AP Physics C Mechanics- 3.3 Potential Energy- Study Notes- New Syllabus
AP Physics C Mechanics- 3.3 Potential Energy – Study Notes
AP Physics C Mechanics- 3.3 Potential Energy – Study Notes – per latest Syllabus.
Key Concepts:
- Potential Energy in a System of Interacting Objects
- Potential Energy as a Scalar Quantity Associated with Position
- Choice of Zero Potential Energy (Reference Point)
- Relationship Between Conservative Forces and Potential Energy
- Potential Energy Graphs and Equilibrium Points
- Potential Energy of Common Physical Systems
- Total Potential Energy of a System with Multiple Objects
Potential Energy in a System of Interacting Objects
A system composed of two or more objects possesses potential energy if the objects interact only through conservative forces such as gravity, electrostatic force, or elastic (spring) force.
- Potential energy represents the stored mechanical energy of a system due to the positions or configurations of its interacting parts.
- It exists only when the force between the objects is conservative, meaning the work done depends only on the initial and final positions, not on the path taken.
- Common examples include:
- Gravitational potential energy: \( \mathrm{U_g = mgh} \)
- Elastic (spring) potential energy: \( \mathrm{U_s = \dfrac{1}{2}kx^2} \)
- Electrostatic potential energy: \( \mathrm{U_e = k\dfrac{q_1q_2}{r}} \)
System Consideration:
- Potential energy belongs to the system as a whole, not to any individual object within it.
- When a conservative force acts, energy is exchanged between kinetic and potential forms, but the total mechanical energy of the system remains constant (in the absence of nonconservative forces).
- Changes in potential energy represent the work done by conservative forces within the system.
Example:
Two objects of masses \( \mathrm{m_1 = 2.0\,kg} \) and \( \mathrm{m_2 = 3.0\,kg} \) are separated vertically by a distance of \( \mathrm{5.0\,m} \). Determine the gravitational potential energy of the system with respect to the lower object.
▶️ Answer / Explanation
Given:
- \( \mathrm{m_1 = 2.0\,kg}, \; m_2 = 3.0\,kg} \)
- \( \mathrm{h = 5.0\,m}, \; g = 9.8\,m/s^2} \)
Gravitational potential energy of the upper object (relative to the lower one):
\( \mathrm{U_g = m_1 g h = (2.0)(9.8)(5.0) = 98\,J} \)
Interpretation:
- The system’s potential energy depends on the configuration — the separation height — not on how the mass was raised.
- If the mass returns to its starting position, \( \mathrm{\Delta U_g = 0} \), showing gravity’s conservative nature.
Result: \( \mathrm{U_g = 98\,J} \)
Potential Energy as a Scalar Quantity Associated with Position
Potential energy is a scalar quantity that depends only on the position or configuration of the objects within a system. It represents the capacity of the system to do work due to the relative arrangement of its components under the influence of a conservative force.
Scalar Nature of Potential Energy:
- Unlike vector quantities such as force or displacement, potential energy has magnitude only — it has no direction.
- Its value at any point in space depends solely on the position of the object(s) relative to a chosen reference configuration.
- Since it is scalar, potential energies from different conservative forces can be added algebraically to find total potential energy of a system.
Relation to Work by Conservative Forces:
For a conservative force \( \mathrm{\vec{F}} \), the work done when moving an object between two positions is equal to the negative change in potential energy:
\( \mathrm{W_c = -\Delta U = U_i – U_f} \)
- If the force does positive work, the potential energy of the system decreases.
- If the force does negative work, the potential energy of the system increases.
- Thus, potential energy acts as a measure of the stored capacity to perform mechanical work.
Dependence on Position:
- Gravitational potential energy depends on height: \( \mathrm{U_g = mgh} \).
- Elastic potential energy depends on displacement: \( \mathrm{U_s = \dfrac{1}{2}kx^2} \).
- Electrostatic potential energy depends on separation: \( \mathrm{U_e = k \dfrac{q_1 q_2}{r}} \).
Physical Meaning:
- Potential energy describes how the configuration of a system determines its ability to perform work when released.
- It is stored energy resulting from position in a force field (gravitational, elastic, or electric).
Example:
A \( \mathrm{0.5\,kg} \) ball is lifted vertically to a height of \( \mathrm{4.0\,m} \) above the ground. Calculate the change in gravitational potential energy of the ball. Take \( \mathrm{g = 9.8\,m/s^2} \).
