Power: The Rate of Energy Transfer — Print · Grade 8 Science

Chapter: Work & Machines

Power: The Rate of Energy Transfer

Learning goals

Readings for this unit

Before this unit

Review the earlier work and efficiency units. The visible language notes explain this unit’s quantities; then connect them to its concrete records and figure.

Use three teaching sessions of about 40–45 minutes: first observe the phenomenon, draft a model and study Parts 1–2; next explain Parts 3–4 with the figures and evidence; then work the application, examples, practice, model revision and exit check. Divide any difficult part into smaller steps and revisit after feedback. Investigations, review and quizzes have separate sessions. Study the four visual parts across the scheduled sessions, then use the illustrated worked case and independent transfer questions. Extend or revisit a session when the learner cannot explain the figure; completion time is a proposed allocation, not a mastery guarantee.

Start with something to explain

What would explain the observations in power: the rate of energy transfer?

Illustrative evidence case. Illustrative lifting devices each deliver 240 J; A takes 8 s and B 12 s. A receives 400 J and B receives 300 J.

The same useful energy at different ratesThe same useful energy at different ratesQuantity (W)A useful power30B useful power20Each delivers 240 J; A takes 8 s, B takes 12 s.Power = energy / time; efficiency needs input energy.Illustrative model • use the labels and stated assumptions
Both devices deliver 240 J, but the different times produce 30 W and 20 W useful power. The bars show rates, not different total delivered energies. Input energy must also be supplied to compare efficiencies; faster transfer alone does not imply a larger useful fraction.

Notice and wonder

If A receives 400 J and B 300 J, which is more efficient? Write two observations from the supplied record without guessing their cause. Include a quantity and its unit when given. An observation is what was recorded; an inference is what you think explains it.

Write one question about why the pattern occurs and one question about a comparison that could distinguish competing explanations. A testable question identifies what could be observed or measured.

Your first model

Before reading the model answer, sketch the named objects or particles and label what you think changes. Compare useful power and efficiency, then explain why the higher-power device is not automatically more efficient. Keep this first version so you can compare it with your later explanation.

Build the language from the idea

power
Rate of energy transfer or work.
watt
Unit of power: one joule per second.
average power
Transferred energy divided by the stated elapsed time.
kilowatt
One thousand watts.
kilowatt-hour
Energy transferred by an average power of one kilowatt over one hour.
model
A representation that explains selected features and makes predictions; it leaves other details out.
evidence
Relevant observations or records used to support or challenge a claim.
system boundary
The chosen limit separating the objects being studied from their surroundings.
assumption
A condition taken as true for a model or calculation and needing to be checked against the case.

Two devices can transfer the same amount of energy but take different amounts of time. Power describes this rate.

The same useful energy at different ratesThe same useful energy at different ratesQuantity (W)A useful power30B useful power20Each delivers 240 J; A takes 8 s, B takes 12 s.Power = energy / time; efficiency needs input energy.Illustrative model • use the labels and stated assumptions
Both devices deliver 240 J, but the different times produce 30 W and 20 W useful power. The bars show rates, not different total delivered energies. Input energy must also be supplied to compare efficiencies; faster transfer alone does not imply a larger useful fraction.

A watt, written W, is one joule per second; a joule, written J, is a unit of energy. Energy answers how much is transferred, while power answers how rapidly.

Same lifting work, different average powerSame lifting work, different average power3 m height gain4 kg object; g = 10 N/kg; ideal steady liftingWork = 4×10×3 = 120 J. If time = 2 s, average power = 60 W.Another device taking 6 s does the same 120 J at only 20 W.
The stated 4 kg object gains 3 m in height, receiving 120 J of gravitational energy in the ideal steady-lift model. Completing that lift in 2 s gives average useful power 60 W; taking 6 s gives 20 W. More power describes a faster energy transfer, not more energy required for the same ideal height change.

Do not confuse W for watts with W sometimes used as a work symbol; label quantities in words when unsure. Household electricity bills usually report energy in kilowatt-hours, not power alone.

