Open by playing a 20-second clip (or describing) a Cedar Point roller coaster dropping from its 90 m first hill. Pose the question aloud: 'The car starts at rest at the top and reaches 38 m/s at the bottom — but a frictionless calculation predicts 42 m/s. Where did the missing 4 m/s go, and how much energy does that represent?' Tell students that today they'll build the accounting tools physicists and engineers use to answer exactly this kind of question — and that by the end of class they will have a spreadsheet that tracks every joule for a real system they measure themselves. Take 2–3 student guesses, write them on the board without judging, and promise to return to this number at the end of class.

1. 10m

   Energy is conserved — the LOL diagram

   Content

   Conservation of energy says the total energy of a closed system is constant. It is NOT a statement that any one form stays constant — kinetic energy, gravitational potential energy, and thermal energy can all change, as long as their sum does not. The LOL diagram is the visual bookkeeping tool: on the left, draw a bar for each form of energy the system has initially; on the right, draw bars for each form it has finally; in the middle, draw the system boundary as a circle. The rule is simple — the total bar height on the left equals the total bar height on the right, unless energy crossed the boundary (work done by an outside agent). For a 2.0 kg ball dropped from 5.0 m, the initial LOL has one bar: PE = mgh = (2.0)(9.8)(5.0) = 98 J, KE = 0, thermal = 0. Just before hitting the ground, PE = 0, thermal ≈ 0 (air resistance small), so KE must equal 98 J. Solving ½mv² = 98 J gives v ≈ 9.9 m/s.

   Delivery

   Emphasize that choosing the system is a decision the student makes — and it determines what counts as 'inside' versus 'crossing the boundary.' Ask: 'If we include Earth in the system, gravity is an internal force and we use PE. If we exclude Earth, gravity does work across the boundary. Which is easier?' (Including Earth.) Head off the misconception that energy 'disappears' when KE drops — reinforce that a shrinking bar on one side must be matched by a growing bar somewhere else. Have students copy the ball-drop LOL into their notebooks so they have a template for the lab.

2. 8m

   Quantifying KE and PE — the equations you'll use all day

   Content

   Two equations do most of today's work. Kinetic energy KE = ½mv² gives the energy of motion — doubling the speed quadruples the energy, which is why a 40 m/s crash is four times as destructive as a 20 m/s crash at the same mass. Gravitational potential energy PE = mgh gives the energy stored by height, measured from whatever reference line you pick (the choice does not matter as long as you use it consistently at initial and final). Work a two-step example live: a 0.50 kg pendulum bob released from h = 0.20 m above its lowest point has PE = (0.50)(9.8)(0.20) = 0.98 J at release. If the pendulum is frictionless, all of that PE converts to KE at the bottom, so ½(0.50)v² = 0.98 J, giving v = √(2·0.98/0.50) = 1.98 m/s. That's the exact prediction students will test in the lab.

   Delivery

   Insist on units in every step — a common honors slip is writing '½(0.5)(v²) = 0.98' without units and then losing the m/s at the end. Solve for v symbolically first (v = √(2gh) for the frictionless case) so students see that mass cancels — this is a great sanity check they can use in the lab. Ask: 'If we doubled the mass, would v at the bottom change?' (No, because mass cancels.) Foreshadow: 'In the real pendulum you'll measure, v will come out slightly less than √(2gh) — the difference IS your thermal energy.'

3. 7m

   Accounting for 'missing' energy — friction and thermal

   Content

   When mechanical energy seems to disappear, it has almost always been transferred to thermal energy inside the system (warmer surfaces, warmer air) or radiated as sound. Track it as an extra bar in the LOL diagram. The bookkeeping equation for a system that starts with only PE and ends with KE plus thermal is: PE(initial) = KE(final) + E(thermal). Rearranging, E(thermal) = PE(initial) − KE(final). Back to the coaster: predicted KE at the bottom for a 500 kg car dropping 90 m is (500)(9.8)(90) = 441 000 J, giving v = √(2·9.8·90) = 42.0 m/s. Measured v = 38 m/s, so measured KE = ½(500)(38)² = 361 000 J. The thermal energy generated (mostly in wheels, axles, and air) is 441 000 − 361 000 = 80 000 J — about 18% of the initial energy. That is a real, calculable number, not a mystery.

