Conservative Forces and Potential Energy

Conservative vs Non-Conservative Forces

Part 1 of 7 — Conservative Forces & Energy

The distinction between conservative and non-conservative forces is one of the most powerful ideas in physics. It determines when we can use energy conservation — a major simplification.

Definition

A force is conservative if and only if:

1. The work done by the force is independent of the path taken between two points.
2. Equivalently: the work done around any closed loop is zero.
3. Equivalently: the force can be written as the negative slope of a potential energy function.

$F_x = -\frac{dU}{dx} \quad \text{(one dimension)}$

$\oint \vec{F} \cdot d\vec{r} = 0 \quad \text{(closed path)}$

Examples

Conservative	Non-Conservative
Gravity ($\vec{F} = -mg\hat{y}$)	Kinetic friction ($f_k = \mu_k N$)
Spring force ($F = -kx$)	Air resistance ($f = bv$)
Any force with a $U(x)$, $F = -dU/dx$	Tension (when string does net work)
Gravitational ($F = GMm/r^2$)	Applied/push forces (generally)

Why Friction Is Non-Conservative

Consider a block sliding from $A$ to $B$ on a rough surface.

Direct path (length $d_1$): $W_1 = -\mu_k mg \cdot d_1$

Longer path via point $C$ (total length $d_2 > d_1$): $W_2 = -\mu_k mg \cdot d_2$

Since $d_2 > d_1$: $|W_2| > |W_1|$. The work depends on the path — friction is non-conservative.

Closed loop test: $\oint \vec{f}_k \cdot d\vec{r} = -\mu_k mg \cdot (\text{total distance}) \neq 0$

Friction always does negative work (opposes motion), so the total around a closed loop is always negative, never zero.

Why Gravity Is Conservative

For gravity, compare two paths from height $h_A$ to height $h_B$:

$W_{\text{grav}} = -mg(h_B - h_A) = -mg\Delta h$

This depends only on the height difference, not the path. Whether you go straight down, zigzag, or take a winding slope — the work is the same.

Formal Proof (1D)

$W = \int_{y_A}^{y_B} (-mg)\,dy = -mg(y_B - y_A)$

This depends only on the endpoints $y_A$ and $y_B$.

Work-Energy Theorem with Both Types

The total work equals the change in kinetic energy:

$W_{\text{total}} = \Delta K$

Splitting into conservative ($W_c$) and non-conservative ($W_{nc}$) work:

$W_c + W_{nc} = \Delta K$

Since $W_c = -\Delta U$ (definition of potential energy):

$-\Delta U + W_{nc} = \Delta K$

$\boxed{W_{nc} = \Delta K + \Delta U = \Delta E_{\text{mech}}}$

Interpretation

• If only conservative forces act ($W_{nc} = 0$): $\Delta E_{\text{mech}} = 0$ — mechanical energy is conserved.
• If non-conservative forces do work: mechanical energy changes by exactly $W_{nc}$.
• Friction: $W_{nc} = -f_k d < 0$, so mechanical energy decreases (converted to thermal energy).

Worked Example

A 5 kg block slides 4 m down a $30°$ rough incline ($\mu_k = 0.25$), starting from rest.

$W_{nc} = -\mu_k mg\cos\theta \cdot d = -0.25(50)(\cos 30°)(4) = -43.3 \text{ J}$

$\Delta K + \Delta U = W_{nc}$ $\frac{1}{2}(5)v^2 + (-5 \times 10 \times 4\sin 30°) = -43.3$ $2.5v^2 - 100 = -43.3$ $v^2 = 22.7 \implies v = 4.76 \text{ m/s}$

Part 1 Summary

Concept	Key Idea
Conservative force	Path-independent work; $\oint \vec{F}\cdot d\vec{r} = 0$
Non-conservative	Path-dependent; $\oint \vec{F}\cdot d\vec{r} \neq 0$
Potential energy	Exists only for conservative forces: $F = -dU/dx$
Energy theorem	$W_{nc} = \Delta K + \Delta U = \Delta E_{\text{mech}}$

Next up: Part 2 — Potential Energy Functions, deriving $U(x)$ for various forces.

