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Conservative Forces and Potential Energy

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Conservative Forces and Potential Energy - Complete Interactive Lesson

Part 1: Conservative vs Non-Conservative

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:

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

$\vec{F} = -\nabla U = -\frac{dU}{dx}\hat{x} \quad \text{(1D)}$

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

Examples

Conservative	Non-Conservative
Gravity ( $\vec{F} = -mg\hat{y}$ )

Why Friction Is Non-Conservative

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

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

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 () work:

Part 1 Summary

Concept	Key Idea
Conservative force	Path-independent work; $\oint \vec{F}\cdot d\vec{r} = 0$

Part 2: Potential Energy Functions

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

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

Part 3: F = −dU/dx

$F = -dU/dx$

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 :

Part 4: Energy Diagrams & Equilibrium

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 $E$ :

$E = K + U = \frac{1}{2} m$

Part 5: Path Independence & Work

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 rigorously and explore its implications using line integrals.

Line Integrals and Work

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

Part 6: Problem-Solving Workshop

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: $W_{n}$

Part 7: Review & Applications

Review & Applications

Part 7 of 7 — Conservative Forces & Energy

Complete Topic Reference

Concept	Formula	Part
Conservative force	Path-independent work	1
Non-conservative work	$W_{nc} = \Delta K + \Delta U$

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.

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

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

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

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

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

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

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).

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

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

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

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

−∫x0xF(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 $U(x)$ :

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

Example: Quartic Force

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

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

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

Example: Inverse-Square Force (Electrostatic)

$F(r) = \frac{kq_1q_2}{r^2} \quad \text{(positive = repulsive)}$

With $U(\infty) = 0$ :

$U(r) = -\int_\infty^r \frac{kq_1q_2}{r'^2}\,dr' = -\left[-\frac{kq_1q_2}{r'}\right]_\infty^r = \frac{kq_1q_2}{r}$

For like charges ( $q_1 q_2 > 0$ ): $U > 0$ — energy decreases as charges separate. For unlike charges (): — energy decreases as charges attract.

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 ( $F = 0$ ): $-2ax + 4bx^3 = 0$ $x (- 2 a + 4 b^{}$

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 $x = \pm\sqrt{a/(2b)}$ : $U^{''} = 2 a - 12 b \cdot \frac{a}{2 b} =$ → (local maximum of )

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
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.

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 intermolecular forces:

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

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

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

Stability: $U''(r_0) > 0$ (minimum of $U$ ) → stable equilibrium at $r = r_{0}$ .

Example 3: Gravitational Potential

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

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

But wait — this is the radial force, pointing 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.

Higher-Order Analysis

Finding Force Extrema

The force has maximum magnitude where $dF/dx = 0$ , or equivalently $d^2U/dx^2 = 0$ (inflection point of $U$ ).

Taylor Expansion Near Equilibrium

Near a stable equilibrium at $x_0$ (where $U'(x_0) = 0$ ):

$U(x) \approx U(x_0) + \frac{1}{2}U''(x_0)(x - x_0)^2 + \frac{1}{6}U'''(x_0)(x - x_0)^3 + \cdots$

The leading term gives: $F \approx -U''(x_0)(x - x_0)$

This is a harmonic restoring force with effective spring constant:

$k_{\text{eff}} = U''(x_0) = \frac{d^2U}{dx^2}\bigg|_{x_0}$

Angular Frequency of Small Oscillations

$\omega = \sqrt{\frac{k_{\text{eff}}}{m}} = \sqrt{\frac{U''(x_0)}{m}}$

Worked Example

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

Equilibrium: $F = -U' = 0$ → $U_0\left(-\frac{2a^2}{x^3} + \frac{a}{x^2}\right) = 0$ →

Small oscillation frequency: $U''(x) = U_0\left(\frac{6a^2}{x^4} - \frac{2a}{x^3}\right)$

$\omega = \sqrt{\frac{U_0}{8ma^2}}$

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	$d^2U/dx^2 > 0$ (minimum of $U$ )
Effective spring constant	$k_{\text{eff}} = U''(x_0)$
Small oscillation frequency	$\omega = \sqrt{U''(x_0)/m}$

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

E = K + U = \frac{1}{2} m v^{2} + U (x)

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

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

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

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.