   Delivery

   This is the beat where the standard's headline misconception dies: energy is not 'used up.' Say it plainly: 'The 80 000 J is still in the universe — it is warming the wheels and the air. We just stopped tracking it as motion.' Ask students to name three places the thermal energy could be found (bearings, wheel–track contact, air behind the car). Then set up the pivot to the lab: 'Your job in a few minutes is to compute the same number for a pendulum you swing yourselves — and it will not be zero.'

4. 7m

   What a computational model looks like

   Content

   A computational model is any procedure — algebraic, spreadsheet, or code — that takes measured inputs and computes energy values while enforcing conservation. Today students will build one in a spreadsheet. The columns are: time t, height h, speed v, PE = mgh, KE = ½mv², E(thermal) = E(total) − PE − KE, and E(total) (which should be constant). The KEY design idea is that E(thermal) is not measured directly — it is computed as the DEFICIT, exactly the standard's target skill: 'calculate the change in the energy of one component of a system when the energy of the other components and the total energy are known.' If E(thermal) comes out negative, you know something is wrong (measurement error, wrong mass, wrong reference height) — the model checks itself.

   Delivery

   Emphasize that this is exactly how engineers do energy audits — you measure what you can, you know the total from initial conditions, and you SOLVE for the unknown component. Preview the spreadsheet template so students see where the formulas live. Ask: 'Which cell would you change if you realized you had the mass in grams instead of kilograms?' — this is a quick check that they understand the model is a chain of dependencies, not a set of isolated numbers.

1. 35m

   Pendulum Energy Lab with PhotogateLab

   Groups of 3 will measure a pendulum's speed at the bottom of its swing and compare it to the frictionless prediction from PE = mgh, computing the thermal energy as the deficit. Setup: Tie the bob to the string, clamp it so the bob hangs freely, and adjust the string so the bob passes cleanly through the photogate at the lowest point of its swing. Measure and record the bob's diameter d with calipers — the photogate measures how long the bob (width d) takes to pass through the beam, so v = d / Δt. Walk each group through the release-height measurement: measure the vertical height h from the bob's lowest point to its release point (NOT the arc length or the string angle). Use a meter stick held vertically. Have them repeat each release-height 3 times and average Δt. Student handout — hand this out or project it while groups work: Pendulum Energy Lab — Student Handout Part 1 — Measured values - Mass of bob m = ______ kg - Diameter of bob d = ______ m - Release height above lowest point h₁ = ______ m - Photogate time at lowest point (trial 1) Δt = ______ s - Photogate time (trial 2) Δt = ______ s - Photogate time (trial 3) Δt = ______ s - Average Δt = ______ s Part 2 — Computed values (show units in every step) 1. Speed at the lowest point: v = d / Δt(avg) = ______ m/s 2. Initial PE (taking the lowest point as h = 0): PE₁ = mgh₁ = ______ J 3. Initial KE: KE₁ = 0 J (released from rest) 4. Final PE at lowest point: PE₂ = 0 J 5. Final KE: KE₂ = ½mv² = ______ J Part 3 — Conservation check 6. Predicted frictionless speed: v(predicted) = √(2gh₁) = ______ m/s 7. Percent difference between measured v and predicted v = ______ % 8. Thermal energy generated: E(thermal) = PE₁ − KE₂ = ______ J 9. Percent of initial energy converted to thermal: (E(thermal) / PE₁) × 100 = ______ % Part 4 — LOL diagram Sketch an LOL diagram for your pendulum: - Left side (release): one bar for PE₁ - Right side (lowest point): bars for KE₂ and E(thermal) - Verify the total heights match. Part 5 — Analysis questions a. If you doubled the mass of the bob but kept h₁ the same, would v at the bottom change? Explain using your equations. b. Where physically is the thermal energy after the swing? Name two specific locations. c. A classmate says 'the pendulum lost energy.' Rewrite this statement so it is physically correct. Do not release the bob at angles greater than 30° from vertical — large angles both invalidate the small-angle approximation AND increase the risk that the bob strikes the photogate. Walk around and check: (1) groups are measuring h vertically, not along the arc; (2) they are averaging three trials; (3) their E(thermal) values are positive and typically 1–8% of PE₁. If a group gets negative E(thermal), the most common cause is measuring d instead of the swept width (the photogate beam sees the maximum chord — for a spherical bob this equals the diameter, so d is correct — but a hooked bob may have a smaller effective width).