Potential Energy Functions

Part 2 of 7 — Conservative Forces & Energy

For any conservative force, we can define a potential energy function $U(x)$ such that the force is the negative derivative of $U$.

Defining Potential Energy

$U(x) - U(x_0) = -\int_{x_0}^{x} F(x')\,dx'$

Or equivalently: $\Delta U = -W_{\text{conservative}}$

The reference point $x_0$ (where $U = 0$) is arbitrary — only differences in $U$ are physically meaningful.

Common Potential Energy Functions

1. Gravity (Near Earth's Surface)

$F = -mg \quad \text{(downward)}$ $U(y) = -\int_0^y (-mg)\,dy' = mgy$

$\boxed{U_{\text{grav}} = mgy}$

2. Spring (Hooke's Law)

$F = -kx$ $U(x) = -\int_0^x (-kx')\,dx' = \frac{1}{2}kx^2$

$\boxed{U_{\text{spring}} = \frac{1}{2}kx^2}$

3. Universal Gravitation

$F = -\frac{GMm}{r^2} \quad \text{(toward center, attractive)}$ $U(r) = -\int_\infty^r \left(-\frac{GMm}{r'^2}\right)dr' = -\frac{GMm}{r}$

$\boxed{U_{\text{grav}} = -\frac{GMm}{r}}$

(Reference: $U = 0$ at $r = \infty$)

Deriving $U$ from $F$: Step-by-Step

General Procedure

Given $F(x)$, find $U(x)$:

1. Write $U(x) = -\int_{x_0}^x F(x')\,dx'$
2. Choose a convenient reference point $x_0$ where $U(x_0) = 0$
3. Evaluate the integral
4. Check: $F = -dU/dx$ should recover the original force

Example: Quartic Force

$F(x) = -6x^2 + 4x$

$U(x) = -\int_0^x (-6x'^2 + 4x')\,dx' = -[-2x'^3 + 2x'^2]_0^x$ $U(x) = 2x^3 - 2x^2$

Check: $-\frac{dU}{dx} = -(6x^2 - 4x) = -6x^2 + 4x$ ✓

Example: Repulsive Inverse-Square Force

$F(r) = \frac{A}{r^2} \quad (A > 0\text{, pointing outward, so repulsive})$

With $U(\infty) = 0$:

$U(r) = -\int_\infty^r \frac{A}{r'^2}\,dr' = -\left[-\frac{A}{r'}\right]_\infty^r = \frac{A}{r}$

$U$ is positive and decreases as $r$ grows, so the force pushes the objects apart, toward lower $U$. Compare gravity, $F = -GMm/r^2$, which is attractive and gives the negative $U = -GMm/r$.

Custom Potential Energy Functions

On the AP exam, you may encounter a given $U(x)$ and be asked to analyze the system.

Example: $U(x) = ax^2 - bx^4$

Force: $F(x) = -\frac{dU}{dx} = -2ax + 4bx^3$

Equilibrium points ($F = 0$): $-2ax + 4bx^3 = 0$ $x(-2a + 4bx^2) = 0$ $x = 0 \quad \text{or} \quad x = \pm\sqrt{\frac{a}{2b}}$

Stability (check $d^2U/dx^2$): $\frac{d^2U}{dx^2} = 2a - 12bx^2$

At $x = 0$: $U'' = 2a > 0$ → stable (local minimum of $U$)

At $x = \pm\sqrt{a/(2b)}$: $U'' = 2a - 12b \cdot \frac{a}{2b} = 2a - 6a = -4a < 0$ → unstable (local maximum of $U$)

Key Insight

• Stable equilibrium: $U$ has a local minimum ($U'' > 0$)
• Unstable equilibrium: $U$ has a local maximum ($U'' < 0$)

We'll explore this further in Part 4 (Energy Diagrams).

