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Module 3: Energy and Work

 

PHYS-2325 M3L2 Potential Energy and Conservation of Energy

"The total energy of the universe is constant; the total entropy is always increasing."
— Rudolf Clausius


In the last lesson, work changed kinetic energy. Now you will see that work done by a conservative force can also be stored and fully recovered — this stored capacity to do work is called potential energy. Gravitational potential energy Ug = mgy and elastic (spring) potential energy Us = ½kx2 are the two forms you will work with most. When only conservative forces act, mechanical energy E = K + U is conserved — an extraordinarily powerful constraint that lets you solve problems that Newton's laws alone would make extremely difficult. You will also extend the framework to include friction: when non-conservative forces act, mechanical energy is not conserved, but total energy still is — the "missing" mechanical energy is converted to thermal energy.

A potential energy curve U(x) plotted against position x, showing turning points, equilibrium positions, and the total mechanical energy level as a horizontal line.
The potential energy curve: where the total energy line intersects U(x) are the turning points of motion.

Course Competencies and Learning Objectives

A ★ indicates that this page contains content related to that LO.

CC3.1 Analyze the energy transformations associated with the work-energy theorem

★ LO3.2.1 Define gravitational potential energy and elastic potential energy; calculate each for a given configuration

★ LO3.2.2 State the principle of conservation of mechanical energy and apply it to solve for unknown speeds or heights when only conservative forces act

★ LO3.2.3 Read a potential energy curve: identify turning points, equilibrium positions, and kinetic energy at any location

★ LO3.2.4 Calculate the work done on a system by an external force in the presence of friction and relate it to the change in total mechanical energy

★ LO3.2.5 Apply the general law of conservation of energy, accounting for thermal energy generated by friction: ΔEmec = −ΔEth

★ LO3.2.6 Choose an appropriate reference level for gravitational potential energy and demonstrate that the choice does not affect the physics

Required Reading

Click the blue buttons to go to the OpenStax reading assignments. Complete all five sections before watching the videos.

8.1 Potential Energy 8.2 Conservation of Mechanical Energy 8.3 Reading a Potential Energy Curve 8.4 Work Done on a System by an External Force 8.5 Conservation of Energy

 

Optional Reading

Explore More

Want additional practice with potential energy, conservation of energy, and energy graphs? These free resources complement the reading.

Media

Work through the pre-lecture videos for each reading section first, then watch the illustration and mini-lectures for deeper examples.

Video 1: Pre-Lecture — 8.1 Potential Energy

Pre-Lecture: Defining Potential Energy — Gravitational and Elastic

This video introduces potential energy as energy stored in a system by virtue of its configuration. It derives both forms from the work done by conservative forces: gravitational potential energy Ug = mgy (with a chosen reference level) and elastic potential energy Us = ½kx2.

  • Potential energy is associated with a system, not a single object
  • Gravitational PE: Ug = mgy — defined relative to a chosen reference level (y = 0 is arbitrary)
  • Elastic (spring) PE: Us = ½kx2 — always ≥ 0; reference level is equilibrium (x = 0)
  • Relationship to conservative force: F = −dU/dx — force points toward lower potential energy

Time: Video pending upload to Canvas

Video 2: Pre-Lecture — 8.2 Conservation of Mechanical Energy

Pre-Lecture: Emec = K + U = Constant (Conservative Forces Only)

When only conservative forces do work, the total mechanical energy Emec = K + U remains constant throughout the motion. This video derives the conservation law from the work-kinetic energy theorem and shows how to apply it: set Ki + Ui = Kf + Uf.

  • Conservative forces: work is path-independent (gravity, spring); non-conservative: friction, drag
  • Conservation of mechanical energy: Ki + Ui = Kf + Uf when Wnc = 0
  • Strategy: identify initial and final states; write energy equation; solve for unknown
  • Energy can transform between K and U, but their sum is constant — illustrated with a pendulum and a roller coaster

Time: Video pending upload to Canvas

Video 3: Pre-Lecture — 8.3 Reading a Potential Energy Curve

Pre-Lecture: Extracting Physics from a U(x) Graph

A plot of potential energy U vs. position x encodes everything about the conservative force and the allowed motion. This video teaches you to read such a curve: find the force at any point, identify turning points, locate equilibrium positions, and determine whether equilibria are stable or unstable.