▶️ Answer / Explanation
Given: \( \mathrm{m = 0.5\,kg}, \; h = 4.0\,m, \; g = 9.8\,m/s^2} \)
Change in potential energy:
\( \mathrm{\Delta U = mgh = (0.5)(9.8)(4.0) = 19.6\,J} \)
Interpretation:
- Since the ball is raised, the gravitational potential energy increases by \( \mathrm{19.6\,J} \).
- The energy is stored as potential energy due to the ball’s elevated position in Earth’s gravitational field.
Result: \( \mathrm{\Delta U = +19.6\,J} \)
Choice of Zero Potential Energy (Reference Point)
The zero of potential energy for any system is an arbitrary reference level chosen by the observer to simplify calculations or analysis. Only changes in potential energy (\( \mathrm{\Delta U} \)) are physically meaningful, not the absolute values of \( \mathrm{U} \).
- Potential energy is always defined relative to a chosen reference position where \( \mathrm{U = 0} \).
- This reference point can be chosen freely — for example, the ground, the surface of a table, or the point of maximum compression of a spring.
- Since only differences in potential energy affect motion, the choice of zero does not alter physical outcomes.
Mathematical Representation:
The change in potential energy is given by:
\( \mathrm{\Delta U = U_f – U_i = -W_c} \)
- \( \mathrm{W_c} \): work done by the conservative force.
- Choosing \( \mathrm{U = 0} \) at a convenient position simplifies the computation of \( \mathrm{U_f} \) and \( \mathrm{U_i} \).
- Only \( \mathrm{\Delta U} \) determines how much energy is converted between potential and kinetic forms.
Typical Reference Choices:
- For gravitational systems near Earth: \( \mathrm{U_g = 0} \) at ground level.
- For springs: \( \mathrm{U_s = 0} \) when the spring is at its natural (unstretched) length.
- For electric systems: \( \mathrm{U_e = 0} \) at infinite separation between charges.
Example:
A ball of mass \( \mathrm{1.0\,kg} \) is lifted to a height of \( \mathrm{2.0\,m} \) above a table. Find the gravitational potential energy of the ball if (a) the table surface and (b) the floor \( \mathrm{0.5\,m} \) below the table are chosen as the reference level (\( \mathrm{U = 0} \)).
▶️ Answer / Explanation
Given: \( \mathrm{m = 1.0\,kg, \; h_{table} = 2.0\,m, \; h_{floor} = 2.0 + 0.5 = 2.5\,m, \; g = 9.8\,m/s^2} \)
(a) Reference: Table surface \( \mathrm{U = 0} \)
\( \mathrm{U_g = mgh = (1.0)(9.8)(2.0) = 19.6\,J} \)
(b) Reference: Floor \( \mathrm{U = 0} \)
\( \mathrm{U_g = (1.0)(9.8)(2.5) = 24.5\,J} \)
Interpretation:
- The potential energy differs because the zero level was chosen differently.
- However, the change in potential energy between two positions is identical regardless of the reference level — confirming that only \( \mathrm{\Delta U} \) matters physically.
Relationship Between Conservative Forces and Potential Energy
For a system in which the forces acting are conservative, the work done by these forces is related to the change in the system’s potential energy. A conservative force can be derived from a potential energy function \( \mathrm{U} \), meaning that the force is the negative gradient of that potential energy.
The change in potential energy between two points \( \mathrm{a} \) and \( \mathrm{b} \) is equal to the negative of the work done by the conservative force as the system moves between those points:
\( \mathrm{\Delta U = U_b – U_a = – \int_a^b \vec{F}_{cf} \cdot d\vec{r}} \)
or equivalently, the potential energy at a point can be defined as:
\( \mathrm{U(r) = – \int \vec{F}_{cf} \cdot d\vec{r}} \)
- \( \mathrm{\vec{F}_{cf}} \): conservative force acting on the system
- \( \mathrm{d\vec{r}} \): infinitesimal displacement vector
- \( \mathrm{U(r)} \): potential energy function corresponding to the conservative force
Example:
The force acting on a particle along the x-axis is given by \( \mathrm{F(x) = -4x} \) N, where \( \mathrm{x} \) is in meters. Determine the expression for the potential energy function \( \mathrm{U(x)} \) if \( \mathrm{U = 0} \) at \( \mathrm{x = 0} \).
