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Study Guide: AP Physics – Potential Energy (Gravitational, Elastic) and Kinetic Energy
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AP Physics – Potential Energy (Gravitational, Elastic) and Kinetic Energy

By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.

⏱️ ~6 min read

AP Physics – Potential Energy (Gravitational, Elastic) and Kinetic Energy

AP Physics: Potential Energy (Gravitational, Elastic) & Kinetic Energy – Exam-Ready Study Guide


What This Is

Potential energy (PE) and kinetic energy (KE) are two forms of mechanical energy that describe how objects store and transfer energy due to their position or motion. These concepts are foundational for understanding conservation of energy, work, and power—key themes in AP Physics 1 and 2. On the exam, you’ll use them to solve problems involving falling objects, springs, roller coasters, and collisions.

Real-world example: A roller coaster at the top of a hill has maximum gravitational potential energy (stored energy due to height). As it descends, this energy converts to kinetic energy (energy of motion), making the ride thrilling. If the coaster compresses a spring at the end, that kinetic energy temporarily becomes elastic potential energy before bouncing back.


Key Terms & Concepts

  • Kinetic Energy (KE): Energy of motion.
  • Formula: ( KE = \frac{1}{2}mv^2 )
    • ( m ) = mass (kg)
    • ( v ) = velocity (m/s)
  • Example: A 2 kg ball moving at 3 m/s has ( KE = \frac{1}{2}(2)(3)^2 = 9 ) J.

  • Gravitational Potential Energy (Ug): Energy stored due to an object’s height in a gravitational field.

  • Formula: ( U_g = mgh )
    • ( m ) = mass (kg)
    • ( g ) = gravitational acceleration (9.8 m/s² on Earth)
    • ( h ) = height (m) relative to a reference point (e.g., the ground).
  • Example: A 5 kg book on a 2 m shelf has ( U_g = (5)(9.8)(2) = 98 ) J.

  • Elastic Potential Energy (Us): Energy stored in stretched or compressed springs (or elastic materials).

  • Formula: ( U_s = \frac{1}{2}kx^2 )
    • ( k ) = spring constant (N/m, measures stiffness)
    • ( x ) = displacement from equilibrium (m).
  • Example: A spring with ( k = 200 ) N/m stretched 0.1 m has ( U_s = \frac{1}{2}(200)(0.1)^2 = 1 ) J.

  • Conservation of Mechanical Energy: In a closed system with no non-conservative forces (e.g., friction, air resistance), the total mechanical energy (KE + PE) remains constant.

  • Formula: ( KE_i + PE_i = KE_f + PE_f )
  • Example: A pendulum swings from its highest point (max ( U_g ), zero ( KE )) to its lowest point (max ( KE ), zero ( U_g )).

  • Work-Energy Theorem: The work done on an object equals its change in kinetic energy.

  • Formula: ( W = \Delta KE = KE_f - KE_i )
  • Example: Pushing a cart from rest to 4 m/s requires work equal to its final ( KE ).

  • Reference Level (for Ug): The arbitrary "zero height" point where ( U_g = 0 ). Always define this in problems!

  • Example: If you measure height from a tabletop, the floor is below the reference level (negative ( h )).

  • Non-Conservative Forces: Forces like friction or air resistance that dissipate mechanical energy as heat or sound.

  • Example: A sliding block slows down due to friction, converting ( KE ) to thermal energy.

  • Power (P): The rate at which work is done or energy is transferred.

  • Formula: ( P = \frac{W}{t} = Fv ) (for constant force parallel to velocity)

    • ( W ) = work (J)
    • ( t ) = time (s)
    • ( F ) = force (N)
    • ( v ) = velocity (m/s).
  • Hooke’s Law: The force exerted by a spring is proportional to its displacement.

  • Formula: ( F_s = -kx )
    • Negative sign indicates the force opposes displacement (restoring force).

Step-by-Step: Solving Energy Problems

Follow this 4-step process for any energy problem on the AP exam:

  1. Identify the System & Reference Level
  2. Draw a diagram. Label the object(s), forces, and define your reference level for ( U_g ).
  3. Example: For a roller coaster, set ( U_g = 0 ) at the lowest point.

  4. List Initial & Final Energies

  5. Write expressions for ( KE ) and ( PE ) (gravitational and/or elastic) at the start and end of the motion.
  6. Example: A ball dropped from height ( h ):

    • Initial: ( KE_i = 0 ), ( U_{gi} = mgh )
    • Final (just before hitting ground): ( KE_f = \frac{1}{2}mv^2 ), ( U_{gf} = 0 ).
  7. Apply Conservation of Energy (or Work-Energy Theorem)

  8. If no non-conservative forces: ( KE_i + PE_i = KE_f + PE_f ).
  9. If friction/air resistance is present: ( KE_i + PE_i + W_{nc} = KE_f + PE_f ), where ( W_{nc} ) is negative work done by non-conservative forces.
  10. Example: A block slides down a ramp with friction: ( mgh = \frac{1}{2}mv^2 + |W_{friction}| ).