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

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

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

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 :

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.

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

For a conservative force $\vec{F} = -\nabla U$ :

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

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

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

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 $-C_2$ :

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

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. ∎

The Curl Test for Conservative Forces

In 2D, a force $\vec{F} = F_x(x,y)\hat{x} + F_y(x,y)\hat{y}$ is conservative if and only if:

$\frac{\partial F_x}{\partial y} = \frac{\partial F_y}{\partial x}$

In 3D, this generalizes to $\nabla \times \vec{F} = \vec{0}$ (curl is zero).

Examples

$\vec{F} = (2xy)\hat{x} + (x^2)\hat{y}$ : , . Equal → .

$U(x,y) = -x^2 y + C$

$\vec{F} = (y)\hat{x} + (-x)\hat{y}$ : , . Not equal → .

This force circulates — it does net work around closed loops.

Finding $U$ from $\vec{F}$ in 2D

Given conservative $\vec{F} = F_x\hat{x} + F_y\hat{y}$ :

$U = -\int F_x\,dx + g(y)$ (treat $y$ as constant)

Example: $\vec{F} = (2xy + 3)\hat{x} + (x^2 - 2y)\hat{y}$

Step 1: $U = -\int(2xy+3)dx = -x^2 y - 3x + g(y)$

Step 2: $\partial U/\partial y = -x^2 + g'(y) = -F_y = -(x^2 - 2y) = -x^2 + 2y$

Step 3: $g'(y) = 2y \implies g(y) = y^2$

$U(x,y) = -x^2 y - 3x + y^2$

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
Curl test (2D)	$\partial F_x/\partial y = \partial F_y/\partial x$
Finding $U$ from $\vec{F}$	Integrate $F_x$ w.r.t. , then match
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.

W_{n c} = Δ K + Δ U

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: $F = -\frac{dU}{dx} = -(10x - 4x^3) = 0$

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

At $x = 0$ : $U'' = 10 > 0$ → stable (minimum)
At $x = \pm\sqrt{2.5}$ : → (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: $5x^2 - x^4 = 4$ → solve for

For $E = 7$ J:

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

Small oscillation frequency about $x = 0$ : $\omega = \sqrt{\frac{U''(0)}{m}} = \sqrt{\frac{10}{1}} = \sqrt{10} \approx 3.16 \text{ rad/s}$

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 $U(x)$
"Period of small oscillations"	$T = 2\pi\sqrt{m/k_{\text{eff}}}$ with
"Is the force conservative?"	Check $\partial F_x/\partial y = \partial F_y/\partial x$

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

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:

Escape velocity is independent of the object's mass (mass cancels)
It depends only on the planet's mass $M$ and radius $R$
This is a direct application of energy conservation with $U = -GMm/r$
If launched with $v < v_{\text{esc}}$ , the object reaches a maximum height and returns (bound orbit)

Application: Molecular Potential Energy

The Morse potential models the bond between atoms:

$U(r) = D_e\left(1 - e^{-a(r - r_e)}\right)^2 - D_e$

where $D_e$ is the bond dissociation energy, $r_e$ is the equilibrium bond length, and $a$ controls the width.

Equilibrium: $U'(r_e) = 0$ . $U(r_e) = -D_e$ (minimum).

Force near equilibrium: $F = -U'(r) = -2D_e a\left(1 - e^{-a(r-r_e)}\right)e^{-a(r-r_e)}$

Effective spring constant: $k_{\text{eff}} = U''(r_e) = 2D_e a^2$

Small oscillation frequency: $\omega = \sqrt{\frac{2D_e a^2}{\mu}}$

where $\mu$ is the reduced mass.

Energy levels:

Bound states: $E < 0$ (particle oscillates between turning points)
Dissociation: $E \geq 0$ (molecule breaks apart)
Total binding energy: $D_e$ (depth of the well)

This directly connects conservative force theory to chemistry and quantum mechanics.

🎉 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.

U(x) = 2x^3 - 2x^2

(positive = repulsive)

−[−r′kq1q2]∞r=

x (- 2 a + 4 b x^{2}) = 0

x = 0 \quad \text{or} \quad x = \pm\sqrt{\frac{a}{2b}}

U^{''} = 2 a - 12 b \cdot \frac{a}{2 b} = 2 a - 6 a = - 4 a < 0

ϵ[r1312r012−r712r06]

61U′′′(x0)(x−

U''(2a) = U_0\left(\frac{6a^2}{16a^4} - \frac{2a}{8a^3}\right) = U_0\left(\frac{6}{16a^2} - \frac{2}{8a^2}\right) = \frac{U_0}{8a^2}

\partial F_x/\partial y = 2x

\partial F_y/\partial x = 2x

\partial F_x/\partial y = 1

\partial F_y/\partial x = -1

non-conservative

Take

\partial U/\partial y

and set equal to

-F_y

2(2)(10)(0.5)800(0.09)=

2(2)(10)(0.5+0.173)72=

272−2(10)(1)(0.5)=

x = 0, \quad x = \pm\sqrt{2.5} \approx \pm 1.58

U'' = 10 - 30 = -20 < 0

Cannot escape since

E = 4 < 6.25

(the barrier height)

k_{\text{eff}} = U''(x_0)

2(6.67×10−11)(5.97×1024)

v = v_{\text{esc}}

: the object barely escapes (

v \to 0

r \to \infty

)

v > v_{\text{esc}}

: the object escapes with kinetic energy remaining

Conservative Forces and Potential Energy

Conservative Forces and Potential Energy - Complete Interactive Lesson

Part 1: Conservative vs Non-Conservative

Conservative vs Non-Conservative Forces

Definition

Examples

Why Friction Is Non-Conservative

Work-Energy Theorem with Both Types

Part 1 Summary

Part 2: Potential Energy Functions

Potential Energy Functions

Defining Potential Energy

Part 3: F = −dU/dx

F=−dU/dxF = -dU/dxF=−dU/dx

Graphical Interpretation

Part 4: Energy Diagrams & Equilibrium

Energy Diagrams and Equilibrium

Reading an Energy Diagram

Part 5: Path Independence & Work

Path Independence and Work

Line Integrals and Work

Part 6: Problem-Solving Workshop

Problem-Solving Workshop

Problem-Solving Strategy for Energy Problems

Part 7: Review & Applications

Review & Applications

Complete Topic Reference

Why Gravity Is Conservative

Formal Proof (1D)

Interpretation

Worked Example

Common Potential Energy Functions

1. Gravity (Near Earth's Surface)

2. Spring (Hooke's Law)

3. Universal Gravitation

Deriving UUU from FFF: Step-by-Step

General Procedure

Example: Quartic Force

Example: Inverse-Square Force (Electrostatic)

Custom Potential Energy Functions

Example: U(x)=ax2−bx4U(x) = ax^2 - bx^4U(x)=ax2−bx4

Key Insight

Part 2 Summary

The Force Points "Downhill" on the UUU Curve

Worked Examples

Example 1: Harmonic Oscillator

Example 2: Lennard-Jones Potential

Example 3: Gravitational Potential

Example 4: Piecewise Potential

Higher-Order Analysis

Finding Force Extrema

Taylor Expansion Near Equilibrium

Angular Frequency of Small Oscillations

Worked Example

Part 3 Summary

Turning Points

Forbidden Regions

Speed at Any Point

Types of Equilibrium

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

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

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

Summary Table

Bounded vs Unbounded Motion

Bounded Motion (Trapped)

Unbounded Motion (Escaping)

Escape Energy

Worked Example

Part 4 Summary

Proof: Path Independence ↔ ∮F⃗⋅dr⃗=0\oint \vec{F} \cdot d\vec{r} = 0∮F⋅dr=0

Forward Direction

Reverse Direction

The Curl Test for Conservative Forces

Examples

Finding UUU from F⃗\vec{F}F in 2D

Applications of Path Independence

Work Done by Gravity on Any Path

Work-Energy with Multiple Conservative Forces

Worked Example: Spring + Gravity

Part 5 Summary

$F = -dU/dx$

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

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

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

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

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

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

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

Finding $U$ from $\vec{F}$ in 2D

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

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