7.5 Energy Diagrams and Qualitative Motion Analysis
The Whole Story in One Curve
Here is a potential energy curve for a marble rolling along a bumpy track. The horizontal axis is position. The vertical axis is potential energy $U(x)$. The curve has hills, valleys, and a few flat spots.
[Video: A smooth potential energy curve $U(x)$ is drawn, featuring two wells (valleys) separated by a hill (local maximum), with the curve rising steeply on both ends. A horizontal dashed line marks a total energy $E$. A marble is placed on the curve and released. It rolls back and forth inside one of the wells, speeding up at the bottom and slowing near the edges, turning around exactly where the curve meets the energy line.]
Without solving a single equation of motion, you can read everything off this graph: where the marble speeds up, where it slows down, where it turns around, and where it is trapped. You can identify the positions where the marble could sit in perfect balance --- and distinguish the ones where it would stay put from the ones where the slightest nudge sends it flying.
All of this from one curve and one horizontal line.
This section is about learning to read that story. It is the most visual, most diagram-heavy section in the chapter, and it ties together everything you have built so far --- energy conservation from Section 7.4, the relationship between force and potential energy, and the calculus of slopes and extrema.
Prediction
Before you read on: Here is a potential energy curve $U(x)$ with two valleys and a hill between them. A horizontal dashed line marks the total mechanical energy $E$.
[Image: A $U(x)$ curve with a deep well on the left (minimum at point B), a local maximum at point C, and a shallower well on the right (minimum at point D). The curve rises steeply for large $|x|$. A horizontal dashed line at energy level $E$ intersects the curve at four points: $x_1$, $x_2$, $x_3$, and $x_4$, from left to right. The energy line sits above the bottom of both wells but below the peak at C.]
(a) In which regions of $x$ can the particle actually exist? Where is it forbidden?
(b) If the particle starts near point B (the left well), can it reach point D (the right well)?
(c) Where exactly does the particle turn around?
Commit to your answers. Draw on the diagram if you can. Then continue.
The Guiding Question
How can a graph of energy reveal the entire story of motion before any equation is solved?
In Section 7.3, the work-energy theorem gave you a scalar equation connecting forces, displacement, and speed. In Section 7.4, energy conservation compressed that further: $KE + U = \text{constant}$. That equation contains everything you need to know about the speed at every point. But writing it as an equation obscures something that a picture makes immediately obvious: the shape of the potential energy function tells you the qualitative behavior of the motion at a glance.
Energy diagrams are the tool that makes this visible.
Exploration: Reading Motion from a Curve
[Interactive: Energy Diagram Explorer. A smooth potential energy curve $U(x)$ is displayed, featuring two wells and a central barrier. A horizontal line represents the total mechanical energy $E$, and students can drag it up and down. A marble moves along the curve according to the energy diagram: it oscillates within allowed regions, speeds up where the gap between $E$ and $U(x)$ is large, slows down where the gap narrows, and turns around where the line meets the curve. The kinetic energy $KE = E - U(x)$ is shown as a shaded region between the energy line and the curve. A velocity indicator on the marble changes size with speed.
Guided prompts appear below the interactive:
Prompt 1: Set the energy line low, so it sits inside the left well but below the central barrier. Watch the marble. Where does it go? Where does it turn around?
Prompt 2: Slowly raise the energy line. What happens to the turning points? How does the allowed region change?
Prompt 3: Raise the energy until it just barely reaches the top of the central barrier. What happens to the marble's motion now?
Prompt 4: Raise the energy above the barrier. Can the marble now reach the right well? What has changed?
Prompt 5: Set the energy very high --- above all features of the curve. Is the marble trapped anywhere? Where is it fastest?]
Spend real time with this. The interactive is the centerpiece of this section. Here is what you should notice as you explore:
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When $E$ is low, the marble is trapped in one well. It bounces back and forth between two turning points. It cannot escape because it would need more kinetic energy than it has to climb over the barrier.