▶️ Answer / Explanation
Given: \( \mathrm{F(x) = -4x} \), and \( \mathrm{U(0) = 0} \)
Step 1: Use the relation \( \mathrm{F(x) = -\dfrac{dU}{dx}} \)
\( \mathrm{-4x = -\dfrac{dU}{dx}} \Rightarrow \dfrac{dU}{dx} = 4x} \)
Step 2: Integrate to find \( \mathrm{U(x)} \):
\( \mathrm{U(x) = \displaystyle \int 4x \, dx = 2x^2 + C} \)
Step 3: Apply the boundary condition \( \mathrm{U(0) = 0} \):
\( \mathrm{C = 0} \)
Final Expression:
\( \mathrm{U(x) = 2x^2 \, J} \)
Conservative Force from the Slope of the Potential Energy Function
In one-dimensional motion, the conservative force acting on a system can be obtained directly from the slope of its potential energy curve. The force always acts in the direction of decreasing potential energy.
Mathematical Relationship:
The magnitude and direction of a conservative force are given by the negative derivative of the potential energy with respect to position:
\( \mathrm{F(x) = -\dfrac{dU(x)}{dx}} \)
- \( \mathrm{F(x)} \): conservative force at position \( \mathrm{x} \)
- \( \mathrm{U(x)} \): potential energy function of the system
- The negative sign shows that the force acts opposite to the increase in potential energy.
Direction of Force:
- If \( \mathrm{\dfrac{dU}{dx} > 0} \): the potential energy increases with position, so \( \mathrm{F < 0} \) — the force acts in the negative x-direction.
- If \( \mathrm{\dfrac{dU}{dx} < 0} \): the potential energy decreases with position, so \( \mathrm{F > 0} \) — the force acts in the positive x-direction.
- The system tends to move naturally toward regions of lower potential energy.
Example:
The potential energy of a particle moving along the x-axis is given by \( \mathrm{U(x) = 3x^2 – 2x + 1} \) J. Determine the expression for the conservative force acting on the particle and find the force at \( \mathrm{x = 2\,m} \).
▶️ Answer / Explanation
Step 1: Use \( \mathrm{F(x) = -\dfrac{dU}{dx}} \)
\( \mathrm{\dfrac{dU}{dx} = 6x – 2} \)
\( \mathrm{F(x) = -(6x – 2) = -6x + 2} \)
Step 2: Evaluate at \( \mathrm{x = 2\,m} \)
\( \mathrm{F(2) = -6(2) + 2 = -10\,N} \)
Interpretation: The negative value indicates the force acts toward decreasing x — that is, opposite to the direction in which potential energy increases.
Result: \( \mathrm{F(x) = -6x + 2 \, N} \), and \( \mathrm{F(2) = -10\,N.} \)
Potential Energy Graphs and Equilibrium Points
Graphs of a system’s potential energy as a function of position are powerful tools for analyzing the system’s stability and motion. The shape and slope of the \( \mathrm{U(x)} \) vs. \( \mathrm{x} \) graph reveal where the system is in equilibrium and whether that equilibrium is stable or unstable.
Equilibrium Condition:
An equilibrium position occurs where the net force on the system is zero.
- From the relation \( \mathrm{F(x) = -\dfrac{dU}{dx}} \), this occurs when:
\( \mathrm{\dfrac{dU}{dx} = 0} \)
- At these points, the potential energy curve has either a local minimum, a local maximum, or a point of inflection.
Stable Equilibrium:
A system is in stable equilibrium if a small displacement produces a restoring force that pushes it back toward equilibrium.
- This occurs at points where the potential energy has a local minimum:
\( \mathrm{\dfrac{d^2U}{dx^2} > 0} \Rightarrow \text{Stable equilibrium} \)
- Physically, the particle tends to remain near this point, oscillating around it if disturbed slightly.
- Example: the bottom of a valley or the lowest point of a pendulum’s swing.
Unstable Equilibrium:
A system is in unstable equilibrium if a small displacement produces a force that moves it farther away from equilibrium.
- This occurs at points where the potential energy has a local maximum:
\( \mathrm{\dfrac{d^2U}{dx^2} < 0} \Rightarrow \text{Unstable equilibrium} \)
- Physically, the particle tends to move away from this point if slightly disturbed.
- Example: a ball balanced at the top of a hill.