Use supplied lifting and device records; there is no need to test a wall-powered appliance or lift a heavy load.

An observation is what a person or instrument records. An inference is a proposed interpretation of those records. Keep the two separate: a graph can show a pattern while more than one mechanism might explain it. State the chosen system, or part of the world being studied, and its boundary before accounting for material or energy transfers.

Use each figure as a reasoning tool. Read its caption and labels, describe the pattern, then explain a cause. If the picture is schematic, use the stated numbers rather than measuring its printed size. Check whether its simplified conditions match the problem before applying a formula.

Build your understanding

Where this fits

This unit helps answer the chapter question: How can a machine make a task easier without creating energy? Keep a notebook table with the question, evidence, explanation and one unresolved issue. Revise a previous model only when you can name the new evidence or idea that changes it.

Use three teaching sessions of about 40–45 minutes: first observe the phenomenon, draft a model and study Parts 1–2; next explain Parts 3–4 with the figures and evidence; then work the application, examples, practice, model revision and exit check. Divide any difficult part into smaller steps and revisit after feedback. Investigations, review and quizzes have separate sessions.

Part 1 · Build the rate from energy and time

Average power equals transferred energy divided by elapsed time. The result must include J/s or W. Average describes the whole stated interval; power at particular instants may vary. A value without the time interval cannot determine the average rate.

The same useful energy at different ratesThe same useful energy at different ratesQuantity (W)A useful power30B useful power20Each delivers 240 J; A takes 8 s, B takes 12 s.Power = energy / time; efficiency needs input energy.Illustrative model • use the labels and stated assumptions
Both devices deliver 240 J, but the different times produce 30 W and 20 W useful power. The bars show rates, not different total delivered energies. Input energy must also be supplied to compare efficiencies; faster transfer alone does not imply a larger useful fraction.

Read the visual model

Does the slower motor necessarily transfer less energy?

Visual reasoning — after your attempt

No. It transfers the same stated 120 J over a longer interval, so its average power is smaller.

A model motor transfers 120 J in 6 s, giving 20 J/s or 20 W. Another takes 12 s for the same transfer and averages 10 W. Both transfer 120 J; their rates differ.

Pause and explain

Does the slower motor necessarily transfer less energy?

Check your reasoning after answering

No. It transfers the same stated 120 J over a longer interval, so its average power is smaller.

Part 2 · Connect lifting work with power

For a constant upward force along the displacement, work equals force × distance. Calculate the work first, then divide by time. A longer lifting time reduces average useful power for the same load and height, without reducing the ideal work needed.

Same lifting work, different average powerSame lifting work, different average power3 m height gain4 kg object; g = 10 N/kg; ideal steady liftingWork = 4×10×3 = 120 J. If time = 2 s, average power = 60 W.Another device taking 6 s does the same 120 J at only 20 W.
The stated 4 kg object gains 3 m in height, receiving 120 J of gravitational energy in the ideal steady-lift model. Completing that lift in 2 s gives average useful power 60 W; taking 6 s gives 20 W. More power describes a faster energy transfer, not more energy required for the same ideal height change.

Read the visual model

A watt is one joule per second. Separate how much useful energy changes from how long the change takes. Do not infer efficiency from timing alone without measuring input energy as well.

If the lift takes 4 s instead, what are work and average useful power?

Visual reasoning — after your attempt

The ideal lifting work stays 120 J; average useful power is 120/4 = 30 W. Conditions and any additional acceleration or losses must be accounted for separately.

A supplied model lifts with 30 N through 2 m in 5 s. Useful work is 60 J and average useful power 12 W. This model does not include starting motion, frictional losses or a person’s metabolism.

Pause and explain

What information is missing to find input power?

Check your reasoning after answering

The input energy or input power and the losses. Useful lifting power alone does not describe all energy entering the device.