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As you raise $E$, the turning points move apart. The marble's allowed region grows. It moves faster at the bottom of the well (where the gap between $E$ and $U$ is largest) and slower near the turning points (where the gap shrinks to zero).
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When $E$ exactly equals the height of the central barrier, something dramatic happens. The marble can just barely reach the top of the barrier. At that point, its kinetic energy is zero --- it arrives with zero speed. It sits there, balanced but unstable.
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When $E$ exceeds the barrier height, the marble is no longer confined to one well. It can cross the barrier and explore both wells --- or escape entirely if $E$ is high enough.
Pause and think: At the exact moment the energy line crosses the top of the barrier, the marble's behavior changes qualitatively --- from trapped to free. What physical quantity determines whether the marble is trapped or free?
Concept Reveal: The Energy Diagram Toolkit
Here is the complete framework for reading motion from a potential energy diagram. Every piece follows from one equation:
$$KE = E_{\text{total}} - U(x)$$
Since kinetic energy can never be negative ($KE = \frac{1}{2}mv^2 \geq 0$), this single relationship constrains everything.
1. Allowed and Forbidden Regions
The particle can only exist where $U(x) \leq E_{\text{total}}$. Wherever $U(x) > E_{\text{total}}$, the kinetic energy would have to be negative, which is physically impossible. These are forbidden regions --- the particle cannot reach them with the given total energy.
On the diagram, the allowed regions are where the $U(x)$ curve lies below the energy line. The forbidden regions are where the curve lies above the energy line.
2. Turning Points
At positions where $U(x) = E_{\text{total}}$, the kinetic energy is exactly zero: $KE = 0$, so $v = 0$. The particle momentarily stops and reverses direction. These are turning points.
On the diagram, turning points are where the $U(x)$ curve intersects the energy line.
3. Speed and Kinetic Energy
The particle's speed at any allowed position is determined by the vertical gap between the energy line and the $U(x)$ curve:
$$v(x) = \sqrt{\frac{2[E_{\text{total}} - U(x)]}{m}}$$
Where the gap is large, the particle moves fast. Where the gap is small, the particle moves slowly. At turning points, the gap closes and the speed is zero.
4. Equilibrium Points
At positions where the potential energy curve has a horizontal tangent ($dU/dx = 0$), the force on the particle is zero:
$$F = -\frac{dU}{dx} = 0$$
These are equilibrium points --- positions where a particle placed at rest would remain at rest. But not all equilibria are created equal.
Stable equilibrium occurs at a minimum of $U(x)$. If the particle is displaced slightly, it experiences a restoring force that pushes it back. The curve is concave up ($d^2U/dx^2 > 0$). A marble placed at the bottom of a valley rolls back if nudged.
Unstable equilibrium occurs at a maximum of $U(x)$. If the particle is displaced slightly, it experiences a force that pushes it further away. The curve is concave down ($d^2U/dx^2 < 0$). A marble placed at the top of a hill rolls away if nudged.
There is also neutral equilibrium, which occurs where $U(x)$ is flat over a region (constant $U$). The particle feels no force anywhere in that region and stays wherever it is placed.
5. Force from the Curve
The force at any point is the negative slope of the potential energy curve:
$$F(x) = -\frac{dU}{dx}$$
Where the curve slopes upward (to the right), the force points to the left (negative). Where the curve slopes downward (to the right), the force points to the right (positive). The steeper the slope, the larger the force. This is the connection between energy diagrams and force: the curve's geometry encodes the force everywhere.
Summary Table
| Feature on $U(x)$ diagram | Physical meaning |
|---|---|
| $U(x) < E$ | Allowed region (particle can be here) |
| $U(x) > E$ | Forbidden region (particle cannot reach) |
| $U(x) = E$ | Turning point ($v = 0$, particle reverses) |
| Large gap $E - U(x)$ | High speed |
| Small gap $E - U(x)$ | Low speed |
| Minimum of $U(x)$ | Stable equilibrium |
| Maximum of $U(x)$ | Unstable equilibrium |
| Slope of $U(x)$ | $F = -dU/dx$ (force is negative of slope) |
Returning to the Prediction
Go back to the prediction diagram. The energy line $E$ sits above both well bottoms but below the central peak at C.