Graphical Interpretation:
| Feature of \( \mathrm{U(x)} \) Graph | Mathematical Condition | Type of Equilibrium | Force Behavior |
|---|---|---|---|
| Local minimum | \( \mathrm{\dfrac{dU}{dx} = 0, \; \dfrac{d^2U}{dx^2} > 0} \) | Stable equilibrium | Force acts toward equilibrium |
| Local maximum | \( \mathrm{\dfrac{dU}{dx} = 0, \; \dfrac{d^2U}{dx^2} < 0} \) | Unstable equilibrium | Force pushes away from equilibrium |
| Flat region (constant \( \mathrm{U} \)) | \( \mathrm{\dfrac{dU}{dx} = 0, \; \dfrac{d^2U}{dx^2} = 0} \) | Neutral equilibrium | No net restoring or repelling force |
Example:
A particle moves along the x-axis under a potential energy function \( \mathrm{U(x) = x^4 – 4x^2} \). Determine the positions of equilibrium and classify each as stable or unstable.
▶️ Answer / Explanation
Step 1: Find equilibrium positions where \( \mathrm{\dfrac{dU}{dx} = 0} \).
\( \mathrm{\dfrac{dU}{dx} = 4x^3 – 8x = 4x(x^2 – 2) = 0} \)
→ \( \mathrm{x = 0, \; x = \pm\sqrt{2}} \)
Step 2: Determine stability using \( \mathrm{\dfrac{d^2U}{dx^2}} \).
\( \mathrm{\dfrac{d^2U}{dx^2} = 12x^2 – 8} \)
- At \( \mathrm{x = 0} \): \( \mathrm{\dfrac{d^2U}{dx^2} = -8 < 0} \) → Unstable equilibrium
- At \( \mathrm{x = \pm\sqrt{2}} \): \( \mathrm{\dfrac{d^2U}{dx^2} = 12(2) – 8 = 16 > 0} \) → Stable equilibrium
Step 3: Interpret graphically.
- The potential energy graph has two valleys (at \( \mathrm{x = \pm\sqrt{2}} \)) and a hill (at \( \mathrm{x = 0} \)).
- The particle tends to rest near the valleys — stable equilibrium points.
Potential Energy of Common Physical Systems
The potential energy of a physical system depends on its configuration and the nature of the conservative forces acting within it. Different types of systems — such as elastic, gravitational (near or far from Earth’s surface), and electrostatic — have potential energies that can be expressed in mathematical form using measurable physical properties of the system.
Elastic Potential Energy (Ideal Spring):
The potential energy stored in an ideal spring due to stretching or compression is given by:
\( \mathrm{U_s = \dfrac{1}{2}k(\Delta x)^2} \)
- \( \mathrm{U_s} \): elastic potential energy (Joules)
- \( \mathrm{k} \): spring constant (N/m), a measure of stiffness
- \( \mathrm{\Delta x} \): displacement from the spring’s equilibrium length (m)
- Work done in stretching or compressing the spring is stored as potential energy.
- This energy can later be released, converting into kinetic or other forms of energy.
- The graph of \( \mathrm{U_s} \) vs. \( \mathrm{x} \) is a parabola opening upward — indicating stability at the equilibrium point \( \mathrm{x = 0} \).
Gravitational Potential Energy (Universal Form):
The potential energy between two spherically symmetric bodies (e.g., Earth and Moon) separated by a distance \( \mathrm{r} \) is given by:
\( \mathrm{U_g = -\dfrac{G m_1 m_2}{r}} \)
- \( \mathrm{G} \): universal gravitational constant (\( \mathrm{6.67 \times 10^{-11} \, N \, m^2 / kg^2} \))
- \( \mathrm{m_1, m_2} \): interacting masses (kg)
- \( \mathrm{r} \): distance between centers of mass (m)
- The negative sign indicates that the gravitational force is attractive.
- \( \mathrm{U_g} \) increases (becomes less negative) as \( \mathrm{r} \) increases — meaning work must be done against gravity to separate the objects.
Near-Earth Gravitational Potential Energy (Approximation):
Near the surface of a planet, the gravitational field \( \mathrm{g} \) is approximately constant. In this case, the change in gravitational potential energy is given by:
\( \mathrm{\Delta U_g = mg\Delta h} \)
- \( \mathrm{m} \): mass of the object (kg)
- \( \mathrm{g} \): acceleration due to gravity (\( \mathrm{9.8 \, m/s^2} \) near Earth’s surface)
- \( \mathrm{\Delta h} \): change in vertical position (m)
- This form is derived as an approximation of the universal gravitational potential energy for small height changes (\( \mathrm{\Delta h \ll R_{planet}} \)).