Part 2 Summary

Force	Potential Energy	Reference
Gravity (near surface)	$U = mgy$	$U = 0$ at $y = 0$
Spring	$U = \frac{1}{2}kx^2$	$U = 0$ at $x = 0$
Universal gravitation	$U = -GMm/r$	$U = 0$ at $r = \infty$
General $F(x)$	$U = -\int F\,dx$	Choose $x_0$

Key Formula: $F(x) = -\frac{dU}{dx}$ (force is the negative slope of $U$)

Next up: Part 3 — $F = -dU/dx$, deep-diving into extracting forces from potential energy graphs and functions.

$F = -dU/dx$

Part 3 of 7 — Conservative Forces & Energy

The relationship $F = -dU/dx$ is arguably the most important equation in AP Physics C energy problems. It connects the force to the slope of the potential energy function.

Graphical Interpretation

Given a graph of $U(x)$:

Feature of $U(x)$	Meaning for $F$
$U$ decreasing (negative slope)	$F > 0$ (force in $+x$)
$U$ increasing (positive slope)	$F < 0$ (force in $-x$)
$U$ at a minimum	$F = 0$ (stable equilibrium)
$U$ at a maximum	$F = 0$ (unstable equilibrium)
Steep slope	Large force
Flat slope	Small force

The Force Points "Downhill" on the $U$ Curve

The negative sign means the force always pushes objects toward lower potential energy. This is a universal principle:

$\text{Force} = -\text{(slope of } U\text{)}$

Worked Examples

Example 1: Harmonic Oscillator

$U(x) = \frac{1}{2}kx^2$

$F = -\frac{dU}{dx} = -kx$

This is Hooke's law — a restoring force proportional to displacement.

Example 2: Lennard-Jones Potential

A model for the force between two neutral molecules. An exam would give you a function like this; you are not expected to know it, only to work with it:

$U(r) = \epsilon\left[\left(\frac{r_0}{r}\right)^{12} - 2\left(\frac{r_0}{r}\right)^6\right]$

$F(r) = -\frac{dU}{dr} = \epsilon\left[\frac{12r_0^{12}}{r^{13}} - \frac{12r_0^6}{r^7}\right]$

Equilibrium: $F = 0$ at $r = r_0$. Check: $\frac{r_0^{12}}{r_0^{13}} = \frac{r_0^6}{r_0^7}$ ✓

Stability: for $r < r_0$ the $r^{-13}$ term wins and $F > 0$ (pushed outward, back toward $r_0$); for $r > r_0$, $F < 0$ (pulled inward, back toward $r_0$). Either small displacement is opposed, so $r = r_0$ is a stable equilibrium, the minimum of $U$.

Example 3: Gravitational Potential

$U(r) = -\frac{GMm}{r}$

$F(r) = -\frac{dU}{dr} = -\frac{GMm}{r^2}$

This is the radial force component, and it points inward (attractive). The negative sign means the force points toward decreasing $r$ (toward the center), which is correct for gravity.

Example 4: Piecewise Potential

$U(x) = \begin{cases} \frac{1}{2}k_1 x^2 & x < 0 \\ \frac{1}{2}k_2 x^2 & x \geq 0 \end{cases}$

$F(x) = \begin{cases} -k_1 x & x < 0 \\ -k_2 x & x \geq 0 \end{cases}$

This describes an asymmetric spring — stiffer on one side than the other.

Equilibrium, Stability and Strongest Force

Classifying an Equilibrium

At an equilibrium $x_0$, $dU/dx = 0$. To decide whether it is stable, use either test:

• Force test: displace slightly to each side. If the force points back toward $x_0$ on both sides, the equilibrium is stable; if it points away, unstable.
• Shape test: a local minimum of $U$ ($d^2U/dx^2 > 0$) is stable; a local maximum ($d^2U/dx^2 < 0$) is unstable.

