Year 11 Physics Module 2: Dynamics 45 min Synthesis Lesson 9 of 15

Energy Synthesis

A car on a hill. Six formulae. One coherent picture. This lesson connects every energy concept from Phase 2 into a single chain — and shows how each calculation feeds the next.

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Think First

A 1200 kg car starts from rest at the top of a 20 m hill. It rolls to the bottom (frictionless), then continues along a flat road. The engine outputs 45 kW and total resistance is 900 N.

Before calculating anything: how many different Phase 2 energy concepts can you identify in this scenario? List every formula you think will be needed and explain what each one does in this situation.

Type your concept inventory below.

Write your concept inventory in your book.

Write your concept inventory in your book
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Come back to this at the end of the lesson.

📐

Formula Reference — This Lesson

$W = Fs\cos\theta$  |  $KE = \tfrac{1}{2}mv^2$  |  $W_{net} = \Delta KE$
ΔU = mgΔh (gravitational PE)  |  Emech = KE + U (conserved when frictionless) P = ΔE/Δt = Fv (power)
Synthesis chain: $\Delta U \to W_{net} = \Delta KE \to$ find $v$ → then $P = Fv$ for power stage   |   Key check: Is friction present? If yes, use $W_{net} = \Delta KE$ (not conservation)

Ph2
Phase 2 Master Formula Reference — All Six Formulae

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$W = Fs\cos\theta$
Work Done
W = work (J) | F = force (N) | s = displacement (m) | θ = angle between F and s
Use when: calculating energy transferred by a force. Foundation of all other energy formulae.
Trap: θ = 90° → W = 0. Normal force, centripetal force, and forces perpendicular to motion never do work.
$KE = \tfrac{1}{2}mv^2$
Kinetic Energy
KE = kinetic energy (J) | m = mass (kg) | v = speed (m/s)
Use when: calculating energy of motion at any instant
Trap: KE ∝ v² — doubling speed quadruples KE
$W_{net} = \Delta KE$
Work-Energy Theorem
Wnet = net work by all forces (J) | ΔKE = KEf − KEi
Use when: friction is present and mechanical energy is not conserved
Trap: must include ALL forces — applied, friction, gravity component along slope
$\Delta U = mg\Delta h$
Gravitational PE Change
ΔU = PE change (J) | m = mass (kg) | g = 9.8 m/s² | Δh = vertical height change (m)
Use when: object changes height in a uniform gravitational field
Trap: Δh = vertical height only — not slope distance
$KE_1 + U_1 = KE_2 + U_2$
Conservation of Mechanical Energy
Subscripts 1 and 2 = any two points on the path
Use when: frictionless — only gravity does work
Trap: NEVER use when friction is present. Use Wnet = ΔKE instead.
$P = \Delta E/\Delta t = Fv\cos\theta$
Power
P = power (W) | ΔE = energy transferred (J) | Δt = time (s) | F = force (N) | v = velocity (m/s)
Use when: finding rate of energy transfer or connecting force, speed, and power
Trap: at constant velocity, Fdrive = Fresistance (Newton 1) before using P = Fv

Know

  • All six Phase 2 formulae and their conditions
  • Three energy categories: stored, transferred, lost
  • How each formula connects to adjacent formulae
  • When NOT to use each formula

Understand

  • Why each calculation in a chain feeds the next
  • How to identify which formula applies before calculating
  • Why energy accounting always balances
  • The difference between ideal and real energy systems

Can Do

  • Draw an energy flow diagram for any scenario
  • Solve multi-step problems using two or more formulae in sequence
  • Identify errors in multi-step energy solutions
  • Select the correct formula from context without being told which to use

Choose how you work — type your answers below or write in your book.

Misconceptions to Fix

✗

Wrong: An object moving at constant velocity has no forces acting on it.

✗

Right: An object at constant velocity has BALANCED forces (net force = 0), not necessarily no forces.

