Forces and potential energy — AP Physics C: Mechanics Study Guide

For: AP Physics C: Mechanics candidates sitting AP Physics C: Mechanics.

Covers: The relation between conservative force and potential energy for 1D, 2D, and 3D systems, equilibrium identification, stability classification from potential energy functions and graphs, and the gradient formula for multi-dimensional force.

You should already know: Conservative vs non-conservative force classification, basic differentiation (including partial derivatives), the work-energy theorem for conservative forces.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics C: Mechanics style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Forces and potential energy?

Forces and potential energy is the core connection between the energy description of mechanical systems and the force description taught earlier in the course. According to the AP Physics C: Mechanics CED, this topic is part of Unit 3: Work, Energy, and Power, which accounts for 14-20% of the total exam score, with this subtopic making up roughly a third of the unit’s exam weight. This topic appears in both multiple-choice (MCQ) and free-response (FRQ) sections, and is often combined with graph analysis, equilibrium problems, and simple harmonic motion to create multi-concept questions. For any conservative force, a corresponding potential energy function can be defined; the relation between the two lets you find force from potential energy, or vice versa, and analyze the stability of equilibrium positions without solving full force equations. This tool is especially useful for non-linear forces, where writing and solving force equations can be more complex than working with energy functions.

2. 1D Relation Between Force and Potential Energy

The relation between conservative force and potential energy comes directly from the definition of potential energy: the change in potential energy of a system is equal to the negative of the work done by the conservative force. Mathematically, this is written as: $$\Delta U=U(x_2)-U(x_1)=-W_c=-{\int }_{x_1}^{x_2}{F}_x(x)dx$$ To get the instantaneous force at a position $x$, we take the derivative of both sides with respect to $x$, which eliminates the integral by the Fundamental Theorem of Calculus. This gives the core formula for 1D motion: $$F_x(x)=-\frac{dU}{dx}$$ The negative sign has physical meaning: a conservative force always points in the direction of decreasing potential energy. If $U$ increases as $x$ increases ($dU/dx>0$), the force points to decreasing $x$, which matches this rule. If $U$ decreases as $x$ increases, the force points to increasing $x$, again consistent with the negative sign.

Worked Example

A particle moving along the x-axis has potential energy given by $U(x)=2x^3-9x^2+12x+5$, where $U$ is in joules and $x$ is in meters. What is the force on the particle at $x=1$ m?

1. Start with the core relation: $F_x(x)=-\frac{dU}{dx}$.
2. Compute the first derivative of $U(x)$: $\frac{dU}{dx}=6x^2-18x+12$.
3. Substitute back into the force formula: $F_x(x)=-(6x^2-18x+12)=-6x^2+18x-12$.
4. Evaluate at $x=1$: $F_x(1)=-6(1{)}^2+18(1)-12=0$ N.

Exam tip: AP MCQs almost always include a distractor option with the correct magnitude but opposite sign. Always confirm your sign matches the rule that force points toward lower potential energy before selecting your answer.

3. Equilibrium and Stability from Potential Energy

Equilibrium occurs when the net force on a particle is zero. Using the force-potential energy relation, this means $F_x=0=-\frac{dU}{dx}$, so the condition for equilibrium is simply $\frac{dU}{dx}=0$: the slope of the $U(x)$ graph is zero at equilibrium. We can classify equilibrium into three types based on the curvature of $U(x)$ at the equilibrium point:

1. Stable equilibrium: $U(x)$ is at a local minimum, so $\frac{d^{2}U}{d{x}^2}>0$. Any displacement from equilibrium creates a force that points back toward equilibrium, so the particle returns to rest at the equilibrium point. The intuitive analogy is a ball at the bottom of a valley.
2. Unstable equilibrium: $U(x)$ is at a local maximum, so $\frac{d^{2}U}{d{x}^2}<0$. Any displacement creates a force that pushes the particle further away from equilibrium. The analogy is a ball at the top of a hill.
3. Neutral equilibrium: $U(x)$ is flat ($\frac{dU}{dx}=0$ for all $x$ around the point), so there is no force for any displacement. The analogy is a ball on flat ground.

Worked Example

For the potential energy function $U(x)=2x^3-9x^2+12x+5$, classify the equilibrium points at $x=1$ m and $x=2$ m.

