Unit 3.3 | Conservation of Energy, Systems, and Problem Solving
- 22 minutes
- 3 formulas
- 23 questions
- 5 videos
Conservation of energy turns pages of force and kinematics math into three lines. This lesson shows you how to choose a system and solve any energy problem.
Unit 3 Breakdown
In this lesson:
- Explain the law of conservation of energy
- Choose a system (open or closed) and know why it matters
- Tell conservative forces from nonconservative forces
- Solve almost any energy problem with a 3 step framework
- The 4 problem types: no friction, friction, springs, and an outside push
Introduction
This lesson covers conservation of energy: the law itself, how to choose a system, and a 3 step framework that uses conservation of energy to solve almost any problem.
You know the types of energy (Unit 3.1) and how work moves energy in and out (Unit 3.2). This lesson ties them into one framework.
This is the most important lesson in the unit. Once it clicks, problems that would take a full page of FBDs and kinematics take three lines.
The Law of Conservation of Energy
Energy cannot be created or destroyed. It can only transfer between objects or transform from one type into another.
A falling ball trades potential energy for kinetic energy. A sliding box trades kinetic energy for thermal energy. In every case the total stays the same. Watch it happen for a [katex] 1\,\text{kg} [/katex] ball dropped from [katex] 10\,\text{m} [/katex]:
A 1 kg ball dropped from 10 m
Written as an equation, the energy at the start equals the energy at the end:
[katex] E_i = E_f [/katex]
That only holds if no energy enters or leaves what you are looking at. Which brings us to systems.
Systems: Open vs. Closed
A system is the object (or group of objects) you choose to analyze. Everything else is the surroundings.
- A closed (isolated) system: no outside force does work on it, so its total energy stays constant: [katex] E_i = E_f [/katex].
- An open system: an outside force does work on it, adding or removing energy.
For an open system, the outside work makes up the difference. This is the equation the whole unit has been building to:
[katex] E_i + W = E_f [/katex]
Positive outside work (a person pushing, a rocket engine) adds energy. Negative outside work (friction, air resistance) removes it.
Choosing a system
The system is your choice, and the choice changes how you describe the problem. Take a falling ball:
Both descriptions give the same answer. You’ll almost always include Earth so you can use [katex] PE_g [/katex], but the AP exam likes to ask you to explain the difference.
▶ Watch: how to pick the system on AP questions (starts at 4:20).
Conservative vs. Nonconservative Forces
| Conservative forces | Nonconservative forces |
| Gravity, spring force | Friction, air resistance, applied forces (pushes, pulls, engines) |
| Energy is stored as potential energy and can be fully recovered | Energy is added from outside or turned into thermal energy |
| Work depends only on the start and end points | Work depends on the path (a longer path means more friction) |
If only conservative forces do work, mechanical energy is conserved: [katex] KE_i + PE_i = KE_f + PE_f [/katex]. As soon as friction or air resistance does work, some mechanical energy becomes thermal energy and you account for it with [katex] W [/katex].
Conceptual Practice
Check your understanding of systems and conserved energy before we start calculating.
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The 3 Step Energy Framework
Just like the framework for force problems, here is one for energy (the full write-up is here):
- Identify the energies. Moving ([katex] KE [/katex])? Changing height ([katex] PE_g [/katex])? Touching a spring ([katex] PE_s [/katex])? Is friction or a push doing work ([katex] W [/katex])?
- Mark two points: a start A and an end B. Pick points where you know the most, or the point the question asks about.
- Apply conservation of energy: [katex] E_A + W = E_B [/katex]. Substitute each equation, then solve for the unknown.
Before solving, sketch a quick energy bar chart for points A and B. If the bars don’t add up to the same total, an outside force must have done work.
Every energy problem on the AP exam runs through the same framework. What changes is what is in it. These are the four types you will see:
| Problem type | The clue | The equation |
| 1. No friction | “frictionless”, “negligible air resistance” | [katex] KE_A + PE_A = KE_B + PE_B [/katex] |
| 2. Friction takes energy | “rough”, “friction”, “lost to air resistance” | [katex] E_A + W_f = E_B [/katex], [katex] W_f \lt 0 [/katex] |
| 3. Springs | a spring launches or stops something | [katex] \frac{1}{2}kx^2 [/katex] at the squeezed point |
| 4. An outside push | an engine, thrust, a person pushing, a pulley system | [katex] E_A + W = E_B [/katex], [katex] W \gt 0 [/katex] |
Problem Type 1: No friction: find a speed or a height
How to spot it: “Frictionless”, “negligible air resistance”. Ramps, drops, pendulums, roller coasters.
Setup: Mechanical energy is conserved: [katex] KE_A + PE_A = KE_B + PE_B [/katex]. Mass usually cancels.
A skateboarder starts from rest at the top of a frictionless ramp [katex] 3\,\text{m} [/katex] tall. How fast is the skateboarder moving at the bottom?
Identify: [katex] PE_g [/katex] at the top, [katex] KE [/katex] at the bottom, no outside work. Mark: A at the top, B at the bottom. Apply:
[katex] PE_A = KE_B \rightarrow mgh = \frac{1}{2}mv^2 \rightarrow v = \sqrt{2gh} = \sqrt{2(9.8)(3)} \approx 7.7 \, m/s [/katex]
The mass canceled out, and the ramp’s shape never mattered: straight, curved, or loopy, only the height drop counts. Try doing that with Newton’s laws!
AP trap: Reaching for kinematics on a curved track. It only works on straight paths. Energy only needs the height drop.
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Problem Type 2: Friction takes energy
How to spot it: “Rough surface”, “coefficient of friction”, “energy lost to air resistance”, or a final speed lower than no-friction would give.
Setup: [katex] E_A + W_f = E_B [/katex], where [katex] W_f = -F_f d [/katex] is negative: friction removes energy as heat.
A [katex] 50\,\text{kg} [/katex] sled starts from rest at the top of a [katex] 10\,\text{m} [/katex] tall hill. It reaches the bottom moving at [katex] 12\,\text{m/s} [/katex]. How much energy did friction remove?
Identify: [katex] PE_g [/katex] at the top, [katex] KE [/katex] at the bottom, friction doing work along the way. Apply: [katex] PE_A + W_f = KE_B [/katex], so [katex] W_f = KE_B – PE_A [/katex].
- [katex] PE_A = (50)(9.8)(10) = 4900 \, J [/katex]
- [katex] KE_B = \frac{1}{2}(50)(12)^2 = 3600 \, J [/katex]
- [katex] W_f = 3600 – 4900 = -1300 \, J [/katex]
Answer: friction removed [katex] 1300 \, J [/katex], which became thermal energy in the sled and the snow.
Show Answer: What was the average friction force if the slope was [katex] 25\,\text{m} [/katex] long?
[katex] W_f = -F_f d [/katex], so [katex] -1300 = -F_f(25) [/katex] and [katex] F_f = 52 \, N [/katex].
▶ Watch: a worked example of mechanical energy lost to friction (starts at 10:53).
AP trap: Giving the friction work a plus sign. Friction always removes mechanical energy, so [katex] W_f [/katex] is negative.
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Problem Type 3: Springs: launch or stop
How to spot it: A compressed spring launches something, or something runs into a spring and squeezes it.
Setup: Put [katex] \frac{1}{2}kx^2 [/katex] in at the squeezed point and [katex] \frac{1}{2}mv^2 [/katex] where it moves: [katex] \frac{1}{2}kx^2 = \frac{1}{2}mv^2 [/katex] (add [katex] mgh [/katex] or friction if the problem has them).
A spring with [katex] k = 400 \, N/m [/katex] is compressed [katex] 0.1\,\text{m} [/katex] and launches a [katex] 0.5\,\text{kg} [/katex] block across a frictionless floor. How fast does the block leave the spring?
[katex] \frac{1}{2}(400)(0.1)^2 = \frac{1}{2}(0.5)v^2 \rightarrow 2 = 0.25v^2 \rightarrow v \approx 2.8 \, m/s [/katex]
AP trap: Using the length of the spring for [katex] x [/katex]. [katex] x [/katex] is how far it is squeezed or stretched from its relaxed length.
Try this one before you watch: a spring launch, then a rough incline.
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▶ Watch: the framework solving this exact problem (starts at 16:36).
Problem Type 4: An outside push adds energy
How to spot it: An engine, a rocket’s thrust, a person pushing, or two objects tied together over a pulley.
Setup: [katex] E_A + W = E_B [/katex], where [katex] W [/katex] is positive for a push along the motion. In a pulley system, put every moving object’s kinetic energy in the sum.
AP trap: Leaving out one object in a pulley problem. Both blocks move, so both have kinetic energy.
Try the rocket before you watch: thrust adds energy while air resistance takes some away.
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▶ Watch: the framework solving this exact problem (starts at 12:43).
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Lesson 3.4 Preview
In the final lesson of this unit we cover power: not just how much energy is transferred, but how fast.
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This article was written by Jason