📚 Core Content

Key Terms
frictionpresent and mechanical energy is not conserved
accounting for energythe foundation of every dynamics problem
Energynever created or destroyed — only converted
your final KEless than the initial PE, the difference went somewhere
The formulaenot separate tools — they are different views of the same energy accounting system
Herethe full calculation chain for the car scenario
01Card 1 — Map the Energy First

Card 1 — Map the Energy First

Before calculating anything, draw the energy flow. Every joule that enters a system must either be stored, converted, or transferred — accounting for energy is the foundation of every dynamics problem.

Every energy problem in Phase 2 involves some combination of three categories. Train yourself to classify each energy transfer before touching a formula:

CategoryWhat it meansFormulaIn the car scenario
Stored energy Energy an object possesses due to position or motion PE = mgh, KE = ½mv² PE at top of hill; KE at bottom; KE on flat road
Energy transferred by a force Work done — energy moving into or out of the system W = Fs cosθ, Wnet = ΔKE Work by engine; work done against friction/drag
Energy lost to environment Mechanical energy converted to heat and sound — not recoverable Elost = initial − final Emech Friction on flat road converts KE to heat in tyres and road
Energy Flow Diagram — Car on a Hill
The Golden Rule Every joule must be accounted for. Energy is never created or destroyed — only converted. If your final KE is less than the initial PE, the difference went somewhere. If you cannot identify where, your energy map is incomplete.
02Card 2 — Quantitative Synthesis

Card 2 — Quantitative Synthesis

The formulae are not separate tools — they are different views of the same energy accounting system. Each answer feeds the next question.

Here is the full calculation chain for the car scenario. Notice how the output of each step becomes the input of the next:

Calculation Chain — Car on a Hill
1
PE = mgh = 1200 × 9.8 × 20 = 235 200 J
Energy stored at the top. Formula: ΔU = mgΔh. Reference level: bottom of hill = h = 0.
Feeds Step 2 — this PE becomes KE at the bottom.
2
Conservation (frictionless): KE = PE → ½mv² = 235 200 J → v = √(2 × 235 200/1200) = 19.8 m/s
No friction on hill — use conservation KE₁ + U₁ = KE₂ + U₂. All PE converts to KE at bottom.
Feeds Step 3 — this speed is now used in P = Fv on the flat road.
3
On flat at v = 19.8 m/s: P needed = Fresist × v = 900 × 19.8 = 17 820 W = 17.8 kW
Constant velocity on flat → Newton 1 → Fdrive = Fresist = 900 N. Power = Fv.
Feeds Step 4 — compare to available engine power to find reserve.
4
Reserve power = Pengine − Pneeded = 45 000 − 17 820 = 27 180 W = 27.2 kW
Engine has 45 kW available; only 17.8 kW needed to maintain constant speed. Remainder can accelerate.
Feeds Step 5 — reserve power drives acceleration.
5
Accelerate from 19.8 → 30 m/s: ΔKE = ½ × 1200 × (30² − 19.8²) = ½ × 1200 × (900 − 392) = 304 800 J
Work needed to change KE = ΔKE = KEf − KEi. Use Wnet = ΔKE.
Feeds Step 6 — use P = W/Δt to find time to accelerate.
6
Time to accelerate: Δt = W/Preserve = 304 800 / 27 180 = 11.2 s
P = ΔE/Δt → rearrange for Δt. The reserve power drives the KE increase over this time.
Chain complete — every Phase 2 formula has been used in sequence.
Notice the chain The answer from conservation (v = 19.8 m/s) became the v in P = Fv. The power calculation gave the reserve, which gave the work available for acceleration, which gave the time via P = W/Δt. Energy accounting is a chain — and every formula is just one link.
03Card 3 — How the Formulae Connect

Card 3 — How the Formulae Connect

If you understand how the formulae connect, you can solve any problem — even ones you have never seen before.

