Year 11 PhysicsModule 2⏱ ~40 min5 MC · 3 Short AnswerLesson 9 of 15
Energy Synthesis
In 2022, a Tesla Model 3 Long Range (1,844 kg) travelling at 100 km/h carried $KE = \frac{1}{2}(1844)(27.8)^2 = 711\text{ kJ}$ of kinetic energy. Tesla's regenerative braking system recovered 90% of this, 640 kJ, on a single stop from highway speed. Engineers used the full synthesis chain: $KE \to W_{regen} = \Delta KE \to P_{output} = Fv$, exactly the multi-step energy chain this lesson teaches.
Today's hook: A 2022 Tesla Model 3 (1,844 kg) brakes from 100 km/h to rest. The car recovers 640 kJ of the 711 kJ of kinetic energy. Before calculating: which two lessons from L06–L08 did Tesla's engineers use to calculate both the energy recovered and the power delivered back to the battery?
0/5TASKS
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Warm up first
Warm up first
Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.
A 1200 kg car starts from rest at the top of a 20 m hill. It reaches the bottom and continues along a flat section. Before calculating anything: how many different energy concepts from L06–L08 can you identify in this scenario?
In the synthesis chain $\Delta U \to W_{net} = \Delta KE \to v \to P = Fv$, which step converts potential energy to kinetic energy?
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Learning Intentions
Learning Intentions
goals
Know
The four-step energy synthesis chain
When to use energy conservation vs work-energy theorem
How to connect KE, GPE, Work, and Power in one problem
Understand
Why the synthesis chain is sequential, each step feeds the next
Why identifying all energy types first prevents errors
How friction breaks energy conservation but not the work-energy theorem
Can Do
Identify all energy types in a multi-step scenario
Apply the synthesis chain to find velocity and then power
Solve multi-concept energy problems at Band 5-6 level
Cross-lesson links: L09 is the synthesis lesson for all of Phase 2. It chains $W = Fs\cos\theta$ (L06), $\Delta U = mgh$ and $KE_1 + U_1 = KE_2 + U_2$ (L07), and $P = \Delta E/\Delta t = Fv$ (L08) into one sequential calculation. The final velocity you calculate here also becomes the momentum $p = mv$ in Phase 3 (L11–L13), making L09 the pivot between the energy and momentum phases of Module 2.
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The Energy Synthesis Chain
Core Content
01
The Energy Synthesis Chain
+5 XP
A ball rolls off a table at 3 m/s and falls 1.2 m to the floor. You want to know how fast it hits the floor and what force the floor must exert to stop it in 0.01 s. This looks like three separate problems, but it is one chain: first use energy conservation to find the speed at impact, then use that speed in the impulse equation. Each answer feeds the next step.
$$\Delta U \to W_{net} = \Delta KE \to v \to P = Fv$$
Step by step:
Find $\Delta U$: Calculate the change in GPE using $\Delta U = mg\Delta h$
Apply $W_{net} = \Delta KE$: If no friction: $W_{net} = \Delta U$ (energy conservation). If friction: $W_{net} = \Delta U - W_{friction}$
Find velocity: $\frac{1}{2}mv^2 = \Delta KE + KE_i$; solve for $v$
Find power (if needed): $P = Fv$ at constant speed, or $P = \Delta E/\Delta t$
Worked ExampleFull Synthesis Chain
A 1200 kg car starts from rest at the top of a 20 m hill and coasts down (no friction). At the bottom, the engine then drives the car at 15 m/s maintaining constant speed. Find: (a) speed at the bottom of the hill, (b) engine power to maintain 15 m/s if drag is 800 N.
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$\Delta U = mgh = 1200 \times 9.8 \times 20 = 235\,200\text{ J}$ (lost by car as it descends)
At constant 15 m/s, engine force = drag = 800 N. $P = Fv = 800 \times 15 = 12\,000\text{ W} = 12\text{ kW}$
Energy synthesis chain: $\Delta U = mg\Delta h \to W_{net} = \Delta KE \to v = \sqrt{2\Delta KE/m} \to P = Fv$; without friction $\Delta KE = \Delta U$ (conservation); with friction $\Delta KE = \Delta U - W_{friction}$.
Pause, copy the highlighted synthesis chain into your book before moving on.
True or false: In a problem with friction, the work-energy theorem still applies, but the total mechanical energy is not conserved.
A 500 kg object drops 10 m from rest (no friction). Its speed at the bottom and its KE there are:
After completing the velocity calculation, the next step to find power (at constant speed) is:
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Full Synthesis Chain Problem
Activities
Full Synthesis Chain Problem
EvaluateBand 6
A 60 kg cyclist starts from rest at the top of a 15 m hill (frictionless). At the bottom, they maintain constant 12 m/s against a 120 N wind drag. Use the full synthesis chain to find: (a) speed at the bottom, (b) power required to maintain 12 m/s. ($g = 9.8$ m/s²)
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Energy with Friction
Energy with Friction
AnalyseBand 5
A 2 kg block slides 5 m down a ramp (30° incline) with $\mu = 0.20$. Find: (a) GPE lost, (b) work done by friction, (c) final KE, (d) final speed. ($g = 9.8$ m/s²)
✓
Show what you have learned
Quick recall, Energy Synthesis
+5 XP
Short Answer, 10 marks
+5 XP
UnderstandBand 3(3 marks) 1. A ball is released from rest at the top of a 12 m hill. Explain using the concept of energy conservation why the ball's final speed at the bottom is the same regardless of the path taken (assuming no friction).
ApplyBand 4(3 marks) 2. A 3 kg ball rolls from rest down a 5 m frictionless slope. At the bottom it rolls along a flat surface at constant speed. If friction on the flat surface is 4.5 N, calculate the power needed to maintain this constant speed. ($g = 9.8$ m/s²)
EvaluateBand 6(4 marks) 3. A 1500 kg car descends a 30 m hill, starting from rest. Friction (brakes + air) does 45 000 J of negative work. (a) Find the speed at the base of the hill. (b) On the flat, the car maintains this speed against 750 N total drag. Find the required engine power. ($g = 9.8$ m/s²)
Show all answers
Short Answer, Model Answers
Q1 (3 marks): By conservation of mechanical energy, $KE_1 + U_1 = KE_2 + U_2$. Starting from rest ($KE_1 = 0$) at height $h$ ($U_1 = mgh$), and reaching the bottom ($U_2 = 0$): $mgh = \frac{1}{2}mv^2$, so $v = \sqrt{2gh}$. This equation depends only on the height change $h$, not the path length or shape. Mechanical energy is conserved as long as no non-conservative forces (like friction) act, so the final speed is always $\sqrt{2gh}$ regardless of path.
The 2022 Tesla Model 3 scenario uses the same synthesis chain taught here. The 1,844 kg car at 100 km/h carries 711 kJ of KE; regenerative braking captures 640 kJ (90%) and delivers it back to the battery via $P = Fv$. The key insight: energy problems with multiple stages are not harder, they just have more sequential steps, each straightforward once the previous result is known.