Conservation of energy — AP Physics C: Mechanics Study Guide

For: AP Physics C: Mechanics candidates sitting AP Physics C: Mechanics.

Covers: Conservative vs non-conservative forces, the general and isolated-system forms of the conservation of energy equation, energy diagrams, equilibrium classification, and problem-solving techniques for motion with changing mechanical energy.

You should already know: Work done by a constant and variable force. Definition of kinetic and gravitational/elastic potential energy. The work-energy theorem for a single particle.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics C: Mechanics style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Conservation of energy?

Conservation of energy is a fundamental law of classical mechanics that states energy cannot be created or destroyed, only converted between different forms. For AP Physics C: Mechanics, this topic is part of Unit 3 Work, Energy, and Power, which makes up ~14-18% of the total exam score per the official CED, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections. Energy conservation is often combined with kinematics, forces, or momentum to solve motion problems that would be far more complex using only Newton’s second law.

Notation conventions used in this chapter: $E_{tot}$ for total energy, $K$ for kinetic energy, $U$ for potential energy, $E_{mech}=K+U$ for total mechanical energy, $W_{nc}$ for work done by non-conservative forces, and $E_{int}$ for internal energy (heat) from dissipative forces. A common synonym used on the exam is "energy conservation", which always refers to this law in a mechanics context. In AP Mechanics, we almost always focus on accounting for mechanical energy, with internal energy only counted when dissipative forces are present.

2. Conservative vs Non-Conservative Forces

A conservative force is a force where the work done moving an object between two points depends only on the starting and ending positions, not the path taken between them. Equivalently, the work done by a conservative force around any closed path (starting and ending at the same point) is exactly zero. This path-independence property allows us to define a unique potential energy function $U$ associated with any conservative force, related by $W_c=-\Delta U$, where $W_c$ is the work done by the conservative force. Common conservative forces tested on the AP exam are near-Earth gravity, universal gravity, and the elastic spring force.

Non-conservative forces have work done that depends explicitly on the path taken between two points, and you cannot define a potential energy function for them. Most non-conservative forces on the exam are dissipative: kinetic friction, air resistance, and rolling friction, which convert mechanical energy to unrecoverable internal heat. Non-dissipative non-conservative forces include applied forces from a person, tension in an external rope, or any external force that adds or removes energy from the system. The key practical distinction is: conservative forces store energy as potential energy that can be converted back to kinetic, while non-conservative forces change the total mechanical energy of the system.

Worked Example

A 2 kg block slides from point A to point B along two different paths on a 30° incline: Path 1 is straight 5 m long, Path 2 is a curved 12 m long path between the same two endpoints. The coefficient of kinetic friction between the block and the surface is 0.2 for both paths. Calculate the work done by gravity and work done by friction along each path, and confirm the classification of each force.

1. First, find the vertical height difference between A and B, which is identical for both paths: $\Delta h=5\sin 30^{\circ }=2.5$ m.
2. Work done by gravity (conservative) depends only on height change: $W_g=mg\Delta h=(2kg)(9.8m/s^2)(2.5m)=49J$ for both paths, confirming path-independence.
3. Work done by friction depends on path length: $W_f=-{\mu }_{k}Nd=-{\mu }_{k}mg\cos 30^{\circ }d$. For Path 1 ($d=5m$): $W_f=-0.2(17N)(5m)=-17J$.
4. For Path 2 ($d=12m$): $W_f=-0.2(17N)(12m)=-40.8J$. Friction does more work over the longer path, confirming it is non-conservative, matching the definitions.

Exam tip: Always sort all forces acting on your system into conservative or non-conservative before writing any energy equation — this immediately tells you which forces go into potential energy terms and which must be counted as $W_{nc}$.

