AP Physics 1- 3.3 Potential Energy- Study Notes- New Syllabus
AP Physics 1-3.3 Potential Energy – Study Notes
AP Physics 1-3.3 Potential Energy – Study Notes -AP Physics 1 – per latest Syllabus.
Key Concepts:
- Potential Energy
- Potential Energy in Common Physical Systems
- Total Potential Energy of a System
Potential Energy
A system composed of two or more objects has potential energy if the objects interact with each other only through conservative forces (e.g., gravity, spring force, electric force). 
- Potential energy is a scalar quantity associated with the relative position or configuration of objects within the system.
- The change in potential energy depends only on the initial and final states, not on the path taken.
- Work done by a conservative force is related to potential energy by: \( W_{cons} = -\Delta U \).
Key Forms of Potential Energy
- Gravitational Potential Energy: \( U_g = m g h \)
- Elastic Potential Energy (Spring): \( U_s = \tfrac{1}{2} k x^2 \)
- Electric Potential Energy: \( U_e = k \dfrac{q_1 q_2}{r} \)
Important Properties
- If a system returns to its original configuration, the net change in potential energy is zero.
- The choice of reference level for potential energy is arbitrary; only differences in potential energy matter physically.
- A decrease in potential energy corresponds to an increase in kinetic energy if no nonconservative forces act.
Example:
A \( 3 \, \text{kg} \) object is moved from a height of \( 2 \, \text{m} \) to a height of \( 7 \, \text{m} \). Take \( g = 9.8 \, \text{m/s}^2 \). Find the change in potential energy.
▶️Answer/Explanation
Step (1): Formula → \( \Delta U_g = m g \Delta h \).
Step (2): Substitute values → \( \Delta U_g = (3)(9.8)(7 – 2) \).
Step (3): \( \Delta U_g = (3)(9.8)(5) = 147 \, \text{J} \).
Final Answer: The potential energy increases by 147 J.
Potential Energy in Common Physical Systems
- The potential energy of physical systems can be described using the physical properties of that system.
- Potential energy is always associated with conservative forces such as gravity, elastic (spring) forces, and electrostatic forces.
Elastic Potential Energy (Spring System)
For an ideal spring, the potential energy depends on how far the spring is stretched or compressed from its equilibrium position.
Equation: \( U_s = \tfrac{1}{2} k x^2 \)
- \( k \) = spring constant (N/m)
- \( x \) = displacement from equilibrium (m)
Gravitational Potential Energy (Two Masses)
For two spherical masses (e.g., planets, moons, or stars), gravitational potential energy is given by:
Equation: \( U_g = – \dfrac{G m_1 m_2}{r} \)
- \( G \) = universal gravitational constant \( (6.67 \times 10^{-11} \, \text{N·m}^2/\text{kg}^2) \)
- \( m_1, m_2 \) = masses of the objects
- \( r \) = center-to-center separation (m)
The negative sign shows that gravitational potential energy is zero at infinite separation and decreases as objects get closer.
Gravitational Potential Energy Near Planet’s Surface
Near the surface of Earth (or any planet), the gravitational field is approximately constant (\( g \)).
The change in gravitational potential energy is approximated as:
- Equation: \( \Delta U_g = m g \Delta h \)
- \( m \) = mass of the object (kg)
- \( g \) = acceleration due to gravity (m/s²)
- \( \Delta h \) = change in vertical height (m)
Example
A spring with \( k = 200 \, \text{N/m} \) is stretched by \( 0.1 \, \text{m} \). Find the stored elastic potential energy.
▶️Answer/Explanation
\( U_s = \tfrac{1}{2} k x^2 = \tfrac{1}{2} (200)(0.1^2) \)
\( U_s = 1.0 \, \text{J} \)
Final Answer: The spring stores 1 J of elastic potential energy.
Example
Find the gravitational potential energy between Earth (\( M = 5.97 \times 10^{24} \, \text{kg} \)) and a \( 1000 \, \text{kg} \) satellite at a distance of \( r = 7.0 \times 10^6 \, \text{m} \) from Earth’s center.
▶️Answer/Explanation
\( U_g = – \dfrac{G M m}{r} \)
\( = – \dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(1000)}{7.0 \times 10^6} \)
\( U_g \approx -5.7 \times 10^{10} \, \text{J} \)
Final Answer: The gravitational potential energy is \(-5.7 \times 10^{10} \, \text{J}\).
Example
A \( 5 \, \text{kg} \) rock is lifted upward by \( 3 \, \text{m} \) near Earth’s surface (\( g = 9.8 \, \text{m/s}^2 \)). Find the change in gravitational potential energy.
▶️Answer/Explanation
\( \Delta U_g = m g h \)
\( = (5)(9.8)(3) = 147 \, \text{J} \)
Final Answer: The change in gravitational potential energy is 147 J.
Total Potential Energy of a System
The total potential energy of a system containing more than two objects is equal to the sum of the potential energy of each interacting pair of objects within the system. This principle applies when the forces between objects are conservative (e.g., gravitational, electric, spring forces).
For gravitational systems:
\( U_{total} = \sum_{i<j} -\dfrac{G m_i m_j}{r_{ij}} \),
where \( m_i \) and \( m_j \) are masses of objects and \( r_{ij} \) is the distance between them.
- This allows analysis of systems such as planetary systems, star clusters, or multiple charges in electrostatics.
Example:
Three masses (\( m_1 = 2 \, \text{kg} \), \( m_2 = 3 \, \text{kg} \), \( m_3 = 4 \, \text{kg} \)) are placed such that each pair is separated by \( 1.0 \, \text{m} \). Find the total gravitational potential energy of the system.
▶️Answer/Explanation
For three objects, the total potential energy is the sum of the three pairs:
\( U_{total} = U_{12} + U_{13} + U_{23} \)
\( = -G \dfrac{m_1 m_2}{r_{12}} – G \dfrac{m_1 m_3}{r_{13}} – G \dfrac{m_2 m_3}{r_{23}} \)
\( = -6.67 \times 10^{-11} \left[ \dfrac{(2)(3)}{1.0} + \dfrac{(2)(4)}{1.0} + \dfrac{(3)(4)}{1.0} \right] \)
\( = -6.67 \times 10^{-11} (6 + 8 + 12) \)
\( = -6.67 \times 10^{-11} (26) \)
\( U_{total} \approx -1.73 \times 10^{-9} \, \text{J} \)
Final Answer: The total gravitational potential energy of the three-mass system is \(-1.7 \times 10^{-9} \, \text{J}\).