1. We already found $\frac{dU}{dx}=6x^2-18x+12$, which confirms both points are equilibrium (roots of $\frac{dU}{dx}=0$).
2. Compute the second derivative to test curvature: $\frac{d^{2}U}{d{x}^2}=12x-18$.
3. Evaluate at $x=1$: $\frac{d^{2}U}{d{x}^2}=12(1)-18=-6<0$. This is a local maximum, so equilibrium is unstable.
4. Evaluate at $x=2$: $\frac{d^{2}U}{d{x}^2}=12(2)-18=6>0$. This is a local minimum, so equilibrium is stable.

Exam tip: For MCQ questions that give you a graph of $U(x)$ (not an algebraic function), use the ball-on-a-hill rule to classify stability instantly, no calculation required.

4. Conservative Forces in Multiple Dimensions

For motion in 2 or 3 dimensions, the force-potential energy relation extends using the gradient operator. The force vector is the negative gradient of the potential energy function, written as: $$\vec{F}(x,y,z)=-\nabla U=-\left(\frac{\partial U}{\partial x}\hat{i}+\frac{\partial U}{\partial y}\hat{j}+\frac{\partial U}{\partial z}\hat{k}\right)$$ Each component of the force is the negative partial derivative of $U$ with respect to that coordinate. When taking the partial derivative with respect to one coordinate, all other coordinates are treated as constants, which makes calculation straightforward. Physically, the negative gradient means the force vector always points in the direction of maximum decrease of potential energy, which matches our 1D intuition extended to multiple dimensions. AP Physics C almost exclusively tests 2D cases for this topic.

Worked Example

A charged particle moving in the xy-plane has electric potential energy $U(x,y)=x^{2}y+3x-2y^2$, where $U$ is in joules and $x,y$ are in meters. What is the force vector on the particle at the point $(x=1,y=2)$?

1. Recall that each force component is the negative partial derivative of $U$: $F_x=-\frac{\partial U}{\partial x}$, $F_y=-\frac{\partial U}{\partial y}$.
2. Compute $F_x$: $\frac{\partial U}{\partial x}=2xy+3$, so $F_x=-(2xy+3)$. Substitute $(1,2)$: $F_x=-(2(1)(2)+3)=-7$ N.
3. Compute $F_y$: $\frac{\partial U}{\partial y}=x^2-4y$, so $F_y=-(x^2-4y)=-x^2+4y$. Substitute $(1,2)$: $F_y=-(1{)}^2+4(2)=7$ N.
4. The final force vector is $\vec{F}=-7\hat{i}+7\hat{j}$ N.

Exam tip: Don’t forget to differentiate linear terms in $U$ (like $3x$ in this example) when calculating partial derivatives. Students often leave out these simple terms, leading to wrong component values.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Forgetting the negative sign in $F_x=-dU/dx$ and writing $F_x=+dU/dx$. Why: Students memorize the relation as "force is the derivative of potential energy" and omit the sign derived from the work-energy relation for conservative forces. Correct move: Always write the full formula with the negative sign at the start of every problem, and verify your sign matches the rule that force points toward lower potential energy.
• Wrong move: Classifying equilibrium based on the sign of the first derivative dU/dx instead of the second. Why: Students confuse the condition for equilibrium (first derivative zero) with the condition for stability (second derivative sign). Correct move: First confirm dU/dx = 0 to confirm equilibrium, then always use the sign of the second derivative (or graph curvature) to classify stability.
• Wrong move: In 2D problems, taking a full derivative dU/dx instead of a partial derivative when finding the x-component of force. Why: Students forget that U depends on multiple variables, so the derivative with respect to x only accounts for variation in x, holding y constant. Correct move: Always explicitly write partial derivatives $\partial U/\partial x$ for each force component when working in multiple dimensions.
• Wrong move: Claiming a point is unstable equilibrium because $U(x)$ is negative there. Why: Students confuse the value of U at equilibrium with its curvature, since potential energy can have any zero reference point. Correct move: Ignore the absolute value of U when classifying stability; only the curvature around the equilibrium point matters, regardless of what U is at that point.
• Wrong move: Using the relation $F=-dU/dx$ for non-conservative forces like friction or air resistance. Why: Students forget that potential energy is only defined for conservative forces. Correct move: Before using any force-potential energy relation, confirm the force is conservative — if it's non-conservative, no potential energy exists, so the relation does not apply.