  11. Solve for the Unknown

  12. Plug in known values and solve algebraically. Check units!
  13. Example: For the dropped ball, ( mgh = \frac{1}{2}mv^2 )? ( v = \sqrt{2gh} ).

Common Mistakes

  • Mistake: Forgetting to define the reference level for ( U_g ).
  • Correction: Always pick a reference point (e.g., ground, tabletop) and stick with it. Negative heights are okay if below the reference!

  • Mistake: Using ( U_g = mgh ) for objects not near Earth’s surface (e.g., satellites).

  • Correction: For large distances (e.g., orbits), use the universal gravitational PE formula: ( U_g = -\frac{Gm_1m_2}{r} ). On the AP exam, this is rare—stick to ( mgh ) unless told otherwise.

  • Mistake: Ignoring the negative sign in Hooke’s Law (( F_s = -kx )).

  • Correction: The negative sign shows the force is restorative (opposes displacement). For energy calculations (( U_s = \frac{1}{2}kx^2 )), the sign doesn’t matter because ( x^2 ) is always positive.

  • Mistake: Assuming energy is always conserved.

  • Correction: Energy is only conserved if no non-conservative forces (e.g., friction) do work. If friction is present, mechanical energy decreases.

  • Mistake: Mixing up ( KE ) and ( U_s ) formulas.

  • Correction:
    • ( KE ) depends on velocity (( v^2 )).
    • ( U_s ) depends on displacement squared (( x^2 )).

AP Exam Insights

  1. Tricky Distinction: ( U_g ) vs. ( U_s )
  2. Gravitational PE (( U_g )) depends on height (( mgh )).
  3. Elastic PE (( U_s )) depends on stretch/compression (( \frac{1}{2}kx^2 )).
  4. Exam trap: A problem might show a spring held at a height—you’ll need both ( U_g ) and ( U_s )!

  5. FRQ Favorite: Energy Bar Charts

  6. The AP exam loves asking you to draw or interpret energy bar charts (initial vs. final energy distributions).
  7. Example: A ball is dropped onto a spring. Sketch bars for ( KE ), ( U_g ), and ( U_s ) at:

    • (a) Release point (max ( U_g ), zero ( KE ), ( U_s )).
    • (b) Lowest point (zero ( U_g ), zero ( KE ), max ( U_s )).
  8. Multiple-Choice Trap: Units & Variables

  9. Watch for units: ( k ) (spring constant) is in N/m, not N/cm.
  10. Watch for hidden variables: A problem might give you mass but ask for velocity—you’ll need to use energy conservation.

  11. Non-Conservative Forces in FRQs

  12. If a problem mentions friction or air resistance, you must account for lost energy: ( KE_i + PE_i = KE_f + PE_f + |W_{nc}| ).
  13. Example: A block slides down a ramp with friction. The final ( KE ) will be less than ( mgh ).

Quick Check Questions

  1. Multiple Choice: A 0.5 kg ball is dropped from a height of 10 m. What is its speed just before hitting the ground? (Ignore air resistance.)
  2. (A) 5 m/s
  3. (B) 10 m/s
  4. (C) 14 m/s
  5. (D) 20 m/s

Answer: (C) 14 m/s. Explanation: Use ( mgh = \frac{1}{2}mv^2 )? ( v = \sqrt{2gh} = \sqrt{2(9.8)(10)} \approx 14 ) m/s.

  1. Short FRQ: A spring with ( k = 100 ) N/m is compressed 0.2 m and used to launch a 0.1 kg block horizontally. What is the block’s speed when it leaves the spring? Answer: 6.3 m/s. Explanation: ( \frac{1}{2}kx^2 = \frac{1}{2}mv^2 )? ( v = x\sqrt{\frac{k}{m}} = 0.2\sqrt{\frac{100}{0.1}} = 6.3 ) m/s.

  2. Multiple Choice: A roller coaster car (mass 500 kg) starts from rest at a height of 40 m. If 20% of its energy is lost to friction, what is its speed at the bottom?

  3. (A) 18 m/s
  4. (B) 25 m/s
  5. (C) 28 m/s
  6. (D) 32 m/s

Answer: (B) 25 m/s. Explanation: Initial ( U_g = mgh = 500(9.8)(40) = 196,000 ) J. 80% remains? ( 0.8(196,000) = \frac{1}{2}(500)v^2 )? ( v = 25 ) m/s.


Last-Minute Cram Sheet

  1. ( KE = \frac{1}{2}mv^2 ) – Depends on velocity squared.
  2. ( U_g = mgh ) – Depends on height; define reference level!
  3. ( U_s = \frac{1}{2}kx^2 ) – Depends on displacement squared.
  4. Conservation of energy: ( KE_i + PE_i = KE_f + PE_f ) only if no friction/air resistance.
  5. Work-energy theorem: ( W = \Delta KE ).
  6. Hooke’s Law: ( F_s = -kx ) (negative sign = restoring force).
  7. Power: ( P = \frac{W}{t} = Fv ).
  8. Non-conservative forces (friction) reduce mechanical energy.
  9. Always check units: ( k ) is in N/m, not N/cm.
  10. Energy bar charts are common on FRQs—practice drawing them!