Part 3 · Distinguish power from efficiency

Efficiency is the fraction of input energy reaching the chosen useful output. Power and efficiency describe different things: a high-power device can have high or low efficiency. Compare input and output over the same interval and define usefulness.

Fast energy transfer and efficient transfer are differentFast energy transfer and efficient transfer are differentDevice A: input 100 JUseful 80 J in 4 s80% efficient; 20 W usefulDevice B: input 100 JUseful 60 J in 2 s60% efficient; 30 W usefulB has greater useful power; A has greater efficiency in these supplied tests.
The supplied A test gives useful average power 80/4 = 20 W and efficiency 80%. B gives 60/2 = 30 W and efficiency 60%. The larger power does not imply the larger fraction of input reaching the intended output. These different useful amounts also need consideration when comparing a fixed real task.

Read the visual model

Power divides an energy transfer by elapsed time. Efficiency divides useful output energy by input energy. Both require a clear job definition, but they answer different questions and cannot be ranked by one number alone.

Which device is more efficient, and which has greater useful average power? Explain both calculations.

Visual reasoning — after your attempt

A is more efficient: 80/100 = 80% versus 60%. B has greater useful power: 60/2 = 30 W versus 20 W.

A modeled device receives 100 J in 5 s and supplies 60 J of useful work. Input power is 20 W, useful power is 12 W and efficiency is 60%. The remaining 40 J is energy in other pathways, not destroyed energy.

Pause and explain

Does 60% efficiency mean the input power is 60 W?

Check your reasoning after answering

No. Efficiency is a ratio. The power also depends on the energy amount and time interval.

Part 4 · Interpret energy units in device comparisons

A kilowatt is 1,000 W. A kilowatt-hour is the energy transferred at 1 kW for one hour. It is not power per hour. Multiplying power by time gives energy when the average power is appropriate for the interval. Actual devices may cycle on and off.

A power rating becomes energy only after time is includedA power rating becomes energy only after time is includedEnergy transferred (kJ)0601201800123Operating time (min)Model constant 1 kW power: 60 kJ each minute; 1 kWh = 3600 kJ.
For the supplied constant 1 kW model, energy increases by 60 kJ each minute, reaching 180 kJ in 3 minutes. A kilowatt is 1000 W, while a kilowatt-hour is an energy unit: 1 kWh = 3.6 million J = 3600 kJ. A real device may cycle or change its power; a rated maximum should not be silently treated as a continuously measured draw.

Read the visual model

The slope of an energy–time graph gives power after units are converted consistently. Minutes on this axis require 60 s per minute. kW and kWh differ because one is a rate and the other includes duration.

How much energy would this ideal device transfer in 6 minutes, in kJ and kWh?

Visual reasoning — after your attempt

360 kJ, equivalent to 0.1 kWh. Six minutes is 0.1 hour, so 1 kW×0.1 h = 0.1 kWh.

A hypothetical 0.5 kW device operating at that average power for 2 h transfers 1 kWh, or 3.6 million J. This is an illustrative arithmetic case, not a current price or an actual appliance guarantee.

Pause and explain

Why can rated power alone fail to predict a daily energy bill?

Check your reasoning after answering

The operating duration, actual average power, cycling and energy price are also needed. A rating describes a condition rather than every day’s use.

Apply the whole unit · Equal energy at different rates

Illustrative case for analysis, not measurements collected by the learner. Illustrative lifting devices each deliver 240 J; A takes 8 s and B 12 s. A receives 400 J and B receives 300 J.

Your task

Compare useful power and efficiency, then explain why the higher-power device is not automatically more efficient. Use the supplied conditions, show intermediate reasoning and distinguish the supported conclusion from a broader claim.

What your response should show

  • Use the specified quantities and units.
  • Explain an intermediate step or comparison.
  • Connect the pattern to the physical mechanism.
  • State an assumption or limitation.
Response guidance — after your attempt

  1. Use useful power = delivered energy / time: A is 240/8=30 W; B is 240/12=20 W.

  2. Use efficiency = useful output / input: A is 240/400=0.60 or 60%.

  3. For B, 240/300=0.80 or 80%; both comparisons use the same useful output definition.

  4. Conclude that A transfers useful energy faster while B directs a larger input fraction to it; power and efficiency differ.

