Potential Energy and Motion: A Student’s Guide to U(x) = 3(x−1) − (x−3)³

Introduction

Imagine you’re sitting in your physics exam, and suddenly there’s a question about a particle moving under a strange-looking potential energy function: U(x) = 3(x−1) − (x−3)³. Your heart races. Does the particle speed up or slow down? Where does it find equilibrium?

This isn’t just another formula to memorize. Understanding how potential energy shapes motion is the secret to predicting everything from a ball rolling down a hill to electrons moving in circuits. Once you grasp how energy and position connect to create acceleration, physics transforms from confusing to crystal clear.

Let’s break down this specific potential energy function and discover exactly how particles behave when governed by these forces.

What is Potential Energy in Physics?

Potential energy represents stored energy based on an object’s position. Think of it as energy waiting to become motion.

Key points about potential energy:

• It depends only on position, not speed
• Higher potential energy means more stored energy at that location
• Particles naturally move toward lower potential energy
• The symbol U(x) means potential energy as a function of position x

When you hold a ball above the ground, it has gravitational potential energy. Release it, and that stored energy converts to motion. The same principle applies to our particle with U(x) = 3(x−1) − (x−3)³.

Understanding the Function U(x) = 3(x−1) − (x−3)³

Let’s decode this specific potential energy function step by step.

The function has two parts:

• Linear term: 3(x−1) = 3x − 3
• Cubic term: −(x−3)³

Combined form: U(x) = 3x − 3 − (x−3)³

This creates a unique energy landscape where the particle experiences different forces at different positions. The cubic term makes it more interesting than simple linear or quadratic potentials you might have seen before.

What makes this function special:

• It’s not symmetric
• It has varying steepness at different positions
• The cubic term dominates at larger distances from x = 3

How Potential Energy Relates to Force and Acceleration

Here’s the fundamental connection every physics student must understand:

Force = negative gradient of potential energy

F(x) = −dU/dx

This means the particle feels a force pushing it from high potential energy toward low potential energy. The steeper the potential energy curve, the stronger the force.

Then acceleration follows from Newton’s Second Law:

F = ma, so a = F/m = −(1/m)(dU/dx)

Think of it like a valley:

• Ball on a hilltop (high U): forces push it down
• Ball in a valley (low U): forces push it back if displaced
• Steep hill: strong force, high acceleration
• Gentle slope: weak force, low acceleration

Finding Force from Potential Energy

Let’s calculate the force for our specific function.

• Given: U(x) = 3(x−1) − (x−3)³
• Step 1: Expand the cubic term (x−3)³ = x³ − 9x² + 27x − 27
• Step 2: Write complete U(x) U(x) = 3x − 3 − x³ + 9x² − 27x + 27 U(x) = −x³ + 9x² − 24x + 24
• Step 3: Take the derivative dU/dx = −3x² + 18x − 24
• Step 4: Apply force formula F(x) = −dU/dx = 3x² − 18x + 24
• Factored form: F(x) = 3(x² − 6x + 8) = 3(x − 2)(x − 4)

This tells us exactly how much force acts on the particle at every position.

Analyzing Particle Motion and Equilibrium Points

Equilibrium points occur where F(x) = 0

Setting our force equation to zero: 3(x − 2)(x − 4) = 0

Equilibrium positions:

• x = 2
• x = 4

But are these stable or unstable?

To determine stability, check the second derivative of U(x): d²U/dx² = −6x + 18

At x = 2: d²U/dx² = −6(2) + 18 = 6 > 0 → Stable equilibrium (minimum of U)

At x = 4: d²U/dx² = −6(4) + 18 = −6 < 0 → Unstable equilibrium (maximum of U)

What this means for the particle:

• Place it at x = 2: it stays there (like a ball in a valley)
• Place it at x = 4: it rolls away (like a ball on a hilltop)
• Between x = 2 and x = 4: force pushes toward x = 2
• Beyond x = 4: force pushes away from x = 4

Position vs Acceleration: What Changes Where?

Acceleration depends on position through the force:

a(x) = F(x)/m = (3/m)(x − 2)(x − 4)

Let’s map out acceleration at different positions:

Key insight: The particle experiences maximum acceleration at positions furthest from equilibrium points.

Real Classroom Examples

Example 1: Finding acceleration at x = 3

Given: U(x) = 3(x−1) − (x−3)³, mass m = 2 kg

Force at x = 3: F(3) = 3(3 − 2)(3 − 4) = 3(1)(−1) = −3 N

Acceleration: a = F/m = −3/2 = −1.5 m/s²

The negative sign means acceleration points left (toward x = 2).

