Calculating Energy Conservation
In 2023, a CSIRO-designed drop tower in Newcastle used a 2 kg mass falling 4 m to test energy harvesting, here's how to predict the speed.
Printable Worksheets
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Q1 · A 1 kg ball and a 5 kg rock both drop from the same height. Before reading, predict: which one hits the ground faster, and which one has more energy when it lands?
Q2 · If you dropped a ball from a 10 m cliff versus a 5 m cliff, how much faster do you think it would land, twice as fast, or some other amount? Why?
Key Relationships, This Lesson
● Know
- The equation: Energy input = Useful output + Waste output
- How to calculate efficiency as a percentage
- That units must be consistent in calculations
● Understand
- Why calculations prove conservation of energy
- How efficiency relates to useful and waste energy
- Why 100% efficiency is impossible in real devices
● Can do
- Calculate waste energy when input and useful output are known
- Calculate efficiency from energy data
- Use calculations to justify conclusions about energy use
Wrong: "If a device is 50% efficient, half the energy disappears."
Right: The "other half" doesn't disappear, it transforms into waste energy, usually heat. A 50% efficient device converts half the input into useful output and the other half into thermal energy that spreads into the surroundings.
Wrong: 50% efficiency means half the input energy becomes useful output, and the other half becomes waste energy (usually thermal). All energy is still accounted for, none disappears.
Right: Efficiency tells you the proportion that does useful work; energy conservation still holds. The waste portion transfers to the surroundings as heat, keeping the total energy budget balanced.
Wrong: "A bigger number for efficiency always means a better device."
Right: Efficiency must be considered alongside total energy consumed. A device with higher efficiency but much larger energy input may still waste more energy overall than a lower-efficiency device that uses very little input energy.
Wrong: Efficiency must be considered alongside the total energy used. An LED bulb at 20% efficiency might use far less total energy than an incandescent bulb at 5% efficiency because the LED needs much less input energy to produce the same light.
Right: A higher efficiency percentage is better only when comparing devices that do the same job with similar input energies. Always compare both efficiency and total energy input together when evaluating which device is truly better.
Watch a ball roll off a table, at the top it is still, at the bottom it is moving fast. All that motion came from the height. Mathematics lets you calculate the exact speed at the bottom before you even let go, because gravitational potential energy (PE) converts entirely into kinetic energy. PE equals mgh: mass times gravitational acceleration (9.8 m/s² on Earth) times height. Kinetic energy (KE) equals ½mv²: half the mass times velocity squared.
If friction and air resistance are negligible, all the gravitational potential energy at the top converts to kinetic energy at the bottom. Setting PE = KE gives us a precise prediction for velocity. This is one of the most elegant applications of the conservation law.
A 1 kg ball dropped from 5 m has PE = 1 × 9.8 × 5 = 49 J at the top. At the bottom, KE = 49 J = ½ × 1 × v², so v² = 98 and v ≈ 9.9 m/s. You can check this with a motion sensor in the lab.
Engineers designing Snowy Hydro 2.0 use these same calculations to predict how fast water will flow through turbines and how much electrical energy can be generated from a given volume and head height.
Complete this energy conservation calculation.
Setting up energy conservation problems requires careful algebra. The most common mistake is forgetting that velocity is squared in the kinetic energy formula. Another is mixing up metres and centimetres, or grams and kilograms.
Always start by writing the formulas clearly: PE = mgh and KE = ½mv². Then state your assumption, usually that energy is conserved (no friction). Only after that should you substitute numbers. This structured approach catches most errors before they happen.
A student calculates the speed of a 0.5 kg ball dropped from 2 m and gets 19.6 m/s. The error? They wrote KE = mv² instead of ½mv², forgetting the half. The correct answer is about 6.3 m/s.
Students often think heavier objects fall faster and therefore hit the ground with more speed. In fact, without air resistance, a feather and a hammer fall at the same rate. Mass affects the energy and momentum, but not the final velocity, that depends only on height.
Here's a student's working. One line has an error, click it.