Useful powers are 30 W for A and 20 W for B. Efficiencies are 240/400=60% and 240/300=80%. A delivers useful energy faster but B directs a larger fraction of input to that output. Power and efficiency answer different questions, and both require the same defined useful output.

Scientific references and adapted teaching resources

Parent / teacher teaching notes

Ask the learner to describe the phenomenon before naming a formula. Discuss each checkpoint after an attempt. Use the supplied records when equipment is unavailable and label them illustrative rather than actual measurements. The final explanation should identify a mechanism and a limit, with different defensible wording accepted. Adaptation credit: OpenSciEd middle-school teaching routines and public task structures by OpenSciEd, CC BY 4.0. This independent version adds original explanatory text, data and figures, terminology support and a home-study schedule. It does not reproduce the complete teacher editions or their classroom validation.

The key idea

Watts measure energy per time; multiply an appropriate average power by operating time to obtain energy. Useful powers are 30 W for A and 20 W for B.

Worked examples

Example 1 · Build the first reasoning step

Does the slower motor necessarily transfer less energy?
  1. Identify what the question asks and what the stated observation establishes.
  2. No. It transfers the same stated 120 J over a longer interval, so its average power is smaller.
Answer: No. It transfers the same stated 120 J over a longer interval, so its average power is smaller.

Example 2 · Equal energy at different rates

Illustrative lifting devices each deliver 240 J; A takes 8 s and B 12 s. A receives 400 J and B receives 300 J. Compare useful power and efficiency, then explain why the higher-power device is not automatically more efficient.
  1. Use useful power = delivered energy / time: A is 240/8=30 W; B is 240/12=20 W.
  2. Use efficiency = useful output / input: A is 240/400=0.60 or 60%.
  3. For B, 240/300=0.80 or 80%; both comparisons use the same useful output definition.
  4. Conclude that A transfers useful energy faster while B directs a larger input fraction to it; power and efficiency differ.
Answer: Useful powers are 30 W for A and 20 W for B. Efficiencies are 240/400=60% and 240/300=80%. A delivers useful energy faster but B directs a larger fraction of input to that output. Power and efficiency answer different questions, and both require the same defined useful output.
Check the reasoning

Explain why each stated condition matters.

Example 3 · Transfer to a different condition

A 50 W model device operates steadily for 20 s. Find the transferred energy and identify an assumption.
  1. Identify the changed condition.
  2. Energy is 50 J/s × 20 s = 1,000 J. This uses a constant or appropriately averaged 50 W over the entire interval; real varying operation requires interval data.
Answer: Energy is 50 J/s × 20 s = 1,000 J. This uses a constant or appropriately averaged 50 W over the entire interval; real varying operation requires interval data.

Example 4 · Find and fix the mistake

A student claims: “Watts measure the total energy a device uses.” Evaluate the claim.
  1. Distinguish what is observed from what is inferred.
  2. Watts measure energy per time; multiply an appropriate average power by operating time to obtain energy.
Answer: Watts measure energy per time; multiply an appropriate average power by operating time to obtain energy.
Check the reasoning

Compare the claim with the stated observations and model assumptions.

Example 5 · Interpret the visual model

Same lifting work, different average powerSame lifting work, different average power3 m height gain4 kg object; g = 10 N/kg; ideal steady liftingWork = 4×10×3 = 120 J. If time = 2 s, average power = 60 W.Another device taking 6 s does the same 120 J at only 20 W.
The stated 4 kg object gains 3 m in height, receiving 120 J of gravitational energy in the ideal steady-lift model. Completing that lift in 2 s gives average useful power 60 W; taking 6 s gives 20 W. More power describes a faster energy transfer, not more energy required for the same ideal height change.
If the lift takes 4 s instead, what are work and average useful power?
  1. Identify the represented objects, axes, labeled quantities and stated conditions before choosing a rule.
  2. A watt is one joule per second. Separate how much useful energy changes from how long the change takes. Do not infer efficiency from timing alone without measuring input energy as well.
  3. The ideal lifting work stays 120 J; average useful power is 120/4 = 30 W. Conditions and any additional acceleration or losses must be accounted for separately.
Answer: The ideal lifting work stays 120 J; average useful power is 120/4 = 30 W. Conditions and any additional acceleration or losses must be accounted for separately.
Check the reasoning