Where Is the Force Strongest?

Because $|F| = |dU/dx|$, the force is largest where the $U(x)$ curve is steepest. Between a minimum and a maximum, that is where the slope stops increasing and starts decreasing (an inflection point, where $d^2U/dx^2 = 0$).

Worked Example

For $x > 0$, a particle has

$U(x) = U_0\left(\frac{a^2}{x^2} - \frac{a}{x}\right)$

Force: $F = -\dfrac{dU}{dx} = U_0\left(\dfrac{2a^2}{x^3} - \dfrac{a}{x^2}\right)$

Equilibrium: $F = 0 \implies 2a^2 = ax \implies x_0 = 2a$

Force test:

• At $x = a$: $F = U_0\left(\dfrac{2}{a} - \dfrac{1}{a}\right) = +\dfrac{U_0}{a}$, pushing toward larger $x$, back toward $2a$.
• At $x = 4a$: $F = U_0\left(\dfrac{2}{64a} - \dfrac{1}{16a}\right) = -\dfrac{U_0}{32a}$, pulling back toward $2a$.

So $x_0 = 2a$ is stable, and $U(2a) = U_0\left(\frac{1}{4} - \frac{1}{2}\right) = -\frac{U_0}{4}$ is the bottom of the well.

Bound or free? As $x \to \infty$, $U \to 0$. A particle with total energy between $-U_0/4$ and $0$ is trapped between two turning points; with $E \geq 0$ it can escape to large $x$.

Part 3 Summary

Concept	Formula
Force from $U$	$F = -dU/dx$
Force = negative slope of $U$	Points toward lower $U$
Equilibrium	$F = 0 \iff dU/dx = 0$
Stable equilibrium	Minimum of $U$; force points back toward $x_0$
Unstable equilibrium	Maximum of $U$; force points away from $x_0$
Strongest force	Where the $U(x)$ curve is steepest

Next up: Part 4 — Energy Diagrams and Equilibrium, using graphs of $U(x)$ to understand motion qualitatively.

Energy Diagrams and Equilibrium

Part 4 of 7 — Conservative Forces & Energy

Energy diagrams are one of the most powerful visual tools in physics. By plotting $U(x)$, we can determine equilibrium positions, stability, turning points, and qualitative motion — all without solving differential equations.

Reading an Energy Diagram

Given a plot of $U(x)$ and a total mechanical energy $E$:

$E = K + U = \frac{1}{2}mv^2 + U(x)$

Since $K = \frac{1}{2}mv^2 \geq 0$:

$K(x) = E - U(x) \geq 0$

$\boxed{U(x) \leq E}$

Turning Points

The object can only exist where $U(x) \leq E$. Points where $U(x) = E$ are turning points — the object momentarily stops ($v = 0$) and reverses direction.

Forbidden Regions

Where $U(x) > E$, the kinetic energy would be negative — this is classically forbidden. The object cannot reach these regions.

Speed at Any Point

$v(x) = \sqrt{\frac{2(E - U(x))}{m}}$

Maximum speed occurs where $U(x)$ is minimum.

Types of Equilibrium

At any point where $F = 0$ (equivalently $dU/dx = 0$), we have equilibrium. The type depends on the curvature:

Stable Equilibrium ($U'' > 0$, local minimum)

If displaced slightly, the force is restoring — the object oscillates around the equilibrium.

Think: a ball at the bottom of a bowl.

Unstable Equilibrium ($U'' < 0$, local maximum)

If displaced slightly, the force pushes the object away from equilibrium.

Think: a ball balanced on top of a hill.

Neutral Equilibrium ($U'' = 0$, flat)

If displaced slightly, there is no restoring force — the object stays in its new position.

Think: a ball on a flat table.