  • Force from the curve: F(x) = −dU/dx — slope of U(x) gives the force (magnitude and direction)
  • Turning points: where Etotal = U(x) — the object momentarily stops (K = 0)
  • Stable equilibrium: local minimum of U(x) — force restores the object if displaced
  • Unstable equilibrium: local maximum of U(x) — force pushes the object further away if displaced

Time: Video pending upload to Canvas

Video 4: Pre-Lecture — 8.4 Work Done on a System by an External Force

Pre-Lecture: External Work, Friction, and the Change in Mechanical Energy

When an external agent does work on a system that also has internal friction, the change in mechanical energy is less than the external work — the difference goes into thermal energy. This video introduces ΔEmec = Wext − ΔEth and the thermal energy term ΔEth = fkd.

  • With no friction: Wext = ΔEmec = ΔK + ΔU
  • With friction: Wext = ΔEmec + ΔEth, where ΔEth = fkd
  • Thermal energy generated: ΔEth = fkd — always positive (friction always increases thermal energy)
  • System boundary: defines what counts as "internal" vs. "external" energy

Time: Video pending upload to Canvas

Video 5: Pre-Lecture — 8.5 Conservation of Energy

Pre-Lecture: The General Law — Total Energy of an Isolated System is Constant

This video states the full conservation of energy law: the total energy of an isolated system (mechanical + thermal + internal) never changes. For non-isolated systems, the change in total energy equals the energy transferred across the system boundary. This is one of the most fundamental laws in all of physics.

  • Isolated system: ΔEtotal = 0 → ΔEmec + ΔEth + ΔEint = 0
  • Non-isolated system: ΔEtotal = W + Q (work and heat transfer across boundary)
  • Conservative vs. non-conservative: friction converts mechanical energy to thermal, not lost — just transformed
  • Practical form with friction: Kf + Uf = Ki + Ui − fkd

Time: Video pending upload to Canvas

Video 6: Illustration — Pendulum Energy

Video Illustration: Energy Transformation in a Swinging Pendulum

This illustration video shows a pendulum swinging back and forth and tracks the continuous exchange between kinetic energy and gravitational potential energy throughout the swing — demonstrating conservation of mechanical energy in one of its most visually intuitive forms.

  • At the lowest point: K is maximum, Ug is minimum (reference level)
  • At the highest point (turning points): K = 0, Ug is maximum
  • At any intermediate angle θ: K + Ug = Emec = constant
  • With air resistance (non-ideal): Emec slowly decreases as energy converts to thermal — amplitude decreases over time

Time: Video pending upload to Canvas

Video 7: Mini-Lecture — Conservation of Mechanical Energy

Mini-Lecture: Applying Ki + Ui = Kf + Uf

This video works a complete example — a ball launched from a height, or a roller coaster car at various points on a frictionless track — using conservation of mechanical energy to find speeds and heights without needing to track forces or use kinematics.

  • Choose a reference level (y = 0); assign Ug accordingly at each point
  • Write Ki + Ui = Kf + Uf: ½mvi2 + mgyi = ½mvf2 + mgyf
  • Solve for the unknown (vf or yf); note that mass often cancels
  • Key check: the reference level choice does not affect the final answer

Time: Video pending — to be recorded by instructor

Video 8: Mini-Lecture — Elastic Potential Energy and Conservation of Mechanical Energy

Mini-Lecture: Spring-Block System — K, Us, and Emec

Combining kinetic energy with elastic potential energy: a block on a frictionless surface attached to a spring is displaced and released. This video tracks K and Us at each stage and finds the speed at any position using energy conservation.

  • At maximum compression/extension: K = 0, Us = ½kxmax2 = Emec
  • At equilibrium (x = 0): Us = 0, K = ½mvmax2 = Emec
  • At any position x: ½mv2 + ½kx2 = ½kxmax2
  • Solve for v: v = √[(k/m)(xmax2 − x2)] — previews simple harmonic motion (Module 7)

Time: Video pending — to be recorded by instructor

Video 9: Mini-Lecture — Energy Accounting: K and Ug

Mini-Lecture: Tracking Kinetic and Gravitational Potential Energy at Multiple Points

A systematic energy accounting exercise: given a projectile or object moving along a curved path under gravity only, this video computes K and Ug at several key positions and verifies that Emec = K + Ug is constant throughout — reinforcing the meaning of energy conservation as a bookkeeping tool.