Keep the model’s stated limits explicit: The stated 4 kg object gains 3 m in height, receiving 120 J of gravitational energy in the ideal steady-lift model. Completing that lift in 2 s gives average useful power 60 W; taking 6 s gives 20 W. More power describes a faster energy transfer, not more energy required for the same ideal height change.

Avoid these mistakes

Check your understanding

Does the slower motor necessarily transfer less energy?

Discuss the answer

No. It transfers the same stated 120 J over a longer interval, so its average power is smaller.

What information is missing to find input power?

Discuss the answer

The input energy or input power and the losses. Useful lifting power alone does not describe all energy entering the device.

Does 60% efficiency mean the input power is 60 W?

Discuss the answer

No. Efficiency is a ratio. The power also depends on the energy amount and time interval.

Why can rated power alone fail to predict a daily energy bill?

Discuss the answer

The operating duration, actual average power, cycling and energy price are also needed. A rating describes a condition rather than every day’s use.

Return to the phenomenon

Illustrative lifting devices each deliver 240 J; A takes 8 s and B 12 s. A receives 400 J and B receives 300 J. Compare useful power and efficiency, then explain why the higher-power device is not automatically more efficient. Revise your first explanation using a named idea from the lesson. Mark the changed part and the evidence that supports it.

Compare explanations after your attempt

Useful powers are 30 W for A and 20 W for B. Efficiencies are 240/400=60% and 240/300=80%. A delivers useful energy faster but B directs a larger fraction of input to that output. Power and efficiency answer different questions, and both require the same defined useful output.

What should we investigate next?

Which condition could be changed to test a remaining uncertainty in this explanation? State the predicted observation and what result would challenge your model.

Investigate, explain and revise

Use these tasks after the three teaching sessions as needed. Discuss an attempted explanation before opening response guidance; a matching sentence is less important than a supported mechanism.

Activity 1 · Explain the new evidence case

Assigned reading and preparation

Use the visible scenario above, Parts 1–4 and your first notebook model. No outside reading or equipment is required.

Your task

Illustrative lifting devices each deliver 240 J; A takes 8 s and B 12 s. A receives 400 J and B receives 300 J. Compare useful power and efficiency, then explain why the higher-power device is not automatically more efficient. Write 100–150 words or give a labeled drawing with an equivalent oral explanation.

Success criteria

  • Use two relevant stated records or conditions.
  • Explain the mechanism rather than only repeat a definition.
  • Label observation and inference.
  • Name a changed model feature or a remaining uncertainty.
Hint

Read the stated boundary and conditions before selecting a relationship. The unit’s worked examples show intermediate steps.

Response guidance — discuss after your attempt

Useful powers are 30 W for A and 20 W for B. Efficiencies are 240/400=60% and 240/300=80%. A delivers useful energy faster but B directs a larger fraction of input to that output. Power and efficiency answer different questions, and both require the same defined useful output.

Activity 2 · Read and extend the figure

Assigned reading and preparation

Study the new labeled figure and its caption, then compare it with the associated teaching part.

Your task

If A receives 400 J and B 300 J, which is more efficient? Explain what each important label represents. Add one missing feature that would be needed for a different question, and explain why the original figure leaves it out.

