and explain its sign (reference at infinity). [2 points]

  • Cue. U=βˆ’GMm/rU = -GMm/r; it is negative because gravity is attractive and the reference (zero) is set at infinite separation.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)5 marksSection II (FRQ, calculus). A conservative force acting along the xx-axis has the potential energy U(x)=3.0x2βˆ’12xU(x) = 3.0x^2 - 12x (SI units). (a) Derive an expression for the force F(x)F(x). (b) Determine the position of equilibrium. (c) State whether the equilibrium is stable or unstable, with justification. (d) Determine the gravitational potential energy of a 2.02.0 kg mass raised 5.05.0 m near the Earth's surface (g=9.8g = 9.8 m/s squared).
Show worked answer β†’

A 5-point FRQ on the force-potential relationship and equilibrium.

(a) Force (1 point): F(x)=βˆ’dUdx=βˆ’(6.0xβˆ’12)=12βˆ’6.0xF(x) = -\dfrac{dU}{dx} = -(6.0x - 12) = 12 - 6.0x.
(b) Equilibrium (1 point): F=0F = 0 when 12βˆ’6.0x=012 - 6.0x = 0, so x=2.0x = 2.0 m.
(c) Stability (2 points): d2Udx2=6.0>0\dfrac{d^2U}{dx^2} = 6.0 > 0, a minimum of UU, so the equilibrium is stable (the force pushes the particle back toward x=2.0x = 2.0 m).
(d) Gravitational PE (1 point): Ξ”U=mgh=(2.0)(9.8)(5.0)=98\Delta U = mgh = (2.0)(9.8)(5.0) = 98 J.

Markers reward differentiating UU with the minus sign and using the second derivative (or the sign of UU's curvature) to classify the equilibrium.

AP 2021 (style)1 marksSection I (multiple choice). The gravitational potential energy of a two-body system, with zero taken at infinite separation, is U=βˆ’GMmrU = -\dfrac{GMm}{r}. As the two masses move farther apart, the potential energy... (A) becomes more negative (B) increases toward zero (C) stays constant (D) becomes infinite. Justify your reasoning.
Show worked answer β†’

A 1-point conceptual MCQ. The answer is (B).

With the reference at infinity, U=βˆ’GMm/rU = -GMm/r is negative and approaches zero as rr increases. Moving the masses apart makes rr larger, so βˆ’GMm/r-GMm/r becomes less negative, that is, it increases toward zero. This matches the idea that you must do positive work against gravity to separate the masses. The trap (A) gets the sign of the change backward.

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