- PE at top = mgh = 3 × 9.8 × 4 = 117.6 J
- KE at bottom = ½mv = ½ × 3 × v
- 117.6 = 1.5v, so v = 78.4 m/s
Calculate efficiency and waste energy for any device
Copy Into Your Books
▼Energy Accounting
- Energy input = Useful output + Waste output
- Waste = Input − Useful
- All energy must be accounted for
Efficiency
- Efficiency (%) = (Useful ÷ Input) × 100
- Or: Useful = Input × (Efficiency ÷ 100)
- No real device is 100% efficient
Australian Efficiencies
- Hydro: ~90% (most efficient)
- Wind: ~45%
- Gas: ~50%
- Coal: ~35%
- Solar PV: ~20%
- LED: ~20% (vs incandescent 5%)
Units
- Joule (J), SI unit of energy
- 1 kJ = 1,000 J
- 1 MJ = 1,000,000 J
- 1 kWh = 3.6 MJ
Energy Accounting Practice
1 A solar panel in Alice Springs receives 2,000 J of light energy and produces 340 J of electrical energy. Calculate the waste energy and the efficiency.
2 A coal power station has an efficiency of 35%. If it burns coal that releases 10,000 MJ of chemical energy, how much electrical energy does it produce, and how much is wasted?
3 An LED bulb is supplied with 60 J of electrical energy. If it is 20% efficient, how much light energy does it produce? How does this compare to an incandescent bulb that produces 3 J of light from the same 60 J input?
4 The Snowy Hydro scheme has turbines that are approximately 90% efficient. If water flowing through delivers 5,000 MJ of gravitational potential energy, calculate the electrical energy output and the waste energy.
Improving Efficiency
At the start of this lesson you were asked about a 1 kg ball and a 10 kg rock dropped from the same height, they hit the ground at the same speed, yet one has ten times more kinetic energy. Did you figure out why before reaching the end?
Now that you've worked through the energy equations, explain in your own words how mass and speed both contribute to kinetic energy. What surprised you most?
Q1. 6. A natural gas power station has an efficiency of 50%. It burns gas that releases 8,000 MJ of chemical energy. Calculate the electrical energy output and the waste energy output. Show all working.
1 mark for correct formula. 1 mark for correct electrical output. 1 mark for correct waste energy.Q2. 7. The following data compares two light bulbs used in Australian homes:
1 mark for calculating waste energy for each bulb. 1 mark for identifying the flaw in the student's reasoning. 1 mark for explaining efficiency in context. 1 mark for a balanced, evidence-based conclusion.Q3. 8. The Snowy 2.0 pumped hydro scheme stores energy by pumping water uphill, then releases it to generate electricity later. For every 100 MJ of electrical energy used to pump water, about 80 MJ of electrical energy can be generated when the water flows back down. Some politicians have claimed this means "20% of energy is lost, so pumped hydro is a waste." Analyse this claim using the law of conservation of energy, the concept of efficiency, and the value of storing energy for later use.
1 mark for explaining where the 20 MJ goes. 1 mark for showing that energy is conserved, not lost. 1 mark for explaining why pumped hydro is valuable despite inefficiency. 1 mark for linking to intermittency of solar/wind. 1 mark for a balanced conclusion.Model answers (click to reveal)
Comprehensive Answers
▼Activity 1, Energy Accounting Practice
1. Alice Springs solar panel: Waste = 2,000 − 340 = 1,660 J. Efficiency = (340 ÷ 2,000) × 100 = 17%.
2. Coal power station: Useful = 10,000 × 0.35 = 3,500 MJ. Waste = 10,000 − 3,500 = 6,500 MJ.
3. LED vs incandescent: LED light = 60 × 0.20 = 12 J. LED waste = 60 − 12 = 48 J. Incandescent waste = 60 − 3 = 57 J. The LED produces 4 times as much light from the same input, or could use far less power to produce the same light.
4. Snowy Hydro: Electrical output = 5,000 × 0.90 = 4,500 MJ. Waste = 5,000 − 4,500 = 500 MJ. The waste energy becomes thermal energy through friction in turbines, generators, and water turbulence.