Success criteria

  • Answer the figure question with its stated labels.
  • Connect the visual representation to the scientific idea.
  • Identify a specific limitation from the caption.
  • Propose a useful change without misrepresenting supplied evidence.
Hint

Both devices deliver 240 J, but the different times produce 30 W and 20 W useful power. The bars show rates, not different total delivered energies. Input energy must also be supplied to compare efficiencies; faster transfer alone does not imply a larger useful fraction.

Response guidance — discuss after your attempt

A is 60% and B 80%. A has higher useful power but lower efficiency for this stated output and input comparison. Both devices deliver 240 J, but the different times produce 30 W and 20 W useful power. The bars show rates, not different total delivered energies. Input energy must also be supplied to compare efficiencies; faster transfer alone does not imply a larger useful fraction.

Words to know

power
Rate of energy transfer or work.
watt
Unit of power: one joule per second.
average power
Transferred energy divided by the stated elapsed time.
kilowatt
One thousand watts.
kilowatt-hour
Energy transferred by an average power of one kilowatt over one hour.
model
A representation that explains selected features and makes predictions; it leaves other details out.
evidence
Relevant observations or records used to support or challenge a claim.
system boundary
The chosen limit separating the objects being studied from their surroundings.
assumption
A condition taken as true for a model or calculation and needing to be checked against the case.

In the real world

Illustrative lifting devices each deliver 240 J; A takes 8 s and B 12 s. A receives 400 J and B receives 300 J.

Exit check

If A receives 400 J and B 300 J, which is more efficient? Explain the relevant mechanism and one model limitation.

Check your explanation after answering

A is 60% and B 80%. A has higher useful power but lower efficiency for this stated output and input comparison. Both devices deliver 240 J, but the different times produce 30 W and 20 W useful power. The bars show rates, not different total delivered energies. Input energy must also be supplied to compare efficiencies; faster transfer alone does not imply a larger useful fraction.

Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.

Problem 1

Does the slower motor necessarily transfer less energy?
Answer and explanation

No. It transfers the same stated 120 J over a longer interval, so its average power is smaller.

No. It transfers the same stated 120 J over a longer interval, so its average power is smaller.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 2

What information is missing to find input power?
Answer and explanation

The input energy or input power and the losses. Useful lifting power alone does not describe all energy entering the device.

The input energy or input power and the losses. Useful lifting power alone does not describe all energy entering the device.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 3

Does 60% efficiency mean the input power is 60 W?
Answer and explanation

No. Efficiency is a ratio. The power also depends on the energy amount and time interval.

No. Efficiency is a ratio. The power also depends on the energy amount and time interval.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 4

Why can rated power alone fail to predict a daily energy bill?
Answer and explanation

The operating duration, actual average power, cycling and energy price are also needed. A rating describes a condition rather than every day’s use.

The operating duration, actual average power, cycling and energy price are also needed. A rating describes a condition rather than every day’s use.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 5

Illustrative lifting devices each deliver 240 J; A takes 8 s and B 12 s. A receives 400 J and B receives 300 J. Compare useful power and efficiency, then explain why the higher-power device is not automatically more efficient.
Answer and explanation

Useful powers are 30 W for A and 20 W for B. Efficiencies are 240/400=60% and 240/300=80%. A delivers useful energy faster but B directs a larger fraction of input to that output. Power and efficiency answer different questions, and both require the same defined useful output.

Useful powers are 30 W for A and 20 W for B. Efficiencies are 240/400=60% and 240/300=80%. A delivers useful energy faster but B directs a larger fraction of input to that output. Power and efficiency answer different questions, and both require the same defined useful output.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 6

A 50 W model device operates steadily for 20 s. Find the transferred energy and identify an assumption.
Answer and explanation

Energy is 50 J/s × 20 s = 1,000 J. This uses a constant or appropriately averaged 50 W over the entire interval; real varying operation requires interval data.

Energy is 50 J/s × 20 s = 1,000 J. This uses a constant or appropriately averaged 50 W over the entire interval; real varying operation requires interval data.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 7

Evaluate and correct: Watts measure the total energy a device uses.
Answer and explanation

Watts measure energy per time; multiply an appropriate average power by operating time to obtain energy.