(a) Allowed regions. The particle can exist in two separate regions: from $x_1$ to $x_2$ (the left well) and from $x_3$ to $x_4$ (the right well). Between $x_2$ and $x_3$, the curve rises above the energy line, creating a forbidden region. The particle cannot be there.
(b) Can the particle reach the right well? No --- not with this energy. If the particle starts in the left well, it is trapped between $x_1$ and $x_2$. The barrier at C is higher than $E$, so the particle does not have enough energy to cross. It would need $E \geq U(C)$ to escape.
(c) Turning points. The particle turns around at $x_1$ and $x_2$ (if in the left well) or at $x_3$ and $x_4$ (if in the right well). These are the positions where $U(x) = E$ and the speed drops to zero.
If your predictions matched, you already know how to read energy diagrams. If they did not, go back to the interactive and set the energy to match this scenario. Watch the marble and connect what you see to the analysis above.
Connection: Energy, Force, and Calculus
This section sits at the intersection of three ideas you have already built.
Energy conservation (Section 7.4): The equation $KE + U = E_{\text{total}}$ is the engine behind every energy diagram. The entire graphical framework is just a visual representation of this equation.
Force and potential energy: The relationship $F = -dU/dx$ connects the energy diagram to forces. The slope of the $U(x)$ curve at any point tells you the force there. A steep downward slope means a strong force pushing the particle in the positive $x$ direction. A gentle upward slope means a weak force pushing in the negative $x$ direction.
Calculus of extrema: Stable equilibria are at minima of $U(x)$, unstable equilibria are at maxima. You identify these using the first derivative test ($dU/dx = 0$) and the second derivative test ($d^2U/dx^2 > 0$ for minima, $< 0$ for maxima). The calculus you learned for finding extrema of functions now has direct physical meaning: minima are where objects naturally settle, and maxima are where they balance precariously.
Notice the chain of reasoning: the shape of the $U(x)$ curve tells you the slope, which tells you the force, which tells you the acceleration, which determines the motion. The energy diagram compresses this entire chain into a single picture. That is its power.
A Narrative Example: The Molecular Bond
Energy diagrams are not just textbook exercises. They describe real physical systems. One of the most important is the potential energy of a diatomic molecule --- two atoms bonded together.
[Video: The Lennard-Jones potential is drawn: $U(r)$ rises steeply for small $r$ (atoms repel when they get too close), drops to a minimum at the equilibrium bond length $r_0$ (the stable resting distance), and gradually approaches zero for large $r$ (atoms are too far apart to interact). A total energy line is drawn at three levels: below zero (bound state --- atoms oscillate around $r_0$), exactly at zero (atoms can just barely separate), and above zero (atoms fly apart).]
When two atoms are bound in a molecule, their potential energy as a function of separation distance looks like a well with a steep inner wall and a gentler outer slope. The minimum of the well is the equilibrium bond length --- the "natural" distance between the atoms.
If the total energy is negative (below the asymptotic value of $U$ at large $r$), the atoms are trapped in the well. They vibrate back and forth around the equilibrium distance, turning around at two points --- one closer than $r_0$, one farther. The molecule is stable.
If the total energy reaches zero (or more precisely, the asymptotic value of $U$ at large separations), the atoms can separate. The molecule dissociates. This is exactly what happens when you heat a gas enough --- you increase the total energy until molecules break apart.
The entire story of molecular bonding, vibration, and dissociation is encoded in one potential energy curve and one energy line. No differential equations needed.
Spaced Retrieval
Before moving to practice, test your recall of earlier material from this chapter.
