Forces and Potential Energy

AP Physics C: Mechanics· AP Physics C: Mechanics CED — Work, Energy, and Power· 14 min read

1. 1D Relation Between Force and Potential Energy★★☆☆☆⏱ 4 min

The relationship between conservative force and potential energy comes directly from the definition of potential energy change: the change in potential energy of a system equals the negative work done by the conservative force.

\Delta U = U(x_2) - U(x_1) = - W_c = - \int_{x_1}^{x_2} F_x(x) dx

To find the instantaneous force at position $x$ , take the derivative of both sides using the Fundamental Theorem of Calculus, which gives the core 1D formula.

F_x(x) = - \frac{dU}{dx}

The negative sign has a clear physical meaning: a conservative force always points in the direction of decreasing potential energy. If $d U / d x > 0$ , force points toward decreasing $x$ , which matches this rule.

📐 Worked Example

A particle moving along the x-axis has potential energy given by $U (x) = 2 x^{3} - 9 x^{2} + 12 x + 5$ , 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 between force and potential energy:
$F_x(x) = -\frac{dU}{dx}$
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 the force at $x = 1$ m:
$F_x(1) = -6(1)^2 + 18(1) - 12 = 0 \text{ N}$

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.

2. Equilibrium and Stability Classification★★★☆☆⏱ 4 min

Equilibrium occurs when the net force on a particle is zero. Using the $F_{x} = - d U / d x$ relation, this means the equilibrium condition is $\frac{d U}{d x} = 0$ : the slope of the $U (x)$ graph is zero at any equilibrium point. We classify equilibrium into three types based on the curvature of $U (x)$ at the equilibrium point:

Stable equilibrium: $U (x)$ is at a local minimum, so $\frac{d ^{2} U}{d x ^{2}} > 0$ . Any displacement creates a restoring force pointing back to equilibrium. Intuitively, this is a ball at the bottom of a valley.
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, like a ball at the top of a hill.
Neutral equilibrium: $U (x)$ is flat around the point, so there is no restoring or repelling force for any displacement, like a ball on flat ground.

📐 Worked Example

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

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

3. Conservative Forces in Multiple Dimensions★★★★☆⏱ 3 min

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.

\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 a partial derivative with respect to one coordinate, treat all other coordinates as constants. Physically, the force vector always points in the direction of maximum decrease of potential energy, extending the 1D intuition 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 $U (x, y) = x^{2} y + 3 x - 2 y^{2}$ , 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 each force component is the negative partial derivative of $U$ :
$F_x = -\frac{\partial U}{\partial x}, \quad F_y = -\frac{\partial U}{\partial y}$
2
Calculate the x-component of force:
$\frac{\partial U}{\partial x} = 2xy + 3 \implies F_x = -(2xy + 3) = -(2(1)(2) + 3) = -7 \text{ N}$
3
Calculate the y-component of force:
$\frac{\partial U}{\partial y} = x^2 - 4y \implies F_y = -(x^2 - 4y) = -(1)^2 + 4(2) = 7 \text{ N}$
4
Write the final force vector:
$\vec{F} = -7\hat{i} + 7\hat{j} \text{ N}$

Exam tip:

Don’t forget to differentiate linear terms (like $3 x$ in this example) when calculating partial derivatives. Students often omit these simple terms, leading to incorrect component values.

4. AP-Style Practice Worked Examples★★★★☆⏱ 3 min

📐 Worked Example

A particle moving along the x-axis has a potential energy curve $U (x)$ with slope magnitudes at four points: $∣ slope ∣_{A} = 2$ J/m, $∣ slope ∣_{B} = 5$ J/m, $∣ slope ∣_{C} = 0$ J/m, $∣ slope ∣_{D} = 1$ 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

1
From the core relation $F_{x} = - d U / d x$ , the magnitude of force equals the magnitude of the slope of the $U (x)$ curve: $∣ F ∣ = ∣ d U / d x ∣ = ∣ slope ∣$ .
2
A steeper slope corresponds to a larger force magnitude, regardless of the slope sign or the absolute value of $U$ . The largest slope magnitude is 5 J/m at point B, so this is the correct answer.

📐 Worked Example

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.

1
Part (a): Use the 1D force-potential relation:
$F(x) = -\frac{dU}{dx} = -\frac{d}{dx}\left(\frac{1}{4}k x^4\right) = -k x^3$
2
Substitute values to find force magnitude at $x = 0.5$ m:
$|F(0.5)| = 10(0.5)^3 = 1.25 \text{ N}$
3
Part (b): Use conservation of mechanical energy. Total energy at release equals total energy at $x = 0$ .
4
At release, kinetic energy $K = 0$ , so total energy equals potential energy:
$E = U(0.5) = \frac{1}{4}(10)(0.5)^4 = 0.15625 \text{ J}$
5
At $x = 0$ , $U = 0$ , so all energy is kinetic: $E = \frac{1}{2} m v^{2}$ . Solve for $v$ :
$v = \sqrt{\frac{2E}{m}} = \sqrt{\frac{2(0.15625)}{2}} \approx 0.40 \text{ m/s}$
6
Part (c): $x = 0$ is a local minimum of $U (x)$ . For any $x \neq = 0$ , $U (x) > U (0)$ , so equilibrium is stable.

5. Common Pitfalls

Wrong move:

Forgetting the negative sign in $F_{x} = - d U / d x$ and writing $F_{x} = + d U / d x$ .

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 $d U / d x$ 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 $d U / d x = 0$ 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 $d U / d x$ 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 $F = - d U / d x$ 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. Quick Reference Cheatsheet

Category	Formula	Notes
1D Conservative Force	$F_{x} (x) = - \frac{d U ( x )}{d x}$	Only applies to conservative forces; force points toward decreasing potential energy.
Force Magnitude from $U (x)$ Graph	$∣ F ∣ = ∣ slope of U (x) vs x ∣$	Steeper slope = larger force magnitude, regardless of slope sign.
Equilibrium Condition (1D)	$\frac{d U}{d x} = 0, F_{x} = 0$	Net force is zero at any equilibrium point.
Stability Classification (1D)	Stable: $\frac{d ^{2} U}{d x ^{2}} > 0$ (local min) Unstable: $\frac{d ^{2} U}{d x ^{2}} < 0$ (local max) Neutral: $\frac{d ^{2} U}{d x ^{2}} = 0$ everywhere	Absolute value of U does not affect stability, only curvature.
2D/3D Conservative Force	$F = - \nabla U = - (\frac{\partial U}{\partial x} \hat{i} + \frac{\partial U}{\partial y} \hat{j} + ...)$	Each component is a negative partial derivative; treat other coordinates as constants.
Rotational/Torsional Analogue	$τ (θ) = - \frac{d U ( θ )}{d θ}$	Same relation as linear motion, replace $x \to θ$ and $F \to τ$ .
Conservation of Mechanical Energy	$E_{t o t a l} = K + U = constant$	Only holds when all forces doing work are conservative.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Stability from U(x) graph
2022 · FRQ
Force from 2D potential energy

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 upcoming topics across the AP Physics C: Mechanics course. Mastering the force-potential energy relation is required to derive restoring forces for non-linear simple harmonic motion, analyze the stability of oscillating systems, and 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. Continue building your knowledge with the following topics: