⚡ Conservation of Energy — No Friction

Part 1 of 7 — Conservation of Energy

The Law of Conservation of Energy is one of the most powerful tools in physics. When only conservative forces (gravity, springs) act on a system, total mechanical energy is conserved.

In this lesson you will learn:

• The conservation of energy principle
• The equation $KE_i + PE_i = KE_f + PE_f$
• How to apply it to frictionless systems

The Conservation Principle

Total Mechanical Energy

$E_{\text{mech}} = KE + PE$

$E_{\text{mech}} = \frac{1}{2}mv^2 + mgh + \frac{1}{2}kx^2$

When no non-conservative forces (friction, air resistance, applied push/pull) do work:

$E_{\text{mech},i} = E_{\text{mech},f}$

$KE_i + PE_i = KE_f + PE_f$

What "Conserved" Means

• Energy is not created or destroyed
• Energy transforms between types (KE ↔ PE)
• The total stays the same throughout the motion

Key Conditions

Conceptual Check 🎯

Worked Example: Free Fall

A 2 kg ball is dropped from 5 m. Find its speed just before hitting the ground ($g = 10$ m/s²).

Define: $i$ = top, $f$ = bottom. Set $h = 0$ at bottom.

$KE_i + PE_i = KE_f + PE_f$

$0 + mgh = \frac{1}{2}mv^2 + 0$

$mgh = \frac{1}{2}mv^2$

$v = \sqrt{2gh} = \sqrt{2(10)(5)} = \sqrt{100} = 10 \text{ m/s}$

Notice: mass canceled! The speed is independent of mass.

Practice Calculations 🧮

Use $g = 10$ m/s².

1. A 3 kg ball is dropped from 20 m. What is its speed just before hitting the ground (in m/s)?

2. A ball is thrown upward at 15 m/s. What maximum height does it reach (in m)?

3. A roller coaster car starts from rest at 45 m. What is its speed at a height of 20 m (in m/s)?

Round all answers to 3 significant figures.

Energy Transformations 🔍

Exit Quiz ✅

🔧 Non-Conservative Forces & Work-Energy

Part 2 of 7 — Conservation of Energy

Real-world systems often involve non-conservative forces like friction, air resistance, and applied forces. These forces add or remove energy from the system.

In this lesson you will learn:

• The generalized energy equation with $W_{\text{nc}}$
• How friction converts mechanical energy to thermal energy
• How applied forces add energy to a system

The Generalized Energy Equation

When non-conservative forces do work on the system:

$KE_i + PE_i + W_{\text{nc}} = KE_f + PE_f$

Or equivalently:

$W_{\text{nc}} = \Delta KE + \Delta PE = \Delta E_{\text{mech}}$

Types of Non-Conservative Forces

Friction Specifically

$W_f = -f_k d = -\mu_k mg d$

The negative sign means friction always removes mechanical energy, converting it to thermal energy:

$E_{\text{thermal}} = f_k d = \mu_k mg d$

Conceptual Check 🎯

Worked Example: Friction on a Ramp

A 4 kg block slides down a 5 m ramp (height 3 m) with $\mu_k = 0.2$. Find the speed at the bottom ($g = 10$ m/s²).

Step 1: Identify energies

• $KE_i = 0$ (starts from rest)
• $PE_i = mgh = 4(10)(3) = 120$ J
• $KE_f = \frac{1}{2}mv^2$ (find this)
• $PE_f = 0$ (at bottom)

Step 2: Find friction work

• Normal force on ramp: $N = mg\cos\theta$
• $\cos\theta = 4/5$ (from the triangle: base 4, height 3, hyp 5)
• $W_f = -\mu_k mg\cos\theta \cdot d = -0.2(4)(10)(0.8)(5) = -32$ J

Step 3: Apply the equation

$KE_i + PE_i + W_f = KE_f + PE_f$

$0 + 120 + (-32) = \frac{1}{2}(4)v^2 + 0$

$88 = 2v^2$

$v = \sqrt{44} \approx 6.63 \text{ m/s}$

Without friction: $v = \sqrt{2(10)(3)} = \sqrt{60} \approx 7.75$ m/s. Friction reduced the speed!

Practice with Non-Conservative Forces 🧮

Use $g = 10$ m/s².

1. A 5 kg block slides down from a height of 8 m on a frictionless ramp, then across a rough flat surface ($\mu_k = 0.4$) for 10 m. What is its speed (in m/s)? (Round to the nearest tenth.)

2. A person pushes a 10 kg box 5 m across a rough floor ($\mu_k = 0.3$) with a force of 50 N. Starting from rest, find the box's final speed (in m/s)? (Round to the nearest tenth.)

3. A 2 kg ball is thrown upward at 10 m/s. If air resistance does $-8$ J of work, what maximum height does it reach (in m)?

Identify $W_{\text{nc}}$ 🔍

Exit Quiz ✅

🌀 Springs and Gravity Combined

Part 3 of 7 — Conservation of Energy

Many AP problems combine gravitational PE, elastic PE, and kinetic energy in a single system. Mastering the full energy equation is essential.

In this lesson you will learn:

• How to combine $PE_g$ and $PE_s$ in one equation
• Vertical spring problems (drop onto a spring)
• Spring launcher + gravity problems

The Full Energy Equation

$\frac{1}{2}mv_i^2 + mgh_i + \frac{1}{2}kx_i^2 = \frac{1}{2}mv_f^2 + mgh_f + \frac{1}{2}kx_f^2$

Setting Up the Problem

1. Choose initial and final states — pick moments where you know the most variables
2. Choose a reference level for $h = 0$ — usually the lowest point
3. List known and unknown energy terms
4. Write the energy equation and solve

Common Combinations

Worked Example: Dropping onto a Spring

A 0.5 kg ball is dropped from 2 m above a vertical spring ($k = 500$ N/m). How far does the spring compress?

Reference: $h = 0$ at the top of the uncompressed spring.

The ball falls 2 m to reach the spring, then compresses it by $x$. Total fall = $2 + x$.

$mgh_{\text{total}} = \frac{1}{2}kx^2$

$0.5(10)(2 + x) = \frac{1}{2}(500)x^2$

$5(2 + x) = 250x^2$

$10 + 5x = 250x^2$

$250x^2 - 5x - 10 = 0$

$50x^2 - x - 2 = 0$

Using the quadratic formula:

$x = \frac{1 \pm \sqrt{1 + 400}}{100} = \frac{1 + 20.02}{100} \approx 0.21 \text{ m}$

(Take the positive root.)

Conceptual Check 🎯

Practice Calculations 🧮

Use $g = 10$ m/s².

1. A spring ($k = 200$ N/m) compressed 0.3 m launches a 0.5 kg ball straight up. What height does it reach (in m)?

2. A 1 kg block slides from a height of 2 m down a frictionless ramp onto a spring ($k = 500$ N/m). What is the maximum compression (in m)? (Round to nearest hundredth.)

3. A spring ($k = 800$ N/m) compressed 0.1 m launches a 0.4 kg ball up a frictionless $30°$ incline. How far along the incline does it travel (in m)?

Energy Identification 🔍

Exit Quiz ✅

🎢 Roller Coasters & Pendulums

Part 4 of 7 — Conservation of Energy

Roller coasters and pendulums are classic conservation of energy scenarios that appear frequently on AP exams. Both involve continuous conversion between KE and gravitational PE.

In this lesson you will learn:

• Roller coaster energy analysis at multiple points
• Minimum speed at the top of a loop
• Pendulum energy problems
• Height-speed relationships along curved paths

Roller Coasters — Energy at Every Point

For a frictionless roller coaster starting from rest at height $h_0$:

$mgh_0 = \frac{1}{2}mv^2 + mgh$

$v = \sqrt{2g(h_0 - h)}$

Key Insights

• Speed depends only on height difference, not the path shape
• The lowest point has the most KE (and maximum speed)
• The highest reachable point is where $v = 0$ ($h = h_0$)

Minimum Speed at the Top of a Loop

At the top of a circular loop of radius $R$, the car needs enough speed to maintain contact:

$mg = \frac{mv^2_{\text{min}}}{R} \Rightarrow v_{\text{min}} = \sqrt{gR}$

To find the minimum starting height for a loop of radius $R$:

$mgh_{\text{min}} = \frac{1}{2}mv^2_{\text{min}} + mg(2R)$

$gh_{\text{min}} = \frac{1}{2}gR + 2gR = \frac{5}{2}gR$

$h_{\text{min}} = \frac{5}{2}R = 2.5R$

Roller Coaster Problems 🎯

Pendulums

A pendulum of length $L$ released from angle $\theta$ to the vertical:

Height Change

$h = L - L\cos\theta = L(1 - \cos\theta)$

Speed at the Bottom

$v = \sqrt{2gL(1 - \cos\theta)}$

At Any Angle $\phi$ (where $\phi < \theta$):

$v = \sqrt{2gL(\cos\phi - \cos\theta)}$

Key Features

• Maximum speed at the bottom ($\phi = 0$)
• Zero speed at the maximum angle ($\phi = \theta$)
• Speed at the bottom is independent of mass
• The pendulum can never swing higher than its release height

Roller Coaster & Pendulum Calculations 🧮

Use $g = 10$ m/s².

1. A roller coaster starts from rest at 50 m. What is its speed at the bottom (in m/s)? (Round to nearest integer.)

2. What is the minimum speed at the top of a circular loop of radius 8 m (in m/s)? (Round to nearest integer.)

3. A pendulum ($L = 2$ m) is released from $60°$. What is the speed at the bottom (in m/s)? Use $\cos 60° = 0.5$. (Round to nearest tenth.)

Conceptual Analysis 🔍

Exit Quiz ✅

🔥 Energy Dissipation by Friction

Part 5 of 7 — Conservation of Energy

Friction is the most common non-conservative force on AP exams. Understanding how friction dissipates mechanical energy is crucial for solving real-world energy problems.

In this lesson you will learn:

• How to calculate energy lost to friction
• Friction on flat surfaces, ramps, and curves
• Where the "lost" energy goes
• Multi-step problems with friction

Energy Dissipation by Friction

The Energy Lost to Friction

$E_{\text{thermal}} = f_k \cdot d = \mu_k N \cdot d$

This energy becomes thermal energy — it heats up the surfaces.

On a Horizontal Surface

$f_k = \mu_k mg$

$E_{\text{thermal}} = \mu_k mg d$

On a Ramp (angle $\theta$)

$N = mg\cos\theta$

$f_k = \mu_k mg\cos\theta$

$E_{\text{thermal}} = \mu_k mg\cos\theta \cdot d$

where $d$ is the distance along the ramp.

The Modified Energy Equation

$KE_i + PE_i = KE_f + PE_f + E_{\text{thermal}}$

$\frac{1}{2}mv_i^2 + mgh_i = \frac{1}{2}mv_f^2 + mgh_f + \mu_k N d$

Stopping Distance

If an object slides to a stop ($KE_f = 0$):

$d = \frac{KE_i}{f_k} = \frac{\frac{1}{2}mv^2}{\mu_k mg} = \frac{v^2}{2\mu_k g}$

Friction & Energy Problems 🎯

Friction Calculations 🧮

Use $g = 10$ m/s².

1. A car ($m = 1000$ kg) moving at 20 m/s brakes on a road ($\mu_k = 0.5$). What is the stopping distance (in m)?

2. A 3 kg block slides down a frictionless ramp from 4 m, then across 5 m of rough floor ($\mu_k = 0.4$). What is its final speed (in m/s)? (Round to nearest tenth.)

3. A 2 kg block sliding at 8 m/s reaches a frictionless ramp. How high does it go (in m)?

Energy Dissipation Concepts 🔍

Multi-Step Friction Problems

Strategy

For problems with multiple segments (ramp → flat → ramp):

1. Don't solve each segment separately — use energy conservation over the whole path
2. Calculate total friction work: $W_f = -\sum f_k \cdot d_i$ for each segment
3. Apply: $E_{\text{mech},i} + W_f = E_{\text{mech},f}$

Example

A 2 kg block starts at rest atop a 3 m ramp ($\mu_k = 0.25$, ramp length 5 m), slides across 4 m of rough floor ($\mu_k = 0.4$), then up a frictionless ramp.

Friction on ramp: $\mu_k mg\cos\theta \cdot d = 0.25(20)(4/5)(5) = 20$ J

Friction on floor: $\mu_k mgd = 0.4(20)(4) = 32$ J

Total friction: $52$ J

Energy equation: $mgh = mgh_f + 52$

$20(3) = 20 h_f + 52$

$h_f = (60 - 52)/20 = 0.4 \text{ m}$

Exit Quiz ✅

🔧 Problem-Solving Workshop

Part 6 of 7 — Conservation of Energy

Time to tackle complex energy conservation problems that combine springs, gravity, and friction in multi-step scenarios. These are the types of problems you'll see on the AP exam.

In this lesson you will:

• Solve problems combining springs, gravity, and friction
• Use energy bar charts to organize solutions
• Apply the full energy equation to realistic scenarios
• Master the "choose your two states" strategy

The Complete Energy Equation

When springs, gravity, and friction are all present:

$\frac{1}{2}mv_i^2 + mgh_i + \frac{1}{2}kx_i^2 = \frac{1}{2}mv_f^2 + mgh_f + \frac{1}{2}kx_f^2 + f_k d$

Problem-Solving Strategy

1. Draw the situation — identify start and end states
2. Choose your reference level for $h = 0$
3. List what you know at each state (v, h, x, friction)
4. Write the energy equation — cross out zero terms
5. Solve for the unknown

Energy Bar Chart Method

Draw vertical bars representing each energy type at the initial and final states. The total height of bars (minus friction loss) must be equal:

Scenario 1: Spring Launcher on a Ramp

A spring ($k = 800$ N/m) is compressed $0.25$ m at the bottom of a frictionless ramp (angle $30°$). A 2 kg block is placed against the spring and released.

Question: How far up the ramp does the block travel?

Solution

• Initial state: block at rest, spring compressed
• Final state: block at maximum height, spring relaxed, block at rest

$\frac{1}{2}kx^2 = mgh = mg(d\sin\theta)$

$d = \frac{kx^2}{2mg\sin\theta} = \frac{800(0.0625)}{2(2)(10)(0.5)} = \frac{50}{20} = 2.5 \text{ m}$

The block travels 2.5 m along the ramp (reaching height $h = 2.5\sin 30° = 1.25$ m).

Spring Launcher Variations 🧮

Use $g = 10$ m/s².

1. Same spring ($k = 800$ N/m), compressed $0.25$ m, launches a 2 kg block up a ramp with friction ($\mu_k = 0.2$, angle $30°$). Find the distance along the ramp (in m, to 3 significant figures). Hint: $N = mg\cos\theta$.

2. A spring ($k = 500$ N/m) compressed $0.30$ m launches a 1 kg block vertically. What maximum height does it reach (in m)?

3. A spring ($k = 1200$ N/m) compressed $0.20$ m launches a 0.5 kg block along a rough horizontal surface ($\mu_k = 0.3$). How far does the block slide before stopping (in m)?

Scenario 2: Block Sliding onto a Spring

A 4 kg block slides down a frictionless ramp from height $h = 5$ m and hits a spring ($k = 2000$ N/m) on a horizontal surface.

Question: What is the maximum compression of the spring?

Solution

$mgh = \frac{1}{2}kx^2$

$x = \sqrt{\frac{2mgh}{k}} = \sqrt{\frac{2(4)(10)(5)}{2000}} = \sqrt{0.2} \approx 0.447 \text{ m}$

What if there's friction on the horizontal surface?

If there's $\mu_k = 0.25$ friction over a 3 m flat section before the spring:

$mgh = \frac{1}{2}kx^2 + \mu_k mg \cdot (3 + x)$

Note: friction acts over 3 m plus the compression distance $x$! This gives a quadratic in $x$.

Spring-Gravity Combination Problems 🎯

Complex Multi-Step Problems 🧮

Use $g = 10$ m/s².

1. A 3 kg block starts at rest at height 8 m, slides down a frictionless ramp, across 2 m of rough floor ($\mu_k = 0.4$), and compresses a spring ($k = 600$ N/m). What is the maximum spring compression (in m, to 3 significant figures)?

2. A spring ($k = 1000$ N/m) compressed 0.50 m launches a 2 kg block up a rough ramp ($\mu_k = 0.15$, angle $45°$). How far along the ramp does it travel (in m, to 3 significant figures)?

3. A 5 kg block moving at 12 m/s on a rough surface ($\mu_k = 0.3$) hits a spring ($k = 2000$ N/m). Find the maximum compression (in m, to 3 significant figures). Hint: friction acts during compression too.

Problem-Solving Strategy Check 🔍

Exit Quiz — Problem-Solving Workshop ✅

🎓 Synthesis & AP Review

Part 7 of 7 — Conservation of Energy

This final lesson brings together every energy concept: kinetic energy, gravitational PE, spring PE, work-energy theorem, friction, and energy bar charts. We'll focus on AP-style questions, common mistakes, and FRQ strategies.

In this lesson you will:

• Tackle AP-style multiple choice and free response questions
• Identify and avoid common energy misconceptions
• Master energy bar chart analysis
• Build a strategy toolkit for AP exam energy problems

Energy Bar Charts (LOL Diagrams)

Energy bar charts are a powerful tool for AP Physics. They show energy storage at different points in a process.

How to Read Them

Conservation Rule

$KE_i + PE_{g,i} + PE_{s,i} = KE_f + PE_{g,f} + PE_{s,f} + \Delta E_{th}$

The sum of all initial bars must equal the sum of all final bars (including thermal energy).

Common AP Scenarios

• Free fall: $PE_g \rightarrow KE$ (bars shift from PE to KE)
• Spring launch: $PE_s \rightarrow KE$ (spring PE becomes KE)
• Friction stop: $KE \rightarrow \Delta E_{th}$ (KE becomes thermal)
• Projectile at max height: $KE_i \rightarrow KE_f + PE_g$ (some KE remains as horizontal KE)

Energy Bar Chart Analysis 🎯

Common AP Mistakes to Avoid

Mistake 1: Forgetting that KE depends on $v^2$

• Doubling speed → 4× the kinetic energy, NOT 2×

Mistake 2: Using the wrong height

• Height is measured from your chosen reference level
• It's the vertical height, not the distance along a ramp
• $h = d\sin\theta$ on a ramp of length $d$

Mistake 3: Forgetting friction acts over the ENTIRE path

• If a block slides 3 m on a floor then compresses a spring 0.2 m, friction acts over $3 + 0.2 = 3.2$ m (if the floor is rough under the spring too)

Mistake 4: Saying "centripetal force does work"

• Centripetal force is always perpendicular to velocity → does zero work
• Energy is NOT gained or lost going around a circular loop (ignoring friction)

Mistake 5: Confusing force and energy

• A large force doesn't mean large energy — work depends on $F \cdot d \cdot \cos\theta$
• Normal force on a flat surface does zero work ($\cos 90° = 0$)

Spot the Mistake 🔍

FRQ Strategy Guide

Energy Conservation FRQ Template

Part (a): "Derive an expression for..."

1. State: "Using conservation of energy between states A and B..."
2. Write the full equation with all terms
3. Cross out zero terms with justification
4. Solve algebraically — keep everything symbolic

Part (b): "Calculate..."

1. Substitute numbers into your Part (a) expression
2. Show units cancellation
3. Circle/box your final answer with units

Part (c): "Explain qualitatively..."

1. Identify the relevant physics principle
2. Connect cause to effect using energy concepts
3. Use phrases like "because energy is conserved..." or "since friction dissipates energy..."

Key Phrases for Full Credit

• "By conservation of energy..."
• "Setting the reference level at..."
• "The work done by friction equals $f_k d = \mu_k N d$"
• "Since only conservative forces act, mechanical energy is conserved"
• "The thermal energy generated equals the loss in mechanical energy"

AP-Style Multiple Choice 🎯

AP-Style Calculations 🧮

Use $g = 10$ m/s².

1. A 0.5 kg ball is dropped from 20 m onto a spring ($k = 500$ N/m). Find the maximum compression of the spring (in m, to 3 significant figures). Hint: the ball falls an extra distance $x$ below the spring's natural length.

2. A 4 kg block slides 5 m down a $37°$ ramp ($\mu_k = 0.25$), then 3 m across a frictionless floor, then up a frictionless $53°$ ramp. What height does it reach on the second ramp (in m, to 3 significant figures)?

3. In an AP FRQ, a 2 kg block is pushed against a spring ($k = 800$ N/m), compressing it 0.3 m. The block is released and slides across a rough surface ($\mu_k = 0.4$) for 5 m before reaching a frictionless ramp. What maximum height does it reach (in m)?

Final Exit Quiz — Conservation of Energy ✅