Formula
Derived from
Connects to
Do NOT use when
W = Fs cosθ
Definition of work — force through displacement
Dividing by Δt gives P = Fv. Setting Wnet = ΔKE gives the work-energy theorem.
θ = 90° — then W = 0 (no work done regardless of F or s)
KE = ½mv²
W = Fs applied to a mass from rest, using kinematics
Paired with ΔU = mgΔh in conservation. Used in Wnet = ΔKE for final speed.
As a shortcut for work — work and KE are related but not identical
Wnet = ΔKE
Newton 2 applied over a displacement
When only gravity: Wnet = −ΔU → leads to conservation. When friction: Elost = −Wfriction.
When only some forces are included — must be NET work of ALL forces
ΔU = mgΔh
W = Fs cosθ applied to vertical lifting (θ = 180° for gravity)
Substituted into conservation equation. Divided by Δt gives P = mgh/Δt.
When Δh is a slope distance — must be the vertical component only
KE₁+U₁ = KE₂+U₂
Wnet = ΔKE when Wnet = work by gravity = −ΔU
Mass cancels → v = √(2gΔh) for frictionless drop from rest. Special case of Wnet = ΔKE.
When ANY friction or air resistance is present — use Wnet = ΔKE instead
P = ΔE/Δt = Fv
P = W/Δt = Fs cosθ / Δt = F(s/Δt) cosθ = Fv cosθ
At constant v → Newton 1 gives Fdrive. P = mgh/Δt is the special case for lifting.
At constant v without first finding Fdrive = Fresist via Newton 1
Part A — from Work to Power and Conservation W = Fs·cosθ Work done — the foundation divide by t set W_net s ÷ t = v P = W ÷ t Average power W_net = ΔKE Work-energy theorem P = Fv·cosθ Instantaneous power only gravity acts KE₁+U₁ = KE₂+U₂ Conservation (frictionless only)
Part B — from Gravitational PE to other formulae ΔU = mgΔh Gravitational PE divide by t from rest, frictionless friction present P = mgh ÷ t Lifting power v = √(2gΔh) From rest — mass cancels E_lost = ΔE_mech F_avg = E_lost ÷ s All paths connect through W_net = ΔKE as the central link
Interactive: Energy Transformation Flow
Interactive: Energy Transfer — Spring Oscillation (3D)
Navigation: ↗ Drag to rotate  ·  Scroll to zoom  ·  Double-click to reset. Watch elastic potential energy convert to KE and back. At maximum compression, KE = 0 and EPE is at its peak — and vice versa at maximum extension.

✏️ Worked Examples

Worked Example 1 Multi-concept Synthesis

Problem Setup

Key insight: The speed from conservation is NOT the speed on the flat — the cyclist must slow down because their power cannot sustain the hill speed against drag.

Scenario: A 500 kg cyclist + bike system starts from rest at the top of a 15 m frictionless hill. At the bottom they travel along a flat road. Drag force = 60 N. Cyclist power output = 300 W. Find: (a) speed at bottom of hill, (b) constant speed on flat road, (c) time to travel 500 m along flat at this speed.

  • m = 500 kg | h = 15 m | frictionless hill
  • Fdrag = 60 N (flat road) | P = 300 W

Solution

1
(a) Conservation (frictionless hill): v = √(2gh) = √(2 × 9.8 × 15) = √294 = 17.1 m/s
Frictionless → use conservation. Mass cancels. This is the speed at the bottom of the hill.
2
Check: at 17.1 m/s, P needed = Fdrag × v = 60 × 17.1 = 1026 W — but cyclist only has 300 W.
Cyclist cannot maintain hill speed on the flat — they decelerate. Conservation gave the entry speed, not the cruise speed.
3
(b) Constant speed when P = Fdrag × v: v = P/Fdrag = 300/60 = 5 m/s
At constant v → Newton 1 → Fdrive = Fdrag. P = Fdrive × v → v = P/Fdrive = P/Fdrag.
4
(c) t = s/v = 500/5 = 100 s
Constant speed on flat — simple kinematics. Distance divided by speed.

What would change if...

The cyclist's power output increased to 1200 W. Would they now be able to maintain 17.1 m/s on the flat? What would their new constant cruise speed be? What is the key check you must do before assuming conservation speed = cruise speed?

Worked Example 2 Multi-concept Synthesis — Full Chain

Problem Setup

Key insight: Start with the known (constant speed → P = Fv gives starting v), then use Wnet = ΔKE for the hill.