Recall prompt 1: What is the work-energy theorem? State it in words and in an equation. (Section 7.3)
Recall prompt 2: What makes a force conservative? Give the defining property and an example of a conservative and a nonconservative force. (Section 7.4)
Recall prompt 3: If a particle moves from point A to point B in the presence of a conservative force, how does the work done by that force relate to the change in potential energy? (Section 7.4)
Practice Layers
Layer 1: Concrete --- Read the Diagram
Problem 1. A particle moves in one dimension under the influence of a conservative force. Its potential energy is given by the curve shown below.
[Image: A $U(x)$ curve with the following features: - A local minimum at $x = 2$ m with $U = -8$ J - A local maximum at $x = 5$ m with $U = 4$ J - A local minimum at $x = 8$ m with $U = -2$ J - The curve rises to $U = 10$ J at both ends ($x = 0$ and $x = 10$ m) - A horizontal energy line at $E = 0$ J]
(a) Find all the turning points for a particle with total energy $E = 0$ J.
(b) Identify all allowed and forbidden regions.
(c) If the particle starts near $x = 2$ m, describe its motion qualitatively. Where is it fastest? Where does it turn around? Can it reach $x = 8$ m?
(d) What minimum total energy would the particle need to travel from the left well to the right well?
Check your answer
**(a)** The turning points are where $U(x) = E = 0$. Reading from the graph, the curve crosses $U = 0$ at four points: approximately $x \approx 1$ m, $x \approx 3.5$ m, $x \approx 6$ m, and $x \approx 9.5$ m. **(b)** The particle can exist wherever $U(x) \leq 0$. The allowed regions are approximately $1 \leq x \leq 3.5$ m (the left well) and $6 \leq x \leq 9.5$ m (the right well). The forbidden regions are $x < 1$ m, $3.5 < x < 6$ m, and $x > 9.5$ m. **(c)** A particle starting near $x = 2$ m oscillates in the left well. It is fastest at $x = 2$ m (the bottom of the well, where $E - U = 0 - (-8) = 8$ J, giving maximum $KE$). It turns around at $x \approx 1$ m and $x \approx 3.5$ m. It *cannot* reach $x = 8$ m because the barrier at $x = 5$ m has $U = 4$ J $> E = 0$ J. The particle is trapped. **(d)** The particle needs $E \geq 4$ J (the height of the barrier at $x = 5$ m) to cross from the left well to the right well. At exactly $E = 4$ J, it could reach the top of the barrier with zero speed, but any energy above 4 J would allow it to cross.Layer 2: Pattern --- Sketch Force from Energy
Problem 2. Using the same $U(x)$ curve from Problem 1, sketch $F(x)$ over the interval $0 \leq x \leq 10$ m. Use the relationship $F = -dU/dx$.
(a) At what positions is $F = 0$?
(b) In which regions does the force point in the positive $x$ direction? In which regions does it point in the negative $x$ direction?
(c) Where is the magnitude of the force greatest?
Check your answer
**(a)** $F = 0$ wherever $dU/dx = 0$ --- at the extrema of $U(x)$. That is at $x = 2$ m (minimum), $x = 5$ m (maximum), and $x = 8$ m (minimum). These are the equilibrium points. **(b)** The force is $F = -dU/dx$, so it points opposite to the slope. - From $x = 0$ to $x = 2$ m: $U$ is decreasing (slope is negative), so $F > 0$ (force pushes to the right). - From $x = 2$ to $x = 5$ m: $U$ is increasing (slope is positive), so $F < 0$ (force pushes to the left). - From $x = 5$ to $x = 8$ m: $U$ is decreasing (slope is negative), so $F > 0$ (force pushes to the right). - From $x = 8$ to $x = 10$ m: $U$ is increasing (slope is positive), so $F < 0$ (force pushes to the left). Notice the pattern: in the neighborhood of every minimum ($x = 2$, $x = 8$), the force always points *toward* the minimum. Near the maximum ($x = 5$), the force always points *away* from the maximum. This is what makes minima stable and maxima unstable. **(c)** The magnitude of the force is greatest where $|dU/dx|$ is largest --- where the $U(x)$ curve is steepest. Looking at the graph, the steepest slopes occur on the walls of the deep left well (between $x = 0$ and $x = 2$ m, and between $x = 2$ and $x = 5$ m near $x \approx 3$ m). The steep outer walls near $x = 0$ and $x = 10$ also have large forces. The force is weakest near the equilibrium points and in regions where $U(x)$ is relatively flat.Layer 3: Structure --- Why Minima Are Stable
Problem 3. Explain, using both physical and mathematical reasoning, why potential energy minima correspond to stable equilibria and potential energy maxima correspond to unstable equilibria.