Example 2: Determining if particle escapes from x = 2

If released from rest at x = 2, the particle stays there (stable equilibrium).

If given small velocity at x = 2, it oscillates back and forth around x = 2, like a marble in a bowl.

To escape past x = 4, it needs enough initial kinetic energy to overcome the potential barrier.

Common Mistakes Students Make

Mistake 1: Confusing force and potential energy

• Force is the negative derivative of U, not U itself
• High U doesn’t mean high force

Mistake 2: Sign errors in acceleration direction

• Negative acceleration means leftward, not necessarily slowing down
• Direction depends on coordinate system

Mistake 3: Forgetting the negative sign in F = −dU/dx

• This negative sign is crucial
• It ensures particles move toward lower potential energy

Mistake 4: Assuming all equilibria are stable

• Check second derivative to determine stability
• Positive d²U/dx² = stable (minimum)
• Negative d²U/dx² = unstable (maximum)

Mistake 5: Mixing up position-dependent vs time-dependent quantities

• U(x) and F(x) depend on where the particle is
• Velocity v(t) and position as a function of time x(t) depend on when

Quick Tips to Remember

Memory aid for force direction: “Particles roll downhill in potential energy”

To find equilibrium points:

1. Calculate F = −dU/dx
2. Set F = 0 and solve for x
3. Check d²U/dx² to determine stability

Acceleration shortcuts:

• Steep potential slope = large acceleration
• Flat potential = zero acceleration
• Potential minimum = stable equilibrium

Visual thinking: Draw U(x) as a landscape. The particle is a ball rolling on this landscape, always seeking the lowest valley.

Exam strategy: Always write F = −dU/dx first, then differentiate carefully. Check your signs twice.

Frequently Asked Questions about Potential Energy and Acceleration

Q. What is the relationship between potential energy and acceleration?

Acceleration equals negative gradient of potential energy divided by mass: a = −(1/m)(dU/dx). Steeper potential energy slopes create larger accelerations. The particle accelerates toward regions of lower potential energy, converting stored energy into kinetic energy.

Q. What is the potential energy of a particle at position?

Potential energy at position U(x) represents stored energy the particle has due to its location in a force field. For our function, U(x) = 3(x−1) − (x−3)³. Each position x has a specific potential energy value determined by this function.

Q. What is the relationship between position x of a particle, acceleration and time?

Position x changes with time according to x(t), determined by integrating velocity. Acceleration a also changes with time as a(t) because it depends on position: a = a(x(t)). The relationship connects through Newton’s laws: acceleration causes velocity changes which cause position changes.

Q. What is U in elastic potential energy?

U represents potential energy, typically measured in joules. In elastic systems like springs, U = ½kx² where k is spring constant and x is displacement. The symbol U is standard notation for any potential energy function regardless of the physical system.

Q. How do you determine if equilibrium is stable or unstable?

Calculate the second derivative of potential energy d²U/dx². If positive at equilibrium, the point is stable (energy minimum). If negative, it’s unstable (energy maximum). If zero, further analysis is needed. Stable equilibria trap particles; unstable ones repel them.

Q. Can potential energy be negative?

Yes, potential energy can be negative. The absolute value doesn’t matter physically; only differences in potential energy between positions matter. Negative U simply means the reference point (where U = 0) was chosen at a higher energy level than the current position.

Q. Why does force equal negative gradient of potential energy?

Forces arise from potential energy gradients because nature minimizes potential energy. The negative sign ensures particles accelerate from high potential toward low potential regions. This fundamental principle applies to gravity, springs, electric fields, and all conservative forces in physics.

Q. How does particle motion differ at stable vs unstable equilibrium?

At stable equilibrium, small displacements create restoring forces that return the particle to equilibrium, causing oscillations. At unstable equilibrium, small displacements create forces that push the particle further away, causing it to accelerate away from that position permanently.

Conclusion

Understanding how potential energy U(x) = 3(x−1) − (x−3)³ governs particle motion reveals the beautiful connection between position, force, and acceleration. The key takeaway is simple: take the negative derivative to find force, then divide by mass for acceleration.

You’ve learned that equilibrium points occur where force vanishes, but stability depends on whether it’s an energy minimum or maximum. Position determines acceleration through the potential energy landscape, with particles naturally moving toward lower energy states.

This knowledge extends far beyond one practice problem. These principles explain planetary orbits, molecular bonds, electrical circuits, and countless other phenomena. Every time you see a potential energy function, you now know exactly how to predict the motion it creates.

Master this concept, and you’ll find physics exams become less about memorizing formulas and more about understanding how energy shapes the world around you. Keep practicing with different potential energy functions, and soon this will become second nature.