  • Choose reference level; calculate Ug = mgy at each point relative to that level
  • Calculate K = ½mv2 at each point; sum K + Ug
  • Verify Emec is the same at every point (within rounding)
  • Key insight: the reference level choice shifts all U values by the same constant — the differences ΔU are unaffected, and Emec is the same regardless of where you set y = 0

Time: Video pending — to be recorded by instructor

Video 10: Mini-Lecture — Potential Energy Curve

Mini-Lecture: Reading F, Turning Points, and Equilibria from U(x)

This video works through a complete potential energy curve problem: given U(x) as a graph or equation, find the force at several positions, identify all turning points for a given Etotal, and classify each equilibrium position as stable or unstable.

  • Force: F(x) = −dU/dx — evaluate as the negative slope of the U(x) graph
  • Turning points: set U(x) = Etotal and solve for x; object cannot go beyond these points
  • Stable equilibrium at local U minimum: F = 0 here; restoring if displaced
  • Unstable equilibrium at local U maximum: F = 0 here; destabilizing if displaced

Time: Video pending — to be recorded by instructor

Practice and Apply - Conceptual

Strategy: Solving a Conservation of Energy Problem

Arrange the following steps in the correct order for applying conservation of energy to a mechanics problem.

Order Items

Arrange the following items in the correct order.

  1. Identify the system and whether any non-conservative forces (friction, drag) act.
  2. Choose a reference level for gravitational potential energy (set y = 0).
  3. Define the initial and final states of the system.
  4. Write the energy equation: Ki + Ui + Wext = Kf + Uf + ΔEth.
  5. Calculate known energy terms and substitute into the equation.
  6. Solve for the unknown (speed, height, compression, etc.).
  7. Check: verify units (joules) and confirm the answer is physically reasonable.

When Is Mechanical Energy Conserved?

Select all scenarios in which the mechanical energy of the object or system is conserved.

Select All

Select all scenarios where mechanical energy is conserved.

  • A ball in free fall (ignoring air resistance).
  • A block sliding down a frictionless ramp.
  • A pendulum swinging in a vacuum.
  • A spring-block system on a frictionless surface oscillating after release.
  • A roller coaster on a frictionless track.
  • A block sliding down a rough ramp (friction present).
  • A skydiver falling through air (drag present).
  • A car braking to a stop (friction converts K to thermal energy).
  • A box pushed across a floor with an applied force against friction.

Match the Energy Term to Its Definition

Match each energy term or symbol on the left to its correct definition or formula on the right.

Match Items

Select each set of two matching items.

Items to Match

Ug = mgy
Gravitational potential energy; depends on height above the chosen reference level.
Us = ½kx2
Elastic potential energy stored in a spring displaced x from equilibrium.
Emec = K + U
Total mechanical energy — the sum of kinetic and potential energies.
ΔEth = fkd
Thermal energy generated by kinetic friction over displacement d.
F(x) = −dU/dx
Conservative force obtained from the negative derivative of the potential energy function.
Turning point
Location where U(x) = Etotal and K = 0; the object momentarily stops and reverses.

Key Concept Check

Click each card to flip it and test your understanding of potential energy and conservation of energy.

A skier starts from rest at height h on a frictionless slope. What is the skier's speed at the bottom?
Answer

Set y = 0 at the bottom. Conservation of energy: Ki + Ui = Kf + Uf → 0 + mgh = ½mvf2 + 0. Solving: vf = √(2gh). Note: mass cancels — all skiers reach the same speed, regardless of mass.

 

How does friction affect mechanical energy? Where does the "lost" mechanical energy go?
Answer

Friction is a non-conservative force — it reduces mechanical energy: ΔEmec = −ΔEth < 0. The mechanical energy is not "lost" — it is converted to thermal energy (heat) in the sliding surfaces: ΔEth = fkd. Total energy (mechanical + thermal) is still conserved.

 

Does it matter where you place the reference level (y = 0) for gravitational potential energy? Explain.
Answer

No — the choice of reference level does not affect the physics. Shifting y = 0 changes U at every point by the same constant, so ΔU = Uf − Ui is unchanged. Since only differences in energy appear in the conservation equation, the final answers for speed, height, etc., are always the same regardless of where you set y = 0. Choose the level that makes the algebra simplest.