- It is most useful for objects moving near the surface of Earth or another planet.
| Condition | Expression for \( \mathrm{U_g} \) | Applicability |
|---|---|---|
| Near a planet’s surface | \( \mathrm{U_g = mgh} \) | Valid for small height changes (\( \mathrm{h \ll R_{planet}} \)) |
| Large distances between masses | \( \mathrm{U_g = -\dfrac{G m_1 m_2}{r}} \) | Valid for orbital and astrophysical systems |
Example:
A \( \mathrm{0.4\,kg} \) ball is dropped from a height of \( \mathrm{10\,m} \) above the ground. Calculate the change in gravitational potential energy of the ball as it falls, and the corresponding increase in kinetic energy. Assume \( \mathrm{g = 9.8\,m/s^2} \).
▶️ Answer / Explanation
Given: \( \mathrm{m = 0.4\,kg, \; \Delta h = 10\,m, \; g = 9.8\,m/s^2} \)
Step 1: Compute the change in potential energy:
\( \mathrm{\Delta U_g = -mg\Delta h = -(0.4)(9.8)(10) = -39.2\,J} \)
Step 2: Apply conservation of mechanical energy:
\( \mathrm{\Delta K = -\Delta U_g = +39.2\,J} \)
Interpretation: The decrease in gravitational potential energy is exactly equal to the increase in kinetic energy as the ball falls, consistent with conservation of energy.
Result: \( \mathrm{\Delta U_g = -39.2\,J}, \quad \Delta K = +39.2\,J} \)
Total Potential Energy of a System with Multiple Objects
In a system consisting of more than two interacting objects, the total potential energy of the system is the sum of the potential energies associated with every unique pair of interacting objects within the system. This principle applies to systems governed by conservative forces such as gravity, electrostatics, or springs.
Mathematical Representation:
For a system of \( \mathrm{N} \) interacting objects, the total potential energy is given by:
\( \mathrm{U_{total} = \sum_{i=1}^{N-1} \sum_{j=i+1}^{N} U_{ij}} \)
- \( \mathrm{U_{ij}} \): potential energy between the \( \mathrm{i^{th}} \) and \( \mathrm{j^{th}} \) objects.
- The double summation ensures that each interaction pair is counted once.
Example for Gravitational Systems:
For gravitational interactions between masses \( \mathrm{m_i} \) and \( \mathrm{m_j} \) separated by a distance \( \mathrm{r_{ij}} \):
\( \mathrm{U_{ij} = -\dfrac{G m_i m_j}{r_{ij}}} \)
- Total gravitational potential energy for a system of three or more masses is found by adding up these pairwise terms.
Generalization for Other Conservative Forces:
- For electrostatic interactions: \( \mathrm{U_{ij} = k\dfrac{q_i q_j}{r_{ij}}} \)
- For spring systems (if multiple springs connect pairs of objects): \( \mathrm{U_{ij} = \dfrac{1}{2}k_{ij}(\Delta x_{ij})^2} \)
Physical Interpretation:
- The total potential energy depends on the configuration (relative positions) of all interacting objects in the system.
- Each pair’s potential energy contributes independently to the overall stored energy.
- Changes in total potential energy correspond to work done by internal conservative forces as the system changes configuration.
Example:
Three point masses are arranged in space: \( \mathrm{m_1 = 2.0\,kg} \), \( \mathrm{m_2 = 3.0\,kg} \), \( \mathrm{m_3 = 4.0\,kg} \). The distances between them are: \( \mathrm{r_{12} = 2.0\,m} \), \( \mathrm{r_{13} = 3.0\,m} \), \( \mathrm{r_{23} = 4.0\,m} \). Determine the total gravitational potential energy of the system.
▶️ Answer / Explanation
Step 1: Use the gravitational potential energy formula for each pair:
\( \mathrm{U_{ij} = -\dfrac{G m_i m_j}{r_{ij}}} \)
Step 2: Write each term:
\( \mathrm{U_{12} = -\dfrac{G(2.0)(3.0)}{2.0} = -3G} \)
\( \mathrm{U_{13} = -\dfrac{G(2.0)(4.0)}{3.0} = -\dfrac{8}{3}G} \)
\( \mathrm{U_{23} = -\dfrac{G(3.0)(4.0)}{4.0} = -3G} \)
Step 3: Add them up:
\( \mathrm{U_{total} = U_{12} + U_{13} + U_{23} = -3G – \dfrac{8}{3}G – 3G = -\dfrac{26}{3}G} \)
Final Result: \( \mathrm{U_{total} = -\dfrac{26}{3}G \approx -8.67G} \)
Interpretation: The total potential energy is negative, indicating that the system is gravitationally bound — energy must be supplied to separate all three masses to infinite distance.