Forces · FO-05 · PAPER 2

Work done by forces.

A complete cover lesson with a clear pupil route, selected PhysicsUK resources and checks that show what to do next.

60 minutesFoundation + HigherSeparate + Combined

Download pupil handout (PDF) ↓Download teacher guide (PDF)

Pupil booklet (PDF) - 8 pages with reading, figures and writing space. No devices or practical apparatus needed.

THE LEARNING

Know what pupils are working towards.

By the end, pupils should be able to…

01Calculate work, force or distance with consistent SI units.
02Distinguish work by a named force from resultant force and perpendicular forces.
03Explain energy transfer through friction without claiming energy is destroyed.

Course and starting points

AQA 8463: 4.5.2
AQA 8464: 6.5.2

Both courses and tiers: work by a constant named force, distance along its line of action, units, rearrangement and frictional energy transfer. Supplied GPE result; no mgh, spring or trigonometric calculations.

Useful prior knowledge

Pupils have met work done, force directions, balanced forces and thermal energy stores.
Multiply/divide and rearrange a two-factor equation; convert centimetres to metres.

ONE CLEAR ROUTE

From recall to independent explanation.

Write directly in the downloaded pupil handout. All reading, selected questions, data and diagrams are included in order. Your covering teacher will direct the checks and any Try again tasks.

010–5 min5 minutes

Recall work, distance units and joules

Answer 1