Scenario: A 900 kg car cruises at constant highway speed with engine power 60 kW and resistance 1500 N. The car then descends a 10 m hill (road distance 80 m, μk = 0.1, engine off). Find the speed at the bottom.

  • m = 900 kg | P = 60 kW | Fresist = 1500 N (highway)
  • Hill: h = 10 m | road distance = 80 m | μk = 0.1 | engine off

Solution

1
Highway: constant v → Fdrive = 1500 N. v = P/Fdrive = 60 000/1500 = 40 m/s
Starting speed = cruise speed from P = Fv. This is v₁ for the hill calculation.
2
On hill: sinθ = h/s = 10/80 = 0.125 → cosθ = √(1 − 0.0156) = 0.992. FN = mg cosθ = 900 × 9.8 × 0.992 = 8744 N
Normal force on slope — needed for friction calculation. cosθ from slope geometry.
3
fk = μk × FN = 0.1 × 8744 = 874 N (up the slope, opposing motion)
Kinetic friction acts up the slope — it does negative work on the descending car.
4
Wgravity = mgh = 900 × 9.8 × 10 = 88 200 J (positive — gravity does work on descending car)
Work by gravity component along slope = mgh (using vertical height, not road distance).
5
Wfriction = −fk × s = −874 × 80 = −69 920 J (negative — opposes motion)
Friction acts opposite to displacement — negative work, removes KE.
6
Wnet = 88 200 − 69 920 = 18 280 J. Wnet = ΔKE: ½mv₂² = ½mv₁² + Wnet = ½×900×1600 + 18 280 = 738 280 J
Wnet = ΔKE — friction is present so no conservation. Sum all work done.
7
v₂ = √(2 × 738 280/900) = √1640.6 = 40.5 m/s
Car speeds up slightly on the descent because gravity work (88 kJ) exceeds friction loss (70 kJ). Engine is off but gravity does more work than friction removes.

What would change if...

The friction coefficient increased to μk = 0.3. Predict qualitatively whether the car would be faster, slower, or the same speed at the bottom compared to μk = 0.1 — then calculate to check.

🏃 Activities

Activity 01 — Pattern A

Diagram and Calculate — Motorcycle on a Road

Draw an energy flow diagram, then work through the full calculation chain.

A 300 kg motorcycle starts from rest on a flat road. The engine applies 1800 N for 100 m against a friction force of 300 N. At the end of this section the engine cuts out, and the motorcycle travels up a 25 m frictionless hill.

  1. Draw an energy flow diagram showing all inputs (stored, transferred, lost) for both sections. Label each arrow with the formula used.
  2. Calculate the net work done over the flat section and hence the speed at the end of the flat.
  3. Using conservation, calculate the maximum height the motorcycle can reach on the hill (engine is off).
  4. Explain why the motorcycle cannot reach the full 25 m height — what would need to be true for it to just reach the top?

Type your working below. Draw the energy flow diagram in your book.

Complete all four parts in your book, including the energy flow diagram.

Complete all four parts in your book
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Activity 02 — Pattern A

True or False — Classify and Connect

For each statement: select True or False, then identify which Phase 2 formula the statement relates to.

Doubling the speed of an object quadruples its kinetic energy.?
A normal force does positive work on a box sliding horizontally across a flat floor.?
Conservation of mechanical energy applies even when friction is present, as long as you account for the heat produced.?
P = Fv only applies when the object is moving at constant velocity.?
When mass cancels from the energy conservation equation, the speed at the bottom of a frictionless drop depends only on the height — not the mass.?
The work-energy theorem requires the NET work done by all forces — not just the applied force.?
To calculate gravitational PE using ΔU = mgΔh, Δh should be the distance along the slope — not the vertical height change.?
Two motors doing the same work in different times have different power outputs — the faster motor is more powerful.?
A stationary 10 000 kg truck has more kinetic energy than a moving 9 g bullet.?
If a force acts at 90° to the direction of motion, the work done by that force is zero — regardless of the magnitude of the force or the distance travelled.?

