Relate mechanical and electrical work to energy transferred
Measure work in joules and identify the selected transfer rather than assuming one final store receives it all.
Cross-referenced from the related Cambridge IGCSE Chemistry 0620 or Physics 0625 teaching to the 0652 syllabus. Exact 0652 Core/Supplement objective coverage and this final-series course’s examiner fit still need specialist review. Written practice is self-checked and does not predict a Cambridge grade or demonstrate supervised laboratory competence.
Before you begin
- Recognise mechanical and electrical transfer pathways.
- Balance distinct energy destinations.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which unit measures mechanical or electrical work?
What you will learn
- Explain work done as energy transferred.
- Distinguish a transfer amount from its distribution among stores.
Work is a transferred energy amount
Mechanical work done by a force during movement is the energy that force transfers. Electrical work done is the energy transferred electrically, for example through a current to a motor or heater. Both are measured in joules. A 150 J electrical transfer to a motor supplies 150 J of energy to the selected motor account over the stated interval.
That amount can be distributed: if the motor increases a load's gravitational store by 120 J and the only other destination is internal stores, they gain 30 J. Do not claim electrical work is only 120 J just because that is the desired output. The transfer at one boundary and the useful share at a later stage are different quantities.
Specify the force and the account
A pulling force can do positive mechanical work on a cart while friction transfers some energy to internal stores. The pull's work need not equal the cart's kinetic-energy increase alone. If a pull supplies 60 J and friction accounts for 18 J, the kinetic increase is 42 J when the remaining stated stores do not change.
A support force holding an ideal stationary object does no mechanical work on it through displacement; movement is needed for this transfer. A person may still expend chemical energy internally. No current-voltage-time formula is required by this work subsection: interpret a supplied electrical amount without inventing circuit values or implying that a paper exercise demonstrates practical competence.
Sources for this lesson
Pause and explain
Does 150 J electrical work become only the 120 J useful gravitational output?
Worked example
A motor receives 150 J by electrical work and lifts a load with a 120 J potential gain. Its only other output increases internal stores. Find that output.
Show the worked solution
- Electrical work is the full supplied energy, 150 J.
- The desired gravitational increase accounts for 120 J.
- Other increase = 150 − 120 = 30 J.
Answer Electrical work 150 J; other internal gains 30 J.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A motor receives 150 J electrically, supplies 120 J gravitational gain and has one other internal-energy destination. Find it.
Give me a hint
Treat electrical work as the full input.
Compare my reasoning
- Input energy transferred = 150 J.
- The desired gain is 120 J.
- The remaining internal increase is 30 J.
30 J; electrical work was 150 J.
Look for these in your work
- Identify full input work.
- Account for the other output.
A pull does 60 J of work, with 18 J transferred through friction and no other store changes except the cart's kinetic store. Find its kinetic increase.
Give me a hint
Subtract the other destination from the pull's transfer.
Compare my reasoning
- The pull supplies 60 J.
- Friction accounts for 18 J.
- Kinetic energy increases by 42 J.
42 J.
Look for these in your work
- Keep the pull and friction accounts distinct.
- Use the no-other-changes condition.
A person holds a motionless ideal load. Explain why their internal energy use does not prove mechanical work on that load through displacement.
Give me a hint
Separate the body's energy account from the stationary load's motion.
Compare my reasoning
- The load has no displacement.
- The support force therefore transfers no mechanical energy to that ideal load through motion.
- The person's chemical and internal stores can still change in a different account.
Zero mechanical work through load displacement; the person's internal energy use is a separate process.
Look for these in your work
- Select the load explicitly.
- Do not equate bodily effort with this mechanical transfer.
Common mistakes
- Work is in joules.
- A useful share is not necessarily the full input work.
- Identify the force, boundary and displacement.
What can you explain now?
A pull transfers 60 J and friction accounts for 18 J. What kinetic increase remains if no other stores change?
Compare with the explanation
42 J.
The complete stated account gives 60 − 18 = 42 J.
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Make it stick
Define mechanical and electrical work, give the unit and explain why the desired output may be less than the full work input.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.