 

On a U(x) graph, how do you find the force on the object at any position x?
Answer

F(x) = −dU/dx. Graphically: F is the negative slope of the U(x) curve at that point. Where the curve slopes downward (dU/dx < 0), F is positive (force in +x direction). Where it slopes upward (dU/dx > 0), F is negative. At a minimum or maximum of U, the slope is zero and F = 0 — these are equilibrium positions.

 

A spring launches a block from compression x0 on a frictionless surface. What is the block's maximum speed?
Answer

At release: Ki = 0, Us,i = ½kx02. At equilibrium (x = 0): Us,f = 0, Kf = ½mvmax2. Conservation: ½kx02 = ½mvmax2. Solving: vmax = x0√(k/m). This is the maximum speed — all spring PE has converted to KE.

 

What is the difference between a stable and unstable equilibrium on a U(x) graph?
Answer

Stable equilibrium is at a local minimum of U(x): if the object is displaced slightly, F = −dU/dx points back toward the equilibrium — it is restored. Unstable equilibrium is at a local maximum: if displaced, F points away — the object accelerates further from equilibrium. Think: a ball in a bowl (stable) vs. a ball balanced on a hill (unstable).

 

Practice and Apply - Computational

Important: Key Equations and Strategy Reminders

Potential energy: Ug = mgy  |  Us = ½kx2

Mechanical energy: Emec = K + U = ½mv2 + U

Conservation (no friction): Ki + Ui = Kf + Uf

With friction: Kf + Uf = Ki + Ui − fkd  (or: Wext = ΔEmec + ΔEth)

Thermal energy from friction: ΔEth = fkd  |  fk = μkN

Force from potential energy: F(x) = −dU/dx

  • Choose y = 0 to minimize algebra — the final answer is independent of this choice.
  • Check whether friction is present — if yes, ΔEmec < 0; use the friction form.
  • Mass often cancels in pure gravitational problems — verify this before solving.
  • Report energy in joules (J), speed in m/s, height in m. Three significant figures.

Sample Problem Videos

Watch each sample problem video, then attempt the corresponding written practice problem below.

Sample Problem 1: Choosing a Reference Level — Sloth

Sample Problem: Gravitational PE with Two Different Reference Levels

A sloth hangs at different heights. This video calculates gravitational potential energy using two different reference levels and shows that while the numerical value of U changes, the difference ΔU — and hence the physics — is identical in both cases.

  • Case 1: y = 0 at the ground; Case 2: y = 0 at the sloth's lowest position
  • Ug differs between cases by a constant offset
  • ΔU = Uf − Ui is the same in both — this is what enters the energy equation
  • Conclusion: reference level choice is a matter of convenience, not physics

Time: Video pending — to be recorded by instructor

Sample Problem 2: Conservation of Mechanical Energy — Water Slide

Sample Problem: Speed at the Bottom of a Frictionless Water Slide

A rider starts from rest at the top of a water slide of height h. This video applies Ki + Ui = Kf + Uf to find the speed at the bottom, and then asks: at what height along the slide does the rider reach half the maximum speed?

  • Initial: Ki = 0, Ui = mgh; Final: Kf = ½mvf2, Uf = 0
  • vf = √(2gh) — independent of mass and path shape
  • Height for v = vf/2: set ½mvf2/4 + mgy = mgh → y = 3h/4

Time: Video pending — to be recorded by instructor

Sample Problem 3: Equivalent Paths — Slippery Cheese

Sample Problem: Path Independence of Conservative Forces

A wedge of cheese slides from the same starting height to the same ending height by two different routes. This video shows that the work done by gravity — and the final speed — is the same for both paths, confirming the path-independence property of conservative forces.

  • Wg = −ΔUg = −mg(yf − yi) — depends only on endpoints
  • Path A (direct) vs. Path B (longer route): identical Δy → identical Wg → identical vf
  • Key distinction from friction: friction work does depend on path length — longer path → more thermal energy

Time: Video pending — to be recorded by instructor

Sample Problem 4: Reading a Potential Energy Graph

Sample Problem: Extracting Force, KE, Turning Points, and Equilibria from U(x)

Given a specific U(x) graph and a total mechanical energy value, this video identifies the turning points, the kinetic energy at each labeled position, the sign and approximate magnitude of the force at each position, and the stability of each equilibrium.

  • K(x) = Etotal − U(x) at each x — read directly from the graph as the vertical gap between E line and U curve
  • Force at each x: negative slope of U — steep downward slope → large positive force
  • Turning points: where U(x) = Etotal line; K = 0 here
  • Stability: local minima of U are stable equilibria; local maxima are unstable

Time: Video pending — to be recorded by instructor

Sample Problem 5: Energy and Friction — Mountain Slope

Sample Problem: Speed on a Rough Slope Using Energy with Friction

A hiker descends a rough mountain slope of known angle and friction coefficient over a given distance. This video applies Kf + Uf = Ki + Ui − fkd to find the speed at the bottom, accounting for the thermal energy generated by friction.

  • Height descended: Δy = d sin θ; normal force: N = mg cos θ; friction force: fk = μkmg cos θ
  • Energy equation: ½mvf2 = mgd sin θ − μkmgd cos θ = mgd(sin θ − μk cos θ)
  • vf = √[2gd(sin θ − μk cos θ)] — same result as Newton's second law approach
  • Connection to M2L3: a = g(sin θ − μk cos θ) and vf2 = 2ad confirms consistency

Time: Video pending — to be recorded by instructor

Sample Problem 6: Work When There Is Friction — Easter Island Statues

Sample Problem: External Work Required to Move a Heavy Object Against Friction

Moving a large stone statue (like those at Easter Island) across a horizontal surface requires an external force to overcome friction. This video calculates the work done by the applied force, the thermal energy generated by friction, and verifies the energy balance Wext = ΔK + ΔEth.

  • If ΔK = 0 (moved at constant speed): Wext = ΔEth = fkd = μkmgd
  • If moving from rest to some speed: Wext = ½mvf2 + μkmgd
  • Energy audit: work from the people → kinetic energy + heat in the ground/statue surface

Time: Video pending — to be recorded by instructor

Sample Problem 7: Work, Friction, Change in Thermal Energy — Cabbage Heads

Sample Problem: Energy Accounting with Friction — Calculating ΔEth

Cabbages slide along a surface with friction. Given the applied work and measured change in kinetic energy, this video works backward using Wext = ΔK + ΔEth to determine the thermal energy generated — an energy audit that highlights what fraction of the input work became heat.

  • Rearrange: ΔEth = Wext − ΔK
  • Verify using ΔEth = fkd = μkNd independently
  • Percentage of input work converted to heat: (ΔEth / Wext) × 100%
  • Real-world context: efficiency of transport, machinery, and energy conversion

Time: Video pending — to be recorded by instructor

Sample Problem 8: Energy, Friction, Spring, Graph

Sample Problem: Block Compresses a Spring on a Rough Surface — Graphical Analysis

A moving block slides along a rough surface and compresses a spring. This video uses an energy bar chart (graph) to track K, Us, and ΔEth at each stage, then solves for the maximum compression of the spring using the full energy equation including friction.

  • Initial: Ki = ½mvi2, Us,i = 0, Eth,i = 0
  • At maximum compression (v = 0): K = 0, Us = ½kxmax2, ΔEth = fkd
  • Energy equation: ½mvi2 = ½kxmax2 + fkd (where d = xmax)
  • Solve for xmax — quadratic in xmax; choose the physically meaningful root

Time: Video pending — to be recorded by instructor

Sample Problem 9: Energy, Friction, Spring — Tamales

Sample Problem: Object Launched by a Spring, Slides with Friction

A spring launches a tamale (or block) horizontally; the object then slides along a rough surface. This video finds the launch speed from spring PE, then uses energy methods with friction to find the stopping distance — or the speed after traveling a given distance.

  • Phase 1 (launch): ½kx02 = ½mvlaunch2 → vlaunch = x0√(k/m)
  • Phase 2 (sliding with friction): ½mvlaunch2 − fkd = ½mvf2
  • To stop: dstop = mvlaunch2 / (2fk) = kx02 / (2μkmg)
  • Entire problem solved with energy — no Newton's law kinematics needed

Time: Video pending — to be recorded by instructor

Practice Problem 1 — Conservation of Mechanical Energy on a Frictionless Ramp

A 2.00-kg block starts from rest at the top of a frictionless ramp h = 4.00 m above the floor. (a) Find the speed of the block at the bottom of the ramp. (b) Find the height at which the block has a speed of 6.00 m/s on the way down.

Show Solution

Given: m = 2.00 kg, h = 4.00 m, vi = 0, g = 9.80 m/s2, frictionless

Reference level: y = 0 at the floor (bottom of ramp).