   Materials

   • Pendulum stand and clamp
   • String (about 1.0 m) with a metal bob (0.20–0.50 kg)
   • Digital scale (0.01 g resolution)
   • Meter stick
   • Photogate with timing software (e.g., Vernier or PASCO) OR two photogates
   • Calipers to measure bob diameter
   • Protractor (for release angle)
   • Lab notebook or handout below

   Example outputs

   • m = 0.200 kg, d = 0.0254 m, h₁ = 0.150 m, Δt(avg) = 0.0148 s → v = 1.72 m/s, PE₁ = 0.294 J, KE₂ = 0.296 J, E(thermal) ≈ 0 J (within measurement error, ~1%). Excellent — nearly frictionless.
   • m = 0.500 kg, d = 0.0320 m, h₁ = 0.250 m, Δt(avg) = 0.0155 s → v = 2.06 m/s, PE₁ = 1.225 J, KE₂ = 1.062 J, E(thermal) = 0.163 J, which is 13.3% of the initial PE — attributed to air drag and pivot friction.
   • Analysis (a): No — mass cancels in ½mv² = mgh, so v = √(2gh) is mass-independent in the frictionless limit.
2. 30m

   Spreadsheet Energy Model — Cart on an Incline with FrictionLab

   Pairs will build a spreadsheet computational model that tracks PE, KE, and E(thermal) for a cart rolling down an incline, using position–time data provided (or, if time allows, their own measurements). The model must enforce E(total) = constant and solve for E(thermal) as the deficit. Student handout — pairs work through this on their laptop: Spreadsheet Model — Cart on an Incline (Student Handout) Inputs (row 1–5): - A1: Mass m (kg) — enter 0.500 - A2: Incline angle θ (°) — enter 20 - A3: g (m/s²) — enter 9.8 - A4: Reference height at the bottom of the track — h = 0 Data table — enter these measured values in columns A–C, rows 8–14: - Row 8: t = 0.00 s, s = 0.000 m (distance along track), v = 0.00 m/s - Row 9: t = 0.20 s, s = 0.067 m, v = 0.67 m/s - Row 10: t = 0.40 s, s = 0.268 m, v = 1.34 m/s - Row 11: t = 0.60 s, s = 0.594 m, v = 1.98 m/s - Row 12: t = 0.80 s, s = 1.032 m, v = 2.58 m/s - Row 13: t = 1.00 s, s = 1.575 m, v = 3.15 m/s - Row 14: t = 1.20 s, s = 2.215 m, v = 3.68 m/s Computed columns (build these formulas in D–G): - Column D — height above bottom: h = (s(max) − s) × sin(θ). Formula in D8: =(MAX($B$8:$B$14)-B8)*SIN(RADIANS($A$2)) - Column E — PE: =$A$1*$A$3*D8 - Column F — KE: =0.5*$A$1*C8^2 - Column G — E(total, expected): =$E$8 (the initial PE, held constant as a reference) - Column H — E(thermal) SOLVED: =G8-E8-F8 Part 1 — Check the model 1. What is E(thermal) in row 8? ______ J. Why must it be zero? 2. What is E(thermal) in row 14? ______ J. 3. Is E(thermal) monotonically increasing as t grows? Circle: Yes / No. Why does this make physical sense? Part 2 — Coefficient of friction 4. E(thermal, final) = f · s(total), where f is the average friction force. Solve for f using row 14: f = ______ N. 5. The normal force on the cart is N = mg·cos(θ) = ______ N. 6. Coefficient of kinetic friction μₖ = f / N = ______. Part 3 — Predict and test 7. Change the incline angle in A2 from 20° to 30°. Report the new E(thermal) at t = 1.20 s: ______ J. 8. Change the mass in A1 from 0.500 kg to 1.000 kg (angle back to 20°). Does v at row 14 change? Explain in one sentence using your equations. Part 4 — Break the model on purpose 9. Change the mass to a NEGATIVE number, say −0.500 kg. What happens to E(thermal)? Why does the model fail — and what real-world check does this correspond to? If your E(thermal) column ever goes negative for real data, stop — that means either your measurements are wrong or your total-energy reference is wrong. A good model checks itself. Walk around and check: (1) pairs are using absolute references ($) correctly so formulas fill down; (2) column D uses sin(θ) not tan(θ); (3) their μₖ answer for the sample data lands near 0.05–0.10 (a typical low-friction cart). If a pair finishes early, have them add a chart: select columns E, F, H and insert a stacked column chart — they'll see the classic energy-bar visualization animated across time.