3. The General Conservation of Energy Equation

The conservation of energy equation is derived directly from the work-energy theorem, which states that total work done on a system equals the change in kinetic energy: $W_{total}=\Delta K$. We split total work into work done by conservative forces and non-conservative forces: $W_c+W_{nc}=\Delta K$. We already know $W_c=-\Delta U$ from the definition of potential energy, so substituting gives: $$-\Delta U+W_{nc}=\Delta K$$ Rearranging to get the general energy conservation equation that works for any system: $$W_{nc}=\Delta K+\Delta U=(K_f+U_f)-(K_i+U_i)=\Delta E_{mech}$$ This is the most useful form for AP problems: the net work done by all non-conservative forces equals the change in total mechanical energy of the system. For a closed (isolated) system with no non-conservative work done, $W_{nc}=0$, so this simplifies to the familiar conservation of mechanical energy: $$K_i+U_i=K_f+U_f$$ If dissipative non-conservative forces like friction are present, $W_{nc}$ is negative, meaning $\Delta E_{mech}$ is negative — mechanical energy is converted to internal energy, so total energy (including internal) is still conserved: $K+U+E_{int}=\text{constant}$.

Worked Example

A 0.5 kg block is pushed against a horizontal spring with spring constant $k=400N/m$, compressing it 0.1 m from equilibrium. The block is released from rest, and slides along a horizontal surface with coefficient of kinetic friction ${\mu }_k=0.3$. How far does the block travel from the spring’s equilibrium position (release point) before coming to rest?

1. Define the system as block + spring + Earth. Initial state (compressed spring): $K_i=0$, $U_i=\frac{1}{2}k{x}^2=0.5(400)(0.1{)}^2=2J$, gravitational potential energy is constant so it cancels out. Final state: $K_f=0$, $U_f=0$.
2. Work done by non-conservative friction is $W_{nc}=-f_{k}d=-{\mu }_{k}mgd$, where $d$ is the unknown distance.
3. Substitute into the general energy equation: $-{\mu }_{k}mgd=(0+0)-(0+2J)=-2J$.
4. Solve for $d$: $d=\frac{2J}{{\mu }_{k}mg}=\frac{2}{(0.3)(0.5)(9.8)}\approx 1.36m$.

Exam tip: Always explicitly write out your initial and final states before plugging into the equation — this catches common mistakes like forgetting to include potential energy at the final state or double-counting work done by gravity.

4. Energy Diagrams and Equilibrium

For an object moving along one dimension with a known potential energy function $U(x)$ and constant total energy (isolated system), we can use an energy diagram to quickly analyze the motion without solving any differential equations. An energy diagram plots $U(x)$ as a function of position $x$, with a horizontal line for total energy $E_{tot}$ (horizontal because it is constant).

From the relation between force and potential energy: $F(x)=-\frac{dU}{dx}$, so the force on the object at any point is the negative slope of the $U(x)$ graph. Equilibrium occurs when $F(x)=0$, so $\frac{dU}{dx}=0$ (slope of $U(x)$ is zero). We classify equilibrium by the curvature of $U(x)$:

• Stable equilibrium: Local minimum of $U(x)$, $\frac{d^{2}U}{d{x}^2}>0$: displacing the object creates a restoring force back to the equilibrium point.
• Unstable equilibrium: Local maximum of $U(x)$, $\frac{d^{2}U}{d{x}^2}<0$: displacing the object creates a force that pushes it further away.
• Neutral equilibrium: Flat $U(x)$, $\frac{d^{2}U}{d{x}^2}=0$: no force acts on the displaced object, so it stays in place.

Turning points of the motion occur where $E_{tot}=U(x)$, because kinetic energy $K=E_{tot}-U(x)=0$ at these points, so the object stops and reverses direction. Any region where $U(x)>E_{tot}$ is forbidden, since kinetic energy cannot be negative.

Worked Example

The potential energy of a 1 kg object moving along the x-axis is $U(x)=x^3-3x^2+2$, where $U$ is in joules and $x$ in meters. The total energy of the object is 2 J. Identify all equilibrium positions, classify their stability, and find the turning points of the motion.