Summary Table

Type	$U$ shape	$U''$	If displaced...
Stable	Valley/minimum	$> 0$	Returns (oscillates)
Unstable	Hill/maximum	$< 0$	Runs away
Neutral	Flat	$= 0$	Stays put

Bounded vs Unbounded Motion

By examining where $E$ intersects $U(x)$, we can classify the motion:

Bounded Motion (Trapped)

If the object is between two turning points with $U > E$ on both sides, the motion is bounded — the object oscillates back and forth.

Example: A mass on a spring with $U = \frac{1}{2}kx^2$. For any $E > 0$, the turning points are at $x = \pm\sqrt{2E/k}$.

Unbounded Motion (Escaping)

If $U(x) < E$ extends to infinity in at least one direction, the object can escape — it never returns.

Example: A particle near $U(r) = -GMm/r$. If $E > 0$, the particle escapes to infinity (hyperbolic orbit). If $E < 0$, the particle is bound.

Escape Energy

The minimum energy for a particle to escape from a potential well of depth $U_0$ is:

$E_{\text{escape}} = 0 \quad \text{(if } U(\infty) = 0\text{)}$

This means the particle needs $K \geq |U|$ to escape.

Worked Example

For $U(x) = U_0\left(\frac{x_0^2}{x^2} - \frac{2x_0}{x}\right)$ with total energy $E$:

Setting $U(x) = E$ gives the turning points. The motion is bounded if $E < 0$ (since $U(\infty) = 0$).

The minimum of $U$ is at $x = x_0$: $U(x_0) = U_0(1 - 2) = -U_0$.

So bounded motion occurs for $-U_0 < E < 0$.

Part 4 Summary

Concept	How to Read from $U(x)$ Diagram
Kinetic energy	Gap between $E$ and $U(x)$
Turning points	Where $U(x) = E$
Forbidden regions	Where $U(x) > E$
Max speed	Where $U(x)$ is minimum
Force magnitude	Slope of $U(x)$
Force direction	Toward decreasing $U$
Stable equilibrium	Local minimum of $U$
Unstable equilibrium	Local maximum of $U$
Bounded motion	Trapped between two turning points

Next up: Part 5 — Path Independence and Work, proving path independence rigorously with line integrals.

Path Independence and Work

Part 5 of 7 — Conservative Forces & Energy

Path independence is the defining property of conservative forces. In this part, we prove it and explore its implications using the work integral along a path.

Line Integrals and Work

The work done by a force $\vec{F}$ along a path $C$ from $A$ to $B$:

$W = \int_C \vec{F} \cdot d\vec{r} = \int_C (F_x\,dx + F_y\,dy + F_z\,dz)$

For a conservative force, $\vec{F}\cdot d\vec{r} = -dU$ along any small step of the path, so:

$W = -\int_C dU = -(U_B - U_A) = U_A - U_B = -\Delta U$

This depends only on the values of $U$ at the endpoints — not the path.

Proof: Path Independence ↔ $\oint \vec{F} \cdot d\vec{r} = 0$

Forward Direction

Suppose the work is path-independent. Take two paths $C_1$ and $C_2$ from $A$ to $B$:

$\int_{C_1} \vec{F}\cdot d\vec{r} = \int_{C_2} \vec{F}\cdot d\vec{r}$

Now form a closed loop: go from $A$ to $B$ via $C_1$, then back from $B$ to $A$ via $-C_2$:

$\oint \vec{F}\cdot d\vec{r} = \int_{C_1} \vec{F}\cdot d\vec{r} + \int_{-C_2} \vec{F}\cdot d\vec{r} = \int_{C_1} - \int_{C_2} = 0$

Reverse Direction

If $\oint \vec{F}\cdot d\vec{r} = 0$ for any loop, then for any two paths $C_1$, $C_2$ from $A$ to $B$:

$\int_{C_1} \vec{F}\cdot d\vec{r} - \int_{C_2} \vec{F}\cdot d\vec{r} = \oint \vec{F}\cdot d\vec{r} = 0$

So $\int_{C_1} = \int_{C_2}$ — path-independent. ∎

Work by a Constant Force: The Dot Product

For a constant force, the work integral reduces to a dot product with the displacement:

$W = \vec{F}\cdot\Delta\vec{r} = F_x\,\Delta x + F_y\,\Delta y = F\,\Delta r\cos\theta$

Only the endpoints enter, so every constant force is conservative. Gravity near Earth's surface ($\vec{F} = -mg\hat{y}$) is the classic example: $W = -mg\,\Delta y$ along any path.

Contrast: Friction on Two Paths

A box is dragged at constant speed from $A$ to $B$, which are $4.0$ m apart, with a kinetic friction force of $5.0$ N.

Path	Length	Work by friction
Straight line	$4.0$ m	$-(5.0)(4.0) = -20$ J
Curved detour	$6.0$ m	$-(5.0)(6.0) = -30$ J

Same endpoints, different work: friction is non-conservative, so no potential energy can be defined for it.

Work from an $F$–$x$ Graph

For a force that varies with position in one dimension, the work is the area under the $F_x$ versus $x$ graph (areas below the axis count as negative work). For a spring, the triangle under $F = kx$ from $0$ to $x$ has area $\tfrac{1}{2}kx^2$, which is the work needed to stretch it, matching $U_{\text{spring}}$.

Applications of Path Independence

Work Done by Gravity on Any Path

A 5 kg object moves from $(0, 0, 10)$ to $(3, 4, 2)$ along an arbitrary path. Work done by gravity:

$W = -mg\Delta y = -5(10)(2 - 10) = 400 \text{ J}$

We don't need to know the path! Gravity is conservative, so the work depends only on the height change $\Delta h = -8$ m.

Work-Energy with Multiple Conservative Forces

If the system has spring PE and gravitational PE:

$E = K + U_{\text{grav}} + U_{\text{spring}} = \text{const}$

$\frac{1}{2}mv^2 + mgy + \frac{1}{2}kx^2 = \text{const}$

Worked Example: Spring + Gravity

A 2 kg block is released from rest at the natural length of a vertical spring ($k = 200$ N/m). How far does it drop?

Let $y$ be the compression (downward positive):

$\frac{1}{2}mv_0^2 + mgy_0 + \frac{1}{2}ky_0^2 = \frac{1}{2}mv^2 + mgy + \frac{1}{2}ky^2$

At max compression ($v = 0$, start and end):

$0 + 0 + 0 = 0 - mg y_{\max} + \frac{1}{2}ky_{\max}^2$

$mg y_{\max} = \frac{1}{2}ky_{\max}^2$

$y_{\max} = \frac{2mg}{k} = \frac{2(2)(10)}{200} = 0.2 \text{ m}$

Note: this is twice the equilibrium compression ($mg/k = 0.1$ m) — the block overshoots and oscillates.

Part 5 Summary

Concept	Key Result
Path independence	$W = U_A - U_B = -\Delta U$ for conservative forces
Closed-loop test	$\oint \vec{F}\cdot d\vec{r} = 0$ ↔ conservative
Constant force	$W = \vec{F}\cdot\Delta\vec{r}$ (path-independent)
Friction	Work grows with path length (non-conservative)
$F$–$x$ graph	Work $=$ area under the curve
Energy conservation	$K + U = \text{const}$ (when only conservative forces act)

Next up: Part 6 — Problem-Solving Workshop with AP-style energy and force problems.

