Find force or distance from a work requirement
Rearrange the aligned-force equation and connect lifting work to a complete energy budget.
Exam scope and assessment notes
Mapped to the named 0973 section using teaching from the corresponding single-science course. A 0973-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade. This qualification reports two grades on the 9–1 scale; confirm the entry route for your zone and exam series.
Before you begin
- Calculate aligned-force work.
- Separate changes in gravitational and kinetic stores.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which rearrangement finds force from work and aligned distance?
What you will learn
- Use F = W/d and d = W/F.
- Check which force and store changes a supplied work amount describes.
Solve for the missing factor
If a constant aligned force must transfer 90 J over 3 m, F = W/d = 30 N. If a 15 N aligned force transfers 60 J, d = W/F = 4 m. These describe the movement of the force's point of application along the force. Keep joules, newtons and metres consistent rather than swapping energy and power units.
For zero displacement, non-zero mechanical work cannot follow from a finite force under this simple model. If both work and displacement are zero, the quotient 0/0 does not determine a unique force. A reported inconsistent pair needs its conditions checked rather than an infinite force asserted as a physical answer.
A lift illustrates specific work and store changes
A load of mass 2 kg raised steadily by 4 m at g = 10 N/kg without friction needs upward force 20 N, so lifting work is 80 J and the gravitational store increases by 80 J. The kinetic store has no net change at constant speed. The upward force's work is not zero just because the resultant force is zero: gravity makes an opposing contribution in the load-only force account.
If the same 4 m rise requires an aligned 25 N lifting force because of a specified 5 N opposing friction, lifting work is 100 J. The gravitational gain is still 80 J and friction accounts for 20 J internal increase, with no net kinetic change. State the steady-motion and force conditions; a lift that speeds up would need its changing kinetic store included as well.
Sources for this lesson
Pause and explain
At constant upward speed, does zero resultant imply zero work by the lifting force?
Worked example
A constant aligned force transfers 90 J over 3 m. Find the force, then distance for 15 N transferring 60 J.
Show the worked solution
- F = 90/3 = 30 N.
- d = 60/15 = 4 m.
- Both rearrangements use a constant force along the stated movement.
Answer 30 N; 4 m.
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.
Find force for 90 J over 3 m and distance for 15 N transferring 60 J, with constant aligned forces.
Give me a hint
Divide by the known factor in each product.
Compare my reasoning
- F = 90/3 = 30 N.
- d = 60/15 = 4 m.
- The required units are N and m, respectively.
30 N; 4 m.
Look for these in your work
- Use the appropriate quotient.
- Retain constant aligned-force conditions.
A 2 kg load is lifted steadily 4 m without friction at g = 10 N/kg. Find lifting force, lifting work and gravitational gain.
Give me a hint
Steady motion gives force balance while the load still moves upward.
Compare my reasoning
- Weight and upward force are each 20 N.
- Lifting work = 20 × 4 = 80 J.
- Gravitational gain is 80 J and the kinetic store does not change.
20 N; 80 J lifting work; +80 J gravitational gain.
Look for these in your work
- Distinguish force balance from individual work.
- State the unchanged kinetic store.
A steady 2 kg lift rises 4 m at g = 10 N/kg using 25 N upward force against 5 N opposing friction. Account for the lifting work.
Give me a hint
Separate gravitational gain and internal-energy destination.
Compare my reasoning
- Lifting work = 25 × 4 = 100 J.
- Gravitational gain = 2 × 10 × 4 = 80 J.
- Friction accounts for 5 × 4 = 20 J; kinetic energy is unchanged.
100 J input = 80 J gravitational gain + 20 J internal increase.
Look for these in your work
- Account for all supplied force contributions.
- Keep the steady-motion condition.
Common mistakes
- Rearrange the product by division.
- Zero resultant does not mean each force does zero work.
- Include other destinations in a lifting input account.
What can you explain now?
How much work lifts 2 kg steadily through 4 m without friction at g = 10 N/kg?
Compare with the explanation
80 J.
The upward force equals weight 20 N, so work = 20 × 4 = 80 J.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Rearrange W = Fd and explain how a steady lift can gain potential energy while its kinetic store remains unchanged.
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.