1. Find equilibrium positions by setting $F=-\frac{dU}{dx}=0$: $\frac{dU}{dx}=3x^2-6x=3x(x-2)=0$, so equilibrium at $x=0m$ and $x=2m$.
2. Classify stability using the second derivative: $\frac{d^{2}U}{d{x}^2}=6x-6$. At $x=0$: $\frac{d^{2}U}{d{x}^2}=-6<0$, so this is a local maximum → unstable equilibrium. At $x=2$: $\frac{d^{2}U}{d{x}^2}=6>0$, local minimum → stable equilibrium.
3. Find turning points where $E_{tot}=U(x)=2J$: $x^3-3x^2+2=2\to x^2(x-3)=0$, so turning points at $x=0m$ and $x=3m$.
4. For $0<x<3$, $U(x)<2J$, so $K>0$: the motion is bounded between the two turning points, which matches our result.

Exam tip: If the problem asks for the maximum speed of the object in an energy diagram, remember maximum speed occurs at the position of minimum potential energy (since $K=E_{tot}-U$), which is always the stable equilibrium point.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Counting the work done by gravity and including gravitational potential energy in the energy equation. Why: Students confuse the two approaches to energy problems; if you treat gravity as an external force doing work, you do not include it in potential energy, and vice versa. Correct move: Always assign all conservative forces to potential energy terms, and only non-conservative forces go into $W_{nc}$ to avoid double-counting.
• Wrong move: Calculating work done by friction over the straight-line distance between start and end, not the total path length. Why: Students extend gravity’s path-independence to friction incorrectly. Correct move: For any non-conservative force, always use the total distance traveled along the actual path when calculating work.
• Wrong move: Calling a local maximum of $U(x)$ a stable equilibrium because the slope is zero. Why: Students mix up slope and curvature when classifying stability. Correct move: Always use curvature to classify: concave up (second derivative positive) = stable, concave down (second derivative negative) = unstable.
• Wrong move: Using the near-Earth gravitational potential $U=mgh$ for orbital motion problems with large changes in distance from Earth’s center. Why: Students memorize the simpler formula and overapply it. Correct move: Use the universal gravitational potential $U(r)=-GMm/r$ for any problem where the object moves more than a few kilometers from Earth’s surface.
• Wrong move: Leaving the negative root when solving for speed from kinetic energy. Why: Students confuse velocity direction (which can be negative) with speed (which is always positive). Correct move: Always take the positive square root when asked for speed; keep the negative root only if asked for a velocity component.
• Wrong move: Forgetting that friction does negative work, so it reduces the total mechanical energy of the system. Why: Students forget to add the negative sign when calculating $W_{nc}$ for friction. Correct move: Explicitly add a negative sign to work done by kinetic friction whenever you write the energy equation.

6. Practice Questions (AP Physics C: Mechanics Style)

Question 1 (Multiple Choice)

A 500 kg roller coaster car starts from rest at the top of a 40 m tall hill, rolls down the hill, and enters a vertical loop-the-loop with radius 15 m. There is a constant kinetic friction force of 100 N acting on the car throughout the motion, and the total distance traveled from the top of the hill to the top of the loop is 100 m. What is the approximate speed of the car at the top of the loop? (Take $g=10m/s^2$ for simplicity.) A) 10 m/s B) 13 m/s C) 20 m/s D) 25 m/s

Worked Solution: We use the general conservation of energy equation $W_{nc}=(K_f+U_f)-(K_i+U_i)$. Initial kinetic energy $K_i=0$, initial potential energy $U_i=mg{h}_i=500\ast 10\ast 40=200000J$. At the top of the loop, height is $2r=30m$, so $U_f=500\ast 10\ast 30=150000J$, $K_f=\frac{1}{2}m{v}^2=250v^2$. Work done by friction is $W_{nc}=-100\ast 100=-10000J$. Substitute: $-10000=(250v^2+150000)-200000$, which simplifies to $250v^2=40000$, so $v^2=160$, $v\approx 12.6m/s$, which is closest to 13 m/s. Correct answer: B.