Problem-Solving Workshop

Part 6 of 7 — Conservative Forces & Energy

Problem-Solving Strategy for Energy Problems

Step	Action
1	Identify all forces; classify as conservative or non-conservative
2	If only conservative forces: use $E = K + U = \text{const}$
3	If non-conservative forces present: $W_{nc} = \Delta K + \Delta U$
4	Choose a reference point for potential energy
5	Write energy equation at two relevant positions
6	Solve for the unknown

Problem 2: Spring-Launched Block on a Ramp

A spring ($k = 800$ N/m) is compressed by $x_0 = 0.3$ m and launches a 2 kg block up a $30°$ frictionless incline.

How far up the incline does the block travel?

Energy conservation (spring PE → gravitational PE):

$\frac{1}{2}kx_0^2 = mgd\sin\theta$

$d = \frac{kx_0^2}{2mg\sin\theta} = \frac{800(0.09)}{2(2)(10)(0.5)} = \frac{72}{20} = 3.6 \text{ m}$

Now add friction ($\mu_k = 0.2$):

$\frac{1}{2}kx_0^2 = mgd\sin\theta + \mu_k mgd\cos\theta$

$d = \frac{kx_0^2}{2mg(\sin\theta + \mu_k\cos\theta)} = \frac{72}{2(2)(10)(0.5 + 0.173)} = \frac{72}{26.93} = 2.67 \text{ m}$

Speed at a specific point

At $d = 1$ m up the frictionless ramp:

$\frac{1}{2}kx_0^2 = \frac{1}{2}mv^2 + mgd\sin\theta$

$v = \sqrt{\frac{kx_0^2}{m} - 2gd\sin\theta} = \sqrt{\frac{72}{2} - 2(10)(1)(0.5)} = \sqrt{36 - 10} = \sqrt{26} = 5.10 \text{ m/s}$

Problem 3: Reading a $U(x)$ Graph

Given $U(x) = 5x^2 - x^4$ (in SI units) for a 1 kg particle:

Equilibrium points: $F = -\frac{dU}{dx} = -(10x - 4x^3) = 0$ $x(10 - 4x^2) = 0$ $x = 0, \quad x = \pm\sqrt{2.5} \approx \pm 1.58$

Stability: $U''(x) = 10 - 12x^2$

• At $x = 0$: $U'' = 10 > 0$ → stable (minimum)
• At $x = \pm\sqrt{2.5}$: $U'' = 10 - 30 = -20 < 0$ → unstable (maxima)

Values at equilibria:

• $U(0) = 0$ (minimum)
• $U(\pm\sqrt{2.5}) = 5(2.5) - (2.5)^2 = 12.5 - 6.25 = 6.25$ J (maxima)

For a particle with $E = 4$ J, starting at $x = 0$:

• It oscillates in the potential well around $x = 0$
• Turning points: $5x^2 - x^4 = 4$. Let $u = x^2$: $u^2 - 5u + 4 = 0$, so $u = 1$ or $u = 4$. Inside the well ($|x| < 1.58$) the root is $u = 1$, so the turning points are $x = \pm 1$ m
• Cannot escape since $E = 4 < 6.25$ (the barrier height)

For $E = 7$ J:

• The particle has enough energy to overcome the barriers at $x = \pm 1.58$
• Motion is unbounded — the particle escapes

Speed at the bottom of the well (for $E = 4$ J): at $x = 0$, $U = 0$, so $K = 4$ J and $v = \sqrt{2K/m} = \sqrt{8} \approx 2.8$ m/s. The particle oscillates back and forth between $x = -1$ m and $x = +1$ m, but the motion is not simple harmonic, because the force $-(10x - 4x^3)$ is not proportional to $x$.

Workshop Summary

Problem Categories

Type	Method
"Find the speed at point X"	Energy conservation: $K_0 + U_0 = K + U + W_{nc}$
"Find the force from $U(x)$"	$F = -dU/dx$
"Find equilibria and stability"	Set $dU/dx = 0$; check sign of $d^2U/dx^2$
"Is the motion bounded?"	Compare $E$ to the barrier heights in $U(x)$
"Is the motion simple harmonic?"	Only if $F = -kx$, i.e. $U$ has the form $\frac{1}{2}kx^2$; then $T = 2\pi\sqrt{m/k}$
"Is the force conservative?"	Check whether its work around a closed path is zero

Next up: Part 7 — Review & Applications, tying everything together.

