# Conservation of Energy

> AP Physics C: Mechanics · Unit 3: Work, Energy, and Power
> Source: https://www.owlsprep.com/study/ap-physics-c-mech-u3-conservation-of-energy/

This guide covers the law of conservation of energy for AP Physics C: Mechanics, including conservative/non-conservative forces, the energy conservation equation, energy diagrams, equilibrium classification, and problem-solving techniques tested on the AP exam.

**Prerequisites:** [Work done by constant and variable forces](https://www.owlsprep.com/study/ap-physics-c-mech-u3-work/); [Kinetic and potential energy definitions](https://www.owlsprep.com/study/ap-physics-c-mech-u3-kinetic-potential-energy/); [Work-energy theorem](https://www.owlsprep.com/study/ap-physics-c-mech-u3-work-energy-theorem/)

## Learning objectives

- Distinguish between conservative and non-conservative forces
- Apply the general and isolated system forms of conservation of energy to solve motion problems
- Interpret energy diagrams and classify equilibrium points
- Avoid common exam pitfalls when solving energy conservation problems

## Conservative vs Non-Conservative Forces

A conservative force is defined by path independence: the work it does moving an object between two points depends only on the start and end positions, not the path taken. Equivalently, work done by a conservative force around any closed path is zero. This property allows us to define a unique potential energy function $U$ for any conservative force.

**Conservative Force** — A force where work done between two points depends only on endpoints, not the path taken, allowing a potential energy function to be defined.

*Notation:* $W_c = -\Delta U$

*Example:* Gravity, elastic spring force, universal gravitation

**Non-Conservative Force** — A force where work done between two points depends on the path taken, so no potential energy function can be defined. Most non-conservative forces dissipate mechanical energy as internal heat.

*Example:* Kinetic friction, air resistance, applied external forces

**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, identical for both paths:

   $$\Delta h = 5 \sin 30^\circ = 2.5 \text{ m}$$
2. Work done by gravity (conservative) depends only on height change, so it is the same for both paths:

   $$W_g = mg\Delta h = (2 \text{ kg})(9.8 \text{ m/s}^2)(2.5 \text{ m}) = 49 \text{ J}$$
3. Work done by friction depends on path length, and is always negative because friction opposes motion. Normal force on the incline is $N = mg \cos 30^\circ \approx 17 \text{ N}$, so:

   $$W_f = -\mu_k N d = -\mu_k mg \cos 30^\circ d$$
4. For Path 1 ($d=5$ m):

   $$W_f = -0.2(17 \text{ N})(5 \text{ m}) = -17 \text{ J}$$
5. For Path 2 ($d=12$ m):

   $$W_f = -0.2(17 \text{ N})(12 \text{ m}) = -40.8 \text{ J}$$
6. Friction does more work over the longer path, confirming it is non-conservative, matching our definitions.

> **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 $W_{nc}$.

## 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: $W_c + W_{nc} = \Delta K$. Substituting $W_c = -\Delta U$ (from the definition of potential energy) gives the general 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 ($E_{mech} = K + U$) of the system. For an isolated (closed) system with no non-conservative work, $W_{nc} = 0$, so mechanical energy is conserved:

$$K_i + U_i = K_f + U_f$$

When dissipative forces like friction are present, $W_{nc}$ is negative, so $\Delta E_{mech}$ is negative. Mechanical energy is converted to internal (heat) energy, but total energy (including internal) remains conserved: $K + U + E_{int} = \text{constant}$.

**Worked example:** A 0.5 kg block is pushed against a horizontal spring with spring constant $k=400 N/m$, 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. Gravitational potential energy is constant, so it cancels out. Initial state (compressed spring, at rest):

   $$K_i = 0, \quad U_i = \frac{1}{2}kx^2 = 0.5(400)(0.1)^2 = 2 \text{ J}$$
2. Final state (block at rest, spring at equilibrium):

   $$K_f = 0, \quad U_f = 0$$
3. Work done by non-conservative friction is negative, equal to:

   $$W_{nc} = -f_k d = -\mu_k mg d$$
4. Substitute into the general energy equation:

   $$-\mu_k mg d = (0 + 0) - (0 + 2 \text{ J}) = -2 \text{ J}$$
5. Solve for the unknown distance $d$:

   $$d = \frac{2 \text{ J}}{\mu_k mg} = \frac{2}{(0.3)(0.5)(9.8)} \approx 1.36 \text{ m}$$

> **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.

## Energy Diagrams and Equilibrium Classification

For an object moving in one dimension with a known potential energy function $U(x)$ and constant total energy (isolated system), an energy diagram plots $U(x)$ vs position $x$, with a horizontal line for constant total energy $E_{tot}$. This lets you analyze motion without solving differential equations. The relation between force and potential energy gives:

$$F(x) = -\frac{dU}{dx}$$

Force on the object is the negative slope of the $U(x)$ graph. Equilibrium occurs when $F(x) = 0$, so the slope of $U(x)$ is zero ($\frac{dU}{dx} = 0$). Equilibrium is classified by the curvature (second derivative) of $U(x)$:

- **Stable equilibrium**: Local minimum of $U(x)$, $\frac{d^2U}{dx^2} > 0$. Displacement creates a restoring force back to equilibrium.
- **Unstable equilibrium**: Local maximum of $U(x)$, $\frac{d^2U}{dx^2} < 0$. Displacement creates a force that pushes the object further away.
- **Neutral equilibrium**: Flat $U(x)$, $\frac{d^2U}{dx^2} = 0$. No force acts on a displaced object.

Turning points of 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 $F = -\frac{dU}{dx} = 0$, so $\frac{dU}{dx} = 0$:

   $$\frac{dU}{dx} = 3x^2 - 6x = 3x(x-2) = 0$$
2. Equilibrium occurs at $x=0 \text{ m}$ and $x=2 \text{ m}$. Classify stability using the second derivative:

   $$\frac{d^2U}{dx^2} = 6x - 6$$
3. At $x=0$: $\frac{d^2U}{dx^2} = -6 < 0$, so this is a local maximum → unstable equilibrium. At $x=2$: $\frac{d^2U}{dx^2} = 6 > 0$, so this is a local minimum → stable equilibrium.
4. Find turning points where $E_{tot} = U(x) = 2 \text{ J}$:

   $$x^3 - 3x^2 + 2 = 2 \rightarrow x^2(x-3) = 0$$
5. Turning points occur at $x=0 \text{ m}$ and $x=3 \text{ m}$. For $0 < x < 3$, $U(x) < 2 \text{ J}$, so $K>0$, meaning motion is bounded between the two turning points.

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

## AP-Style Worked Practice Examples

**Worked example:** 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? (Use $g=10 \text{ m/s}^2$ for simplicity.)

1. Use the general conservation of energy equation $W_{nc} = (K_f + U_f) - (K_i + U_i)$. Initial state (car at rest): $K_i=0$, initial potential energy:

   $$U_i = mgh_i = 500 \cdot 10 \cdot 40 = 200000 \text{ J}$$
2. Final state (car at top of loop, height $2r = 30 \text{ m}$): $U_f = 150000 \text{ J}$, $K_f = \frac{1}{2}mv^2 = 250v^2$. Work done by friction:

   $$W_{nc} = -100 \cdot 100 = -10000 \text{ J}$$
3. Substitute and solve for $v$:

   $$-10000 = (250v^2 + 150000) - 200000 \\ 250v^2 = 40000 \\ v \approx 12.6 \text{ m/s}$$
4. This result is closest to 13 m/s, the correct answer.

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

1. (a) No non-conservative work, so initial energy equals final energy: $K_i + U_{i,elastic} + U_{i,gravity} = K_f + U_{f,elastic} + U_{f,gravity}$

   $$v(y) = \sqrt{\frac{1}{m}\left(k(y_0^2 - y^2) + 2mg(y_0 - y)\right)}$$
2. (b) Maximum speed occurs at equilibrium where net force is zero:

   $$y = -\frac{mg}{k} \approx -0.10 \text{ m}, \quad v_{max} \approx 6.0 \text{ m/s}$$
3. (c) Maximum displacement occurs when $v=0$. Solving for the non-trivial root gives:

   $$y \approx -0.70 \text{ m}$$
4. So the maximum displacement below $y=0$ is 0.70 m.

## Common pitfalls

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

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Conservative force-work relation | $W_c = -\Delta U$ | Applies to all conservative forces; work done = negative change in potential energy |
| General conservation of energy | $W_{nc} = \Delta K + \Delta U = \Delta E_{mech}$ | Applies to all systems; $W_{nc}$ = net work done by non-conservative forces |
| Isolated system mechanical energy conservation | $K_i + U_i = K_f + U_f$ | Applies when $W_{nc}=0$, no non-conservative work done on the system |
| Force from potential energy (1D) | $F(x) = -\frac{dU}{dx}$ | Force equals negative slope of the $U(x)$ energy diagram |
| Equilibrium condition | $\frac{dU}{dx} = 0$ | Equilibrium occurs where net force from potential energy is zero |
| Stable equilibrium | $\frac{d^2U}{dx^2} > 0$ | Occurs at local minimum of $U(x)$; displacement creates restoring force |
| Unstable equilibrium | $\frac{d^2U}{dx^2} < 0$ | Occurs at local maximum of $U(x)$; displacement pushes object away |
| Turning point condition | $E_{tot} = U(x)$ | Kinetic energy $K=0$ at turning points; object reverses direction |
| Total energy with internal energy | $K + U + E_{int} = \text{constant}$ | Accounts for heat energy lost to dissipative non-conservative forces |

## What's next

Conservation of energy is one of the most powerful problem-solving tools in AP Physics C: Mechanics, and it will be used repeatedly in subsequent topics including rotational motion, oscillations, and orbital mechanics. Mastery of energy conservation allows you to solve complex motion problems much faster than applying Newton’s laws directly, especially when forces are variable or you only need information about initial and final states rather than acceleration as a function of time. Next, you will build on this foundation to study power, which is the rate of energy transfer, and then apply energy conservation to rotational systems and simple harmonic motion.

- [Power](https://www.owlsprep.com/study/ap-physics-c-mech-u3-power/)
- [Systems of Particles and Linear Momentum Overview](https://www.owlsprep.com/study/ap-physics-c-mech-u4-overview/)
- [Center of Mass](https://www.owlsprep.com/study/ap-physics-c-mech-u4-center-of-mass/)

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