Question 2 (Free Response)

A 2 kg block is attached to a vertical ideal spring with spring constant $k=200N/m$. The spring is uncompressed when the block is at $y=0m$, with upward defined as the positive $y$ direction. The block is released from rest at $y=0.5m$. (a) Using conservation of energy, derive an expression for the speed of the block as a function of position $y$. (b) Find the maximum speed of the block, and the position where it occurs. (c) Find the maximum displacement of the block below $y=0$ at its lowest point.

Worked Solution: (a) Define the system as block + spring + Earth, no non-conservative work so $E_i=E_f$. Initial energy: $K_i=0$, $U_{elastic,i}=\frac{1}{2}k{y}_0^2$, $U_{gravity,i}=mg{y}_0$ where $y_0=0.5m$. For any position $y$: $E_f=\frac{1}{2}m{v}^2+\frac{1}{2}k{y}^2+mgy$. Equate and solve for $v$: $$\frac{1}{2}m{v}^2=\frac{1}{2}k{y}_0^2+mg{y}_0-\frac{1}{2}k{y}^2-mgy$$ $$v(y)=\sqrt{\frac{1}{m}\left(k(y_0^2-y^2)+2mg(y_0-y)\right)}$$

(b) Maximum speed occurs at equilibrium where net force is zero, so $\frac{dU}{dy}=0\to ky+mg=0\to y=-mg/k=-(2\ast 9.8)/200\approx -0.10m$. Substitute into $v(y)$: $v_{max}=\sqrt{\frac{1}{2}(200(0.5^2-0.1^2)+2\ast 2\ast 9.8(0.5+0.1))}\approx 6.0m/s$.

(c) Maximum displacement occurs when $v=0$, so set the expression inside the square root to zero: $(y_0-y)(k(y_0+y)+2mg)=0$. The non-trivial solution gives $y=-(2mg/k+y_0)\approx -0.70m$, so the maximum displacement below $y=0$ is 0.70 m.

Question 3 (Application / Real-World Style)

A cyclist and bike have a combined mass of 80 kg. The cyclist starts from rest at the top of a 3.0 km long road that descends 500 m vertically from start to finish. The cyclist does not pedal, and maintains a constant speed of 8 m/s throughout the entire descent. What is the average power dissipated by air resistance and rolling friction for this trip?

Worked Solution: Since speed is constant, $\Delta K=0$. The change in gravitational potential energy is $\Delta U=U_f-U_i=-mg\Delta h=-(80kg)(9.8m/s^2)(500m)=-392000J$. No work is done by the cyclist, so all lost potential energy is dissipated by non-conservative friction: $W_{nc}=\Delta K+\Delta U=-392000J$. The total time of the trip is $t=d/v=3000m/8m/s=375s$. Average power is $P=|W_{nc}|/t=392000J/375s\approx 1050W=1.05kW$. In context: This matches real-world values for a cyclist descending a steep hill at constant speed, with roughly one kilowatt of power lost to air resistance and friction.

7. Quick Reference Cheatsheet

8. What's Next

This chapter is the foundation for all energy-based problem solving in AP Physics C: Mechanics. Immediately after mastering energy conservation, you will apply it to power calculations, collisions, and orbital motion, where it simplifies problems that would be very difficult to solve with only Newton’s laws. Without mastering the accounting of energy and the distinction between conservative and non-conservative forces, you will not be able to correctly solve multi-part FRQ problems that combine energy with other topics like momentum or circular motion. This topic also feeds into the broader theme of conservation laws, which unify all of classical physics, and allow you to avoid complex integration for many motion problems. Follow-on topics to study next: Power and variable force work Conservation of linear momentum Gravitational energy and orbital motion