Activity 2, Improving Efficiency
Current: 100 bulbs × 60 W = 6,000 W total. Each bulb produces 3 W light (5% of 60 W). Total light = 300 W. Waste = 5,700 W as thermal energy [1 mark]. With LEDs: to produce 300 W of light at 20% efficiency, need 300 ÷ 0.20 = 1,500 W input. Or using 10 W LEDs: each produces 2 W light, so need 150 bulbs × 10 W = 1,500 W for same light [1 mark]. Energy saved = 6,000 − 1,500 = 4,500 W. Over 8 hours = 36 kWh saved per day [1 mark]. Annual saving ≈ 36 × 200 school days = 7,200 kWh. At $0.30/kWh = ~$2,160/year [1 mark]. Recommendation: LED replacement is the most cost-effective first step because it has immediate impact, low installation cost, and rapid payback. Solar panels and AC improvements are also valuable but have higher upfront costs [1 mark].
Multiple Choice
1. C500 − 150 = 350 J waste. Option A gives useful output. Option B incorrectly says energy disappears. Option D adds instead of subtracts.
2. B(900 ÷ 2,000) × 100 = 45%. Option A reverses the division. Option C is the waste percentage. Option D is not supported.
3. AInput = Useful ÷ Efficiency = 12 ÷ 0.20 = 60 J. Option B doubles incorrectly. Option C divides by 5 instead of 0.20. Option D multiplies by 20 instead of dividing.
4. D35% of 1,000 = 350 MJ electrical. 1,000 − 350 = 650 MJ waste thermal. Options A, B, and C all contain errors in accounting or description.
5. CBoth improve efficiency, but B saves more total energy because less input is needed. Option A is true but incomplete. Option B is true but incomplete. Option D is false, any improvement below 100% still helps.
Short Answer Model Answers
Q6 (3 marks): Useful output = Input × Efficiency = 8,000 × 0.50 = 4,000 MJ [1 mark]. Waste = 8,000 − 4,000 = 4,000 MJ [1 mark]. The 4,000 MJ of waste energy becomes thermal energy through exhaust gases, friction, and heat loss to the surroundings [1 mark].
Q7 (4 marks): Incandescent waste = 60 − 3 = 57 W [0.5 mark]. LED waste = 10 − 2 = 8 W [0.5 mark]. The student's conclusion is flawed because they only compared light output without considering the input energy required [1 mark]. The LED uses only 10 W to produce 2 W of light, while the incandescent uses 60 W to produce 3 W. The LED is four times more efficient and uses one-sixth the power [1 mark]. A better conclusion: "The LED is superior because it produces almost the same light using far less electrical energy, dramatically reducing waste thermal energy and electricity costs." [1 mark]
Q8 (5 marks): The 20 MJ becomes thermal energy through friction in pipes, turbines, and generators, plus turbulence in the water [1 mark]. Energy is conserved because 100 MJ input = 80 MJ useful electrical output + 20 MJ waste thermal energy. Nothing has disappeared [1 mark]. Pumped hydro is valuable because it stores energy when supply exceeds demand (e.g., midday solar peak) and releases it when demand exceeds supply (evening peak) [1 mark]. Solar and wind are intermittent, they do not generate on demand. Without storage, excess renewable energy would be curtailed (wasted). Pumped hydro captures energy that would otherwise be lost [1 mark]. Conclusion: While 20% energy transformation to waste is a real inefficiency, the ability to time-shift renewable energy makes pumped hydro economically and environmentally valuable. The claim ignores the context of grid stability and renewable intermittency [1 mark].
Marking Criteria Summary
Q6 (3 marks): (1) Correct formula used. (2) Correct electrical output calculated. (3) Correct waste energy calculated.
Q7 (4 marks): (1) Calculates waste for both bulbs. (2) Identifies flaw in reasoning. (3) Explains efficiency difference. (4) Balanced conclusion.
Q8 (5 marks): (1) Accounts for 20 MJ waste. (2) Explains conservation. (3) Values storage. (4) Links to intermittency. (5) Balanced evaluation.
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