Part (a) — Speed at the bottom:

Ki + Ui = Kf + Uf

0 + mgh = ½mvf2 + 0 → vf = √(2gh) = √(2 × 9.80 × 4.00) = √(78.4)

vf = 8.85 m/s  (note: mass cancels — same answer for any mass)

Part (b) — Height where v = 6.00 m/s:

Ki + Ui = K + U → mgh = ½mv2 + mgy

y = h − v2/(2g) = 4.00 − (6.00)2/(2 × 9.80) = 4.00 − 36.0/19.6 = 4.00 − 1.84

y = 2.16 m

Practice Problem 2 — Energy with Friction on a Rough Surface

A 3.00-kg block slides along a horizontal surface with μk = 0.250. The block has an initial speed of 8.00 m/s. (a) How far does the block travel before stopping? (b) How much thermal energy is generated? (c) What is the block's speed after traveling 4.00 m?

Show Solution

Given: m = 3.00 kg, vi = 8.00 m/s, μk = 0.250, horizontal surface, g = 9.80 m/s2

On horizontal surface: N = mg → fk = μkmg = (0.250)(3.00)(9.80) = 7.35 N

Part (a) — Stopping distance:

At rest: Kf = 0. Energy equation: Ki = ΔEth = fkd

d = Ki / fk = ½mvi2 / fk = ½(3.00)(8.00)2 / 7.35 = 96.0 / 7.35 = 13.1 m

Part (b) — Thermal energy generated (stopping):

ΔEth = fkd = (7.35)(13.1) = 96.0 J  (= Ki ✓)

Part (c) — Speed after 4.00 m:

Kf = Ki − fkd = 96.0 − (7.35)(4.00) = 96.0 − 29.4 = 66.6 J

vf = √(2Kf/m) = √(2 × 66.6 / 3.00) = √(44.4) = 6.66 m/s

Practice Problem 3 — Spring Compressed by a Moving Block with Friction

A 1.50-kg block slides at 5.00 m/s across a rough horizontal surface (μk = 0.200) and compresses a spring (k = 600 N/m). The block travels d = xmax from its initial position before stopping at maximum compression. Find the maximum compression xmax of the spring.

Show Solution

Given: m = 1.50 kg, vi = 5.00 m/s, μk = 0.200, k = 600 N/m, g = 9.80 m/s2

Friction force: fk = μkmg = (0.200)(1.50)(9.80) = 2.94 N

Energy equation at maximum compression (vf = 0):

Ki = Us + ΔEth

½mvi2 = ½kxmax2 + fkxmax

½(1.50)(5.00)2 = ½(600)xmax2 + (2.94)xmax

18.75 = 300xmax2 + 2.94xmax

Rearrange: 300xmax2 + 2.94xmax − 18.75 = 0

Quadratic formula: xmax = [−2.94 + √(2.942 + 4 × 300 × 18.75)] / (2 × 300)

= [−2.94 + √(8.64 + 22500)] / 600 = [−2.94 + √(22508.6)] / 600

= [−2.94 + 150.0] / 600 = 147.1 / 600 = 0.245 m

Note: the negative root is rejected (compression must be positive).

Check (no friction): xmax,0 = vi√(m/k) = 5.00√(1.50/600) = 0.250 m. With friction, xmax = 0.245 m < 0.250 m ✓ — friction reduces compression slightly.

Ready to Move On?

Lesson Checklist

Before moving on to the next module, confirm you can do each of the following:

  • ☐ Calculate gravitational potential energy Ug = mgy and elastic potential energy Us = ½kx2 for a given system configuration.
  • ☐ Choose an appropriate reference level for Ug and explain why the choice does not affect ΔU or the physics.
  • ☐ Apply Ki + Ui = Kf + Uf to find unknown speeds or heights when no friction acts.
  • ☐ Read a U(x) graph: determine the force F(x) = −dU/dx, identify turning points, and classify equilibria as stable or unstable.
  • ☐ Calculate the thermal energy generated by friction: ΔEth = fkd.
  • ☐ Apply the full energy equation with friction: Kf + Uf = Ki + Ui − fkd.
  • ☐ Calculate the work done by an external agent on a system with friction: Wext = ΔK + ΔU + ΔEth.
  • ☐ State the general law of conservation of energy and explain what happens to mechanical energy when friction is present.

Conservation of energy is the most powerful single tool in classical mechanics — and in all of physics. You will use it in every remaining module of this course. The ideas here also lay the groundwork for thermodynamics in Module 8.