Review & Applications

Part 7 of 7 — Conservative Forces & Energy

Complete Topic Reference

Concept	Formula	Part
Conservative force	Path-independent work	1
Non-conservative work	$W_{nc} = \Delta K + \Delta U$	1
Gravity PE	$U = mgy$	2
Spring PE	$U = \frac{1}{2}kx^2$	2
Gravitational PE	$U = -GMm/r$	2
Force from PE	$F = -dU/dx$	3
Stable equilibrium	$U'' > 0$ (local min)	3, 4
Turning points	$U(x) = E$	4
Closed loop	$\oint \vec{F}\cdot d\vec{r} = 0$	5

Application: Escape Velocity

The minimum launch speed for an object to escape a planet's gravity (starting from the surface):

$\frac{1}{2}mv_{\text{esc}}^2 + \left(-\frac{GMm}{R}\right) = 0 + 0$

$v_{\text{esc}} = \sqrt{\frac{2GM}{R}}$

For Earth: $v_{\text{esc}} = \sqrt{\frac{2(6.67 \times 10^{-11})(5.97 \times 10^{24})}{6.37 \times 10^6}} \approx 11.2$ km/s

Key Observations:

1. Escape velocity is independent of the object's mass (mass cancels)
2. It depends only on the planet's mass $M$ and radius $R$
3. This is a direct application of energy conservation with $U = -GMm/r$
4. If launched with $v < v_{\text{esc}}$, the object reaches a maximum height and returns (bound orbit)
5. If $v = v_{\text{esc}}$: the object barely escapes ($v \to 0$ as $r \to \infty$)
6. If $v > v_{\text{esc}}$: the object escapes with kinetic energy remaining

Application: Bungee Jump (Gravity + Spring)

A $60$ kg jumper steps off a bridge with a bungee cord of natural length $20$ m that acts like a spring with $k = 300$ N/m once taut ($g = 10$ m/s²). Air resistance is negligible.

Lowest point. Let $x$ be the maximum stretch. The jumper starts and ends at rest, so all the gravitational energy lost goes into the cord:

$mg(20 + x) = \tfrac{1}{2}kx^2 \implies 600(20 + x) = 150x^2 \implies x^2 - 4x - 80 = 0$

$x = \frac{4 + \sqrt{16 + 320}}{2} \approx 11.2 \text{ m}$

so the jumper falls about $31.2$ m in total.

Maximum speed. Speed is greatest where the net force is zero, at a stretch of $mg/k = 2.0$ m (22 m below the bridge):

$\tfrac{1}{2}mv^2 = mg(22) - \tfrac{1}{2}k(2.0)^2 \implies v^2 = 440 - 20 = 420 \implies v \approx 20.5 \text{ m/s}$

Both answers come from $K + U_{\text{grav}} + U_{\text{spring}} = \text{constant}$, without ever solving for the motion in time.

🎉 Topic Complete: Conservative Forces & Energy

You've mastered the full AP Physics C treatment of conservative forces:

Part	Topic	Status
1	Conservative vs non-conservative forces	✅
2	Potential energy functions	✅
3	$F = -dU/dx$	✅
4	Energy diagrams and equilibrium	✅
5	Path independence and work	✅
6	Problem-solving workshop	✅
7	Review & applications	✅

Key Takeaway: The concept of conservative forces enables energy conservation — the most powerful problem-solving tool in mechanics. On the AP exam, master three things: (1) deriving $U$ from $F$ and vice versa, (2) reading energy diagrams for equilibrium and turning points, and (3) applying $W_{nc} = \Delta E_{\text{mech}}$ when non-conservative forces are present.