✅ Check Your Understanding

Revisit Your Thinking

Earlier you were asked: how many Phase 2 energy concepts can you identify in the car scenario — and which formulae will you need?

The full answer: all six. PE at the top (ΔU = mgΔh), conservation on the frictionless hill (KE₁+U₁=KE₂+U₂), KE at the bottom (KE = ½mv²), power to maintain constant speed (P = Fv), the work-energy theorem to find acceleration time (Wnet = ΔKE), and power = energy/time to connect the reserve power to the KE increase (P = ΔE/Δt).

The deeper insight: these are not six separate formulas. They are one idea — energy accounting — viewed from six different angles. Work is the mechanism of transfer. KE and PE are the forms of storage. The work-energy theorem is the accounting rule. Conservation is the special case when nothing leaks. Power is the rate. If you can see that one picture, you can solve any problem in Phase 2.

Look back at your initial inventory. How many did you identify? What connections were you missing?

Annotate your initial inventory in your book — add any connections you missed.

Annotate your initial inventory in your book
Saved
Revisit Your Initial Thinking

Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?

MC

Multiple Choice

5 random questions from a replayable lesson bank — feedback shown immediately

AnalyseBand 3

7. Draw and label a complete energy flow diagram for the following scenario: a 400 kg object slides from rest down a rough 20 m ramp, then travels along a flat surface until it stops. Include all energy inputs, stores, transfers, and losses. State which formula applies at each stage. 3 MARKS

Draw and label the energy flow diagram in your book
Saved
ApplyBand 5

8. A 750 kg car starts from rest at the top of a frictionless 25 m hill. At the bottom it travels along a flat road at constant speed with engine power 30 kW. (a) Find the speed at the bottom of the hill. (b) Find the total resistance force on the flat road. (c) How long does the car take to travel 1.2 km along the flat road at this speed? 3 MARKS

Answer in your book showing all three parts
Saved
EvaluateBand 6

9. A 60 kg skateboarder starts from rest at the top of a 4 m high half-pipe ramp. (a) Using conservation, find the speed at the bottom. (b) In reality, the skateboarder arrives at the bottom at 7 m/s. Calculate the energy lost and identify at least two physical mechanisms responsible. (c) On the next run, the skateboarder pushes off the top edge with an initial speed of 2 m/s. Find the new speed at the bottom (assume the same 30% energy loss as in part b). (d) Explain why the same percentage energy loss applies even at this higher initial speed — what physical insight does this reveal? 4 MARKS

Answer in your book — show all four parts
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Comprehensive Answers

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Activity 01 — Motorcycle on a Road

Part 2 — Flat section: Wengine = 1800 × 100 × cos0° = 180 000 J. Wfriction = −300 × 100 = −30 000 J. Wnet = 150 000 J. v = √(2 × 150 000/300) = √1000 = 31.6 m/s.

Part 3 — Frictionless hill: KE at base = 150 000 J. Use conservation: mgh = KE → h = KE/mg = 150 000/(300 × 9.8) = 51.0 m. The motorcycle can reach 51 m — well above the 25 m hill, so it makes it over with energy to spare.

Note: If the hill were 51 m exactly, the motorcycle would just reach the top with zero speed. It can reach the full 25 m because 150 kJ > mgh(25m) = 300 × 9.8 × 25 = 73 500 J.

Part 4: If the hill were higher than 51 m, the motorcycle could not reach the top — its KE at the base would be less than the PE needed. It would decelerate, stop, and slide back. For it to just reach the top: KE at base = mghtop, which requires enough net work done on the flat.