   Materials

   • Laptop or Chromebook with a spreadsheet application (Google Sheets or Excel), one per pair
   • Low-friction cart on a track (or an actual cart used earlier this unit)
   • Meter stick and stopwatch (for a quick verification run, optional)
   • Handout below

   Example outputs

   • Row 14 with sample data: h = 0 m, PE = 0 J, KE = ½(0.5)(3.68)² = 3.39 J, E(total) = ½(0.5)(g)(2.215·sin20°) = 1.855 J·(from initial)… corrected sample: initial PE at t=0 with h₀ = 2.215·sin(20°) = 0.758 m gives PE₀ = (0.5)(9.8)(0.758) = 3.71 J; final KE = 3.39 J; E(thermal) = 0.32 J.
   • f = 0.32 J / 2.215 m = 0.145 N; N = (0.5)(9.8)cos(20°) = 4.60 N; μₖ = 0.145/4.60 = 0.032.
   • Part 3 Q8: v at row 14 is unchanged when mass doubles (mass cancels in gh = ½v² for the frictionless part, and friction force scales with mass so acceleration is mass-independent) — the model shows this by returning the same v when only m changes.

1. A 1200 kg car traveling at 25 m/s brakes to a stop on a level road. Treat the car + road + air as the system. (a) What is the initial KE? (b) What is the final KE? (c) How much thermal energy is generated in the brakes, tires, and air? (d) State in one sentence why this scenario does NOT violate conservation of energy.

   calculation(a) KE(initial) = ½(1200)(25)² = 375 000 J (b) KE(final) = 0 J (c) E(thermal) = KE(initial) − KE(final) = 375 000 J (d) The 375 000 J of kinetic energy was not destroyed — it was transferred to thermal energy in the brake rotors, tire–road contact, and surrounding air, so the total energy of the system is unchanged.

2. A 0.30 kg pendulum bob is released from a height of 0.40 m above the lowest point. At the lowest point it is measured to be moving at 2.6 m/s. Using the LOL/conservation method, calculate the thermal energy generated during the swing.

   calculationPE(initial) = mgh = (0.30)(9.8)(0.40) = 1.176 J KE(final) = ½mv² = ½(0.30)(2.6)² = 1.014 J E(thermal) = PE(initial) − KE(final) = 1.176 − 1.014 = 0.162 J (About 14% of the initial PE became thermal energy.)

3. A student claims: 'On the second hill of a roller coaster the car speeds up, so the total energy of the system must have increased.' Which of the following is the best physics response?

   multiple choiceCorrect choice: 'The total energy did not increase — as the car went down between the hills, PE converted into KE, so the car speeds up while the total stays constant (minus any thermal loss to friction).' Incorrect alternatives to include as distractors: 'Yes, the car gained energy from momentum'; 'The engine added energy at the top of the second hill'; 'Total energy can increase if the car goes faster.' The student's error is the misconception that adding motion adds total energy — within a closed system energy is only redistributed.

4. You build a spreadsheet model of a falling ball. At t = 0.50 s the model shows PE = 4.2 J, KE = 5.6 J, and E(total, initial) = 9.8 J. What value must the E(thermal) cell display, and what does that value tell you about air resistance in this drop?

   short answerE(thermal) = E(total) − PE − KE = 9.8 − 4.2 − 5.6 = 0.0 J. Because the deficit is zero, the model is telling us air resistance transferred negligible energy to thermal form over this 0.50 s of fall — the drop is effectively frictionless within the precision of the model.

• ☐Pendulum stands, clamps, and string (one per group of 3)
• ☐Metal pendulum bobs of varying mass (0.20–0.50 kg)
• ☐Digital scales (0.01 g resolution)
• ☐Meter sticks and protractors
• ☐Calipers (for bob diameter)
• ☐Photogates with timing software (Vernier LabQuest, PASCO, or equivalent) — one per group
• ☐Low-friction cart and track (for demo or optional student measurement)
• ☐Laptops/Chromebooks with Google Sheets or Excel (one per pair)
• ☐Printed student handouts for both activities
• ☐Projector for slide deck
• ☐Whiteboard markers for LOL diagram practice