Watts measure energy per time; multiply an appropriate average power by operating time to obtain energy.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 8

Choose one prediction from this unit. Explain a comparison that could test it, specifying observations, changed and matched conditions, and a result that would challenge it.
Answer and explanation

A hypothetical 0.5 kW device operating at that average power for 2 h transfers 1 kWh, or 3.6 million J. This is an illustrative arithmetic case, not a current price or an actual appliance guarantee. The operating duration, actual average power, cycling and energy price are also needed. A rating describes a condition rather than every day’s use.

A hypothetical 0.5 kW device operating at that average power for 2 h transfers 1 kWh, or 3.6 million J. This is an illustrative arithmetic case, not a current price or an actual appliance guarantee. The operating duration, actual average power, cycling and energy price are also needed. A rating describes a condition rather than every day’s use.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 9

A model device has average input power 80 W over 10 s and supplies 600 J useful work. Find input energy and efficiency; state a condition behind the arithmetic.
Hints

Distinguish the observed quantities, the proposed mechanism and its conditions.

Answer and explanation

Input is 800 J and useful efficiency 600/800=75%. The 80 W must be the applicable average over the interval; the other 200 J needs an energy pathway or stored-change account.

Input is 800 J and useful efficiency 600/800=75%. The 80 W must be the applicable average over the interval; the other 200 J needs an energy pathway or stored-change account.

Problem 10

A 50 W model device operates steadily for 20 s. Find the transferred energy and identify an assumption. Explain a condition that would change this conclusion.
Hints

Distinguish the observed quantities, the proposed mechanism and its conditions.

Answer and explanation

Energy is 50 J/s × 20 s = 1,000 J. This uses a constant or appropriately averaged 50 W over the entire interval; real varying operation requires interval data.

Energy is 50 J/s × 20 s = 1,000 J. This uses a constant or appropriately averaged 50 W over the entire interval; real varying operation requires interval data.

Problem 11

Same lifting work, different average powerSame lifting work, different average power3 m height gain4 kg object; g = 10 N/kg; ideal steady liftingWork = 4×10×3 = 120 J. If time = 2 s, average power = 60 W.Another device taking 6 s does the same 120 J at only 20 W.
The stated 4 kg object gains 3 m in height, receiving 120 J of gravitational energy in the ideal steady-lift model. Completing that lift in 2 s gives average useful power 60 W; taking 6 s gives 20 W. More power describes a faster energy transfer, not more energy required for the same ideal height change.
For this unit’s illustrated case, if the lift takes 4 s instead, what are work and average useful power?
Answer and explanation

The ideal lifting work stays 120 J; average useful power is 120/4 = 30 W. Conditions and any additional acceleration or losses must be accounted for separately.

A watt is one joule per second. Separate how much useful energy changes from how long the change takes. Do not infer efficiency from timing alone without measuring input energy as well.

The ideal lifting work stays 120 J; average useful power is 120/4 = 30 W. Conditions and any additional acceleration or losses must be accounted for separately.

Response criteria

  • Use the named labels, directions or supplied quantities accurately.
  • Connect the visual evidence to the scientific mechanism or calculation.
  • State the relevant model conditions or limitation.

Problem 12

A device transfers a useful 180 J in 3 s while receiving 240 J in that interval. Find useful average power and efficiency, then explain which changes if the same energy transfers take 6 s.
Answer and explanation

Useful power is 60 W and efficiency 75%. Taking 6 s reduces useful power to 30 W while the same energy ratio stays 75%.

Use energy/time for power and useful/input for efficiency.

Change only the interval in the second comparison.

Useful power is 60 W and efficiency 75%. Taking 6 s reduces useful power to 30 W while the same energy ratio stays 75%.

Response criteria

  • Use the named labels, directions or supplied quantities accurately.
  • Connect the visual evidence to the scientific mechanism or calculation.
  • State the relevant model conditions or limitation.

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