6. Practice Questions (AP Physics C: Mechanics Style)

Question 1 (Multiple Choice)

A particle moving along the x-axis has a potential energy curve $U(x)$ with slope magnitudes at four points: $|\text{slope}{|}_A=2$ J/m, $|\text{slope}{|}_B=5$ J/m, $|\text{slope}{|}_C=0$ J/m, $|\text{slope}{|}_D=1$ J/m. Total mechanical energy is constant, and the particle moves freely across the entire region. At which point is the magnitude of the force on the particle greatest? A) Point A B) Point B C) Point C D) Point D

Worked Solution: From the core relation $F_x=-dU/dx$, the magnitude of force is equal to the magnitude of the slope of the $U(x)$ curve, $|F|=|dU/dx|=|\text{slope}|$. A steeper slope corresponds to a larger force magnitude, regardless of the sign of the slope or the value of $U$. The largest slope magnitude is 5 J/m at point B, so the force magnitude is largest here. The zero slope at point C corresponds to zero force, which matches the equilibrium condition. Correct answer: B

Question 2 (Free Response)

A block of mass $m=2$ kg attached to a nonlinear spring has potential energy given by $U(x)=\frac{1}{4}k{x}^4$, where $k=10$ N/m³ and $x$ is displacement from the origin (equilibrium at $x=0$). (a) Derive an expression for the force $F(x)$ exerted by the spring, and calculate the magnitude of the force at $x=0.5$ m. (b) The block is released from rest at $x=0.5$ m. Assuming no non-conservative forces, find the speed of the block when it passes through $x=0$. (c) Classify the equilibrium at $x=0$ as stable, unstable, or neutral, and justify your answer.

Worked Solution: (a) Use $F(x)=-dU/dx=-\frac{d}{dx}\left(\frac{1}{4}k{x}^4\right)=-k{x}^3$. Substituting values: $|F(0.5)|=10(0.5{)}^3=10(0.125)=1.25$ N. (b) By conservation of mechanical energy, total energy at release equals total energy at $x=0$. At release, $K=0$, so $E=U(0.5)=\frac{1}{4}(10)(0.5{)}^4=0.15625$ J. At $x=0$, $U=0$, so $E=\frac{1}{2}m{v}^2$. Solving for $v$: $v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(0.15625)}{2}}=\sqrt{0.15625}\approx 0.40$ m/s. (c) Equilibrium at $x=0$ is stable. For any displacement $x\ne 0$, $U(x)=\frac{1}{4}k{x}^4>0=U(0)$, so $x=0$ is a local minimum of potential energy. Any displacement creates a restoring force pointing back to $x=0$, so equilibrium is stable.

Question 3 (Application / Real-World Style)

A digital micromirror device (used in most consumer projectors) has each mirror connected to a base via a small torsional spring. The torsional potential energy of the mirror as a function of rotation angle $\theta$ is $U(\theta )=-k\cos \theta$, where $k=2.5\times 10^{-10}$ J and $\theta \in (-\pi /2,\pi /2)$. Find the equilibrium angle in this interval, classify its stability, and calculate the magnitude of the torque on the mirror when $\theta =\pi /4$ (use the relation for rotation: $\tau =-dU/d\theta$, analogous to linear force).

Worked Solution: Find equilibrium by setting $dU/d\theta =0$: $dU/d\theta =k\sin \theta =0$, so $\theta =0$ is the only equilibrium in the interval. Classify stability with the second derivative: $d^{2}U/d{\theta }^2=k\cos \theta$, so at $\theta =0$, $d^{2}U/d{\theta }^2=k(1)>0$, so $\theta =0$ is a stable equilibrium. Calculate torque magnitude: $|\tau |=|-dU/d\theta |=|k\sin \theta |$. Substituting values: $|\tau |=(2.5\times 10^{-10})\sin (\pi /4)\approx 1.8\times 10^{-10}$ N·m. This small torque acts as a restoring force that returns the mirror to its neutral stable position after it is rotated by an external electrostatic force to switch between on and off states.

7. Quick Reference Cheatsheet

8. What's Next

This topic connects the energy framework of Unit 3 to the force analysis you learned in Unit 2, and is a non-negotiable prerequisite for the next topics in this unit and the rest of the course. Without mastering the force-potential energy relation, you will not be able to derive restoring forces for non-linear simple harmonic motion, analyze the stability of oscillating systems, or derive gravitational force from gravitational potential energy in orbital mechanics — all common multi-concept FRQ topics on the AP exam. This topic also forms the foundation for energy-based approaches to dynamics that you will use in college-level mechanics. Next topics to study after this: Conservation of Energy Power Simple Harmonic Motion Gravitational Potential Energy