Activity 02 — True/False Answers

  1. True — KE = ½mv². If v doubles, v² quadruples, so KE quadruples.
  2. False — W = Fs cos90° = 0. Normal force is perpendicular to horizontal motion — zero work done.
  3. False — When friction is present, conservation of mechanical energy does NOT apply. Use Wnet = ΔKE instead. The total energy (including heat) is conserved, but mechanical energy is not.
  4. False — P = Fv cosθ gives instantaneous power at any speed — not just constant velocity. At constant velocity it is most useful because Newton 1 gives Fdrive.
  5. True — From ½mv² = mgh, mass cancels to give v = √(2gh). Speed depends only on height dropped.
  6. True — Wnet = ΔKE. Using only applied force ignores friction and gives wrong ΔKE.
  7. False — ΔU = mgΔh uses the vertical height change — not the slope distance. Δh = h, never the road/ramp length.
  8. True — P = ΔE/Δt. Same work (ΔE the same), smaller Δt → larger P. The faster motor is more powerful.
  9. False — KE = ½mv². A stationary truck has v = 0, so KE = 0. A bullet at 900 m/s: KE = ½ × 0.009 × 810 000 = 3645 J. The bullet has more KE.
  10. True — W = Fs cos90° = 0. Any force perpendicular to displacement does zero work, regardless of F or s.

Multiple Choice

1. B — Flat surface, constant speed → Newton 1 → Fapplied = Ffriction. P = Fv. Option A (mgh/Δt) is for vertical lifting only.

2. C — Frictionless ramp: PE → KE (conservation). Rough flat: KE → heat (friction does negative work — Wnet = ΔKE).

3. D — 18.8 m/s. v = √(2gh) = √(2 × 9.8 × 18) = √352.8 = 18.8 m/s. Option A is wrong — mass is included but cancels. Option C divides by 2 instead of multiplying.

4. A — 40 m/s. Cruise speed from P = Fv: v = P/F = 20 000/500 = 40 m/s. The hill speed (v = √(2 × 9.8 × 30) = 24.2 m/s) is irrelevant to the cruise speed on the flat — this is a common synthesis error.

5. B — With friction, energy is lost to heat. Final KE must be less than initial PE. The student applied conservation (which ignores friction losses) rather than Wnet = ΔKE.

6. C — Friction present + engine on → cannot use conservation or v = √(2gh). Must use Wnet = ΔKE with all three work terms: Wengine (positive), Wgravity (positive, going up = negative, wait — descending is positive → going UP means gravity does negative work), Wfriction (negative). Start from cruise speed via P = Fv.

Short Answer — Model Answers

Q7 (3 marks): Stage 1 (ramp): Gravitational PE stored at top → converts to KE + heat as object descends rough ramp. Formula: Wnet = ΔKE (friction present — cannot use conservation). Wnet = Wgravity + Wfriction = mgh − fks. Stage 2 (flat): KE at base → all KE converts to heat (friction decelerates object to rest). Formula: Wnet = ΔKE (Wnet = −fks = −KE, object stops).

Q8 (3 marks):

(a) v = sqrt(2gh) = sqrt(2 × 9.8 × 25) = sqrt(490) = 22.1 m/s
(b) Constant v: P = F_resist × v → F_resist = P/v = 30 000/22.1 = 1357 N
(c) t = s/v = 1200/22.1 = 54.3 s

Q9 (4 marks):

(a) v = sqrt(2 × 9.8 × 4) = sqrt(78.4) = 8.85 m/s
(b) Ideal KE_bottom = mgh = 60 × 9.8 × 4 = 2352 J. Actual KE = 1/2 × 60 × 49 = 1470 J. E_lost = 882 J (37.5%). Mechanisms: rolling friction in wheels/bearings; air resistance; internal deformation of skateboard/body.
(c) Initial KE = 1/2 × 60 × 4 = 120 J. Total initial E = KE + PE = 120 + 2352 = 2472 J. After 30% loss: E_final = 0.70 × 2472 = 1730 J. v = sqrt(2 × 1730/60) = sqrt(57.7) = 7.59 m/s

(d) Physical insight: assuming the same percentage energy loss means the resistive forces (friction, air resistance) scale with the total energy being transferred. In reality, the same percentage loss is an approximation — but it captures the idea that energy dissipation is proportional to the forces acting over the path length, and those forces remain roughly the same regardless of initial speed. The insight: adding more initial KE does not change the physical environment — the same fraction of energy is "taxed" by friction each run.

Consolidation Game

Energy Synthesis

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