(a) Start with the physical argument: if a particle is displaced slightly from a minimum of $U(x)$, which direction does the force push it? What about from a maximum?
(b) Now give the mathematical argument using $F = -dU/dx$. At a minimum, $dU/dx = 0$ and $d^2U/dx^2 > 0$. If the particle is displaced to the right (positive $dx$), what sign is $dU/dx$ at the displaced position? What sign is $F$? Repeat for displacement to the left.
(c) A student claims: "Equilibrium just means $F = 0$, so the particle stays put regardless of whether it's at a minimum or maximum." What is wrong with this reasoning?
Check your answer
**(a) Physical argument.** At a minimum of $U(x)$, the potential energy increases in both directions away from the equilibrium point. If the particle is displaced to the right, it climbs the potential energy curve; this means $U$ increases, $KE$ decreases, and the particle slows and is pushed back (the force points left, toward the minimum). If displaced to the left, the same thing happens in reverse. The particle always gets pushed back --- this is a restoring force, and the equilibrium is stable. At a maximum, the potential energy decreases in both directions. If displaced to the right, $U$ decreases, $KE$ increases, and the force pushes the particle further to the right (away from the maximum). The particle accelerates away from the equilibrium. This is an anti-restoring force, and the equilibrium is unstable. **(b) Mathematical argument.** At a minimum, $dU/dx = 0$ and $d^2U/dx^2 > 0$ (curve is concave up). If displaced to the right by a small amount, the first derivative becomes positive ($dU/dx > 0$ because the curve is rising to the right of the minimum). The force is $F = -dU/dx < 0$ --- it points to the left, back toward the minimum. If displaced to the left, $dU/dx < 0$ (curve is falling to the right of the displaced position, which means rising to the left), so $F = -dU/dx > 0$ --- pointing right, again back toward the minimum. At a maximum, $d^2U/dx^2 < 0$ (curve is concave down). Displacement to the right gives $dU/dx < 0$ (curve is falling), so $F = -dU/dx > 0$ --- pushing further to the right, away from the maximum. The logic reverses: the force always pushes the particle *away* from a maximum. **(c)** The student is correct that $F = 0$ at both minima and maxima. But equilibrium is about what happens after a *small disturbance*. A ball on the top of a hill is technically in equilibrium --- if perfectly placed, it stays. But any infinitesimal perturbation causes it to roll away. The statement "$F = 0$" describes only the instantaneous force at the equilibrium point itself. It says nothing about the restoring or anti-restoring behavior for nearby positions. Stability is determined by the curvature ($d^2U/dx^2$), not by the value of the force at the equilibrium point.Layer 4: Creation --- Design a Potential Energy Curve
Problem 4. Design a potential energy curve $U(x)$ that has exactly two stable equilibria and one unstable equilibrium. Your design should include:
(a) A sketch of $U(x)$, with the positions of the three equilibria clearly labeled.
(b) An explanation of why each equilibrium is stable or unstable, based on the shape of the curve.
(c) A total energy level that would trap a particle in the left stable equilibrium but not the right one. Explain why this energy level works.
(d) A total energy level for which the particle can access both stable equilibria. Explain what the motion looks like.
