Energy stores, work and efficiency
| English | 中文 | Pinyin |
|---|---|---|
| power/ˈpaʊə/ | 功率 | gōng lǜ |
| efficiency/ɪˈfɪʃənsi/ | 效率 | xiào lǜ |
What would explain this observation?
- A motor can transfer some input energy to lifting and the rest to heating. Useful output is part of the total energy transfer.
- Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.
Build the model
- Work done by a constant force parallel to displacement is force multiplied by distance. Kinetic energy depends on speed squared. Energy is conserved when all transfers and stores are included.
- power 功率: Energy transferred per unit time; efficiency 效率: Useful output divided by total input.
What happens to kinetic energy when speed doubles at fixed mass?
Define the system and useful output before calculating efficiency. Doubling speed quadruples kinetic energy at constant mass. Power describes transfer per time, not total energy.
Match each technical term to its precise meaning.
Use the definitions to distinguish related quantities and processes.
Choose evidence that can test it
- Define the system and useful output before calculating efficiency. Doubling speed quadruples kinetic energy at constant mass. Power describes transfer per time, not total energy.
- Measure a lifting height and load, time the lift, and record electrical input with suitable instruments. Repeat trials and account for heating or friction as transfers, not missing energy.
Which two habits make the investigation or model in this case more defensible?
Measure a lifting height and load, time the lift, and record electrical input with suitable instruments. Repeat trials and account for heating or friction as transfers, not missing energy.
Work from known quantities
- State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
- Known: a device receives 600 J and transfers 420 J usefully. Efficiency = useful output / total input. Efficiency = 420/600 = 0.70 = 70%. Over 3 s, useful power = useful energy/time = 420/3 = 140 W.
A device receives 250 J and gives 150 J useful output. Find efficiency as a percentage. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.
A device receives 250 J and gives 150 J useful output. Find efficiency as a percentage.
The result is 60 %. Known: a device receives 600 J and transfers 420 J usefully. Efficiency = useful output / total input. Efficiency = 420/600 = 0.70 = 70%. Over 3 s, useful power = useful energy/time = 420/3 = 140 W.
Check the conclusion and its limits
- Efficiency cannot exceed 100% for a properly defined energy balance. Energy dissipated by heating is still conserved.
- Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
Dissipated energy is destroyed. This claim is false: Efficiency cannot exceed 100% for a properly defined energy balance. Energy dissipated by heating is still conserved.
Energy stores, work and efficiency: Define the system and useful output before calculating efficiency. Doubling speed quadruples kinetic energy at constant mass. Power describes transfer per time, not total energy.
Dissipated energy is destroyed.
Efficiency cannot exceed 100% for a properly defined energy balance. Energy dissipated by heating is still conserved.
Energy transferred per unit time: write the technical term.
power means Energy transferred per unit time.