Physics • Year 11 • Module 2: Dynamics • Lesson 9
Energy Synthesis
Lock in the six Phase 2 energy formulae, the three energy categories, and the conditions for choosing each formula before tackling harder problems.
1. Term–definition match
The definitions below are shuffled. In the right-hand column write the matching term from this list: work done, kinetic energy, work-energy theorem, gravitational PE, conservation of mechanical energy, power, net work, stored energy, energy lost, mechanical energy. 10 marks (1 each)
| # | Definition | Matching term |
|---|---|---|
| 1.1 | The energy an object possesses due to its motion; equal to ½mv². | |
| 1.2 | The energy transferred when a force acts through a displacement; W = Fs cosθ. | |
| 1.3 | The principle that the total work done by all forces equals the change in kinetic energy; Wnet = ΔKE. | |
| 1.4 | The energy stored by an object due to its height above a reference level in a gravitational field; ΔU = mgΔh. | |
| 1.5 | The principle that in a frictionless system the sum of KE and PE remains constant; KE1 + U1 = KE2 + U2. | |
| 1.6 | The rate of energy transfer; P = ΔE/Δt = Fv. | |
| 1.7 | The vector sum of all work contributions from every force acting on an object. | |
| 1.8 | The sum of kinetic energy and gravitational potential energy at any point. | |
| 1.9 | An energy category covering KE and PE — energy held by an object due to its motion or position. | |
| 1.10 | Mechanical energy converted to heat and sound by friction — not recoverable as mechanical energy. |
2. Formula conditions — true or false with correction
Circle T or F for each statement. If the statement is false, write the corrected version on the line below it. 12 marks (1 T/F + 1 correction each)
2.1 Conservation of mechanical energy (KE1 + U1 = KE2 + U2) can be used whenever friction is present as long as you include the heat produced. T / F
2.2 In the formula ΔU = mgΔh, the symbol Δh represents the vertical height change — not the distance along a slope. T / F
2.3 A normal force acting on a box sliding across a horizontal floor does positive work because the box moves forward. T / F
2.4 When using P = Fv at constant velocity, the driving force equals the resistance force (Newton’s first law). T / F
2.5 Doubling the speed of an object doubles its kinetic energy because KE is proportional to v. T / F
2.6 The work-energy theorem requires the NET work done by ALL forces — not just the applied force. T / F
3. Fill-in-the-blank paragraph
Use the word bank to complete the passage. Each word is used once. 8 marks (1 per blank)
Word bank:
conservation · converted · destroyed · friction · height · net · power · rate
Energy is never created or ___________ — only ___________ from one form to another. When a frictionless hill is involved, we use ___________ of mechanical energy. When ___________ is present, mechanical energy is not conserved, and we must use the work-energy theorem with the ___________ work done by all forces. The formula ΔU = mgΔh uses the vertical ___________ change, not the slope distance. ___________ is the ___________ of energy transfer and is calculated as P = ΔE/Δt or P = Fv.
4. Function recall
Answer each question in 1–2 sentences using precise terms from the lesson. 8 marks (2 each)
4.1 State the condition that determines whether you should use conservation of mechanical energy or the work-energy theorem to solve a dynamics problem.
4.2 Explain why P = Fv is most useful when an object moves at constant velocity, and state what Newton’s first law tells you in this situation.
4.3 What is the “synthesis chain” described in Card 2 of the lesson, and why is it important for multi-step energy problems?
4.4 Explain what happens to mechanical energy “lost” to friction — where does it go?
5. Complete the formula reference table
Fill in the missing cells. Each row refers to one of the six Phase 2 energy formulae. 12 marks (1 per cell)
| Formula | What it calculates | Use when … | Do NOT use when … |
|---|---|---|---|
| W = Fs cosθ | A force acts through a displacement at angle θ | ||
| KE = ½mv² | Kinetic energy of a moving object | As a shortcut for calculating work done | |
| Net work done = change in kinetic energy | Friction is present; mechanical energy is not conserved | Only some forces are included (must use ALL forces) | |
| ΔU = mgΔh | When Δh is slope distance, not vertical height | ||
| KE1 + U1 = KE2 + U2 | Total mechanical energy is the same at any two points | System is frictionless; only gravity does work | |
| P = ΔE/Δt = Fv | Without first finding Fdrive = Fresist via Newton’s first law (at constant v) |
Q1 — Term–definition match
1.1 kinetic energy • 1.2 work done • 1.3 work-energy theorem • 1.4 gravitational PE • 1.5 conservation of mechanical energy • 1.6 power • 1.7 net work • 1.8 mechanical energy • 1.9 stored energy • 1.10 energy lost.
Q2 — True / false with correction
2.1 False. Conservation of mechanical energy does NOT apply when friction is present. When friction acts, use Wnet = ΔKE instead. The total energy (including heat) is conserved, but mechanical energy decreases.
2.2 True.
2.3 False. The normal force acts perpendicular (90°) to the direction of motion. W = Fs cos90° = 0. The normal force does zero work.
2.4 True. At constant velocity, Newton’s first law gives net force = 0, so Fdrive = Fresist. This driving force is then used in P = Fv.
2.5 False. KE = ½mv², so KE is proportional to v². Doubling speed quadruples kinetic energy.
2.6 True. The work-energy theorem requires net work — the sum of work done by every force acting on the object (applied, friction, gravity component, etc.).
Q3 — Cloze paragraph
In order: destroyed / converted / conservation / friction / net / height / Power / rate.
Q4.1 — When to use conservation vs work-energy theorem
Use conservation of mechanical energy (KE1 + U1 = KE2 + U2) only when the system is frictionless and only gravity does work. Use the work-energy theorem (Wnet = ΔKE) whenever friction or other non-conservative forces are present.
Q4.2 — P = Fv at constant velocity
P = Fv gives the instantaneous power at any speed, but it is most useful at constant velocity because Newton’s first law tells us that net force = 0, so the driving force exactly equals the resistance force (Fdrive = Fresist). This known force is used directly in P = Fv to find the power needed to maintain that speed.
Q4.3 — The synthesis chain
The synthesis chain is the process of solving multi-step energy problems by feeding the output of one formula as the input to the next. For example: calculate PE at the top → use conservation to find speed at the bottom → use P = Fv to find power needed on the flat. It is important because no single formula solves a real-world scenario — the formulae are links in a chain.
Q4.4 — Where does “lost” mechanical energy go?
Mechanical energy “lost” to friction is converted into thermal energy (heat) in the surfaces in contact, and sometimes sound energy. It is not destroyed — total energy is always conserved — but it is transferred to the environment in a form that cannot easily be recovered as mechanical energy.
Q5 — Formula reference table (missing cells)
Row 1 (W = Fs cosθ): What it calculates: Energy transferred by a force through a displacement. Do NOT use when: the force is perpendicular to displacement (θ = 90°) — work done is zero.
Row 2 (KE = ½mv²): Use when: calculating the energy of motion at any instant (or finding speed from a known KE value).
Row 3 (Wnet = ΔKE): Formula: Wnet = ΔKE.
Row 4 (ΔU = mgΔh): What it calculates: Change in gravitational potential energy. Use when: an object changes height in a uniform gravitational field.
Row 5 (conservation): Do NOT use when: any friction or air resistance is present — use Wnet = ΔKE instead.
Row 6 (P = ΔE/Δt = Fv): What it calculates: Rate of energy transfer (power). Use when: finding the rate of energy transfer or connecting force, speed, and power.