Check your answer
There are many correct designs. Here is one example: **(a)** Draw a curve with a minimum at $x = 2$ (say $U = 0$), rising to a local maximum at $x = 5$ (say $U = 6$ J), then dropping to a second minimum at $x = 8$ (say $U = 2$ J), and rising steeply on both ends. The three equilibria are: - $x = 2$: stable (minimum, $d^2U/dx^2 > 0$) - $x = 5$: unstable (maximum, $d^2U/dx^2 < 0$) - $x = 8$: stable (minimum, $d^2U/dx^2 > 0$) **(b)** At $x = 2$ and $x = 8$, the curve is concave up (bowl-shaped). A displaced particle experiences a restoring force: $F = -dU/dx$ points back toward the minimum. At $x = 5$, the curve is concave down (hilltop-shaped). A displaced particle experiences a force pointing *away* from the maximum. **(c)** Set $E = 4$ J. This is above the left well minimum ($U = 0$) and below the barrier ($U = 6$ J), so the particle is trapped in the left well. But $E = 4$ J is above the right well minimum ($U = 2$ J) as well --- so why is the particle not in the right well? Because it cannot *reach* the right well. The barrier at $x = 5$ has $U = 6 > E = 4$, so the particle is blocked. If the particle *started* in the right well, it would also be trapped there (between two turning points around $x = 8$), but it could not cross the barrier to reach the left well either. **(d)** Set $E = 7$ J. This is above the barrier at $x = 5$ ($U = 6$ J), so the particle can cross from one well to the other. The motion would look like this: the particle oscillates broadly, sweeping through both wells, slowing down as it climbs the barrier and at the outer turning points, speeding up at the bottoms of both wells (fastest at $x = 2$, where $KE = 7 - 0 = 7$ J, and somewhat slower at $x = 8$, where $KE = 7 - 2 = 5$ J). The key design requirement is exactly two minima (stable) and one maximum (unstable) between them. Any smooth curve with this topology works. You could make the wells different depths, different widths, symmetric or asymmetric. The physics is determined by the number and type of critical points.Reflection
What can an energy diagram tell you that a force diagram cannot --- and vice versa?
An energy diagram gives you the global picture at a glance: where the particle can and cannot go, where it speeds up and slows down, where it turns around, and where it is in equilibrium. You see the entire landscape of possible motion simultaneously. You can change the total energy and immediately see how the motion changes qualitatively --- whether the particle is trapped or free, how many turning points it has, how fast it moves at each position. This kind of overview is extremely difficult to extract from a force diagram.
A force diagram, on the other hand, tells you the instantaneous story: how large is the force right now, in which direction, and therefore what is the acceleration. It gives you the local, moment-by-moment dynamics. If you need to know when the particle arrives at a certain position, or the precise time dependence of the motion, the energy diagram cannot help --- it tells you speed but not time. Force-based methods (with Newton's second law and integration) are needed for that.
The two representations are complementary. The energy diagram is derived from the force ($U$ comes from integrating $F$), and the force is derived from the energy diagram ($F = -dU/dx$). They contain the same information, but they make different aspects of the physics visible. Learning to switch between them --- to see when the global energy picture is more useful than the local force picture, and vice versa --- is part of developing physical fluency.
Looking Ahead
You have now seen the most powerful qualitative tool in the energy framework: the potential energy diagram. With just a curve and a horizontal line, you can predict turning points, allowed regions, equilibrium positions, stability, speeds, and forces --- all without solving a differential equation.
But there is one more concept remaining in this chapter before we integrate everything. You know how much energy is transferred (work) and where that energy goes (kinetic and potential). The next question is: how fast is energy transferred? A car climbing a hill and a rocket climbing a hill may do the same work, but one does it in seconds and the other in minutes. The rate of energy transfer --- power --- is the subject of Section 7.6. It is what determines the performance of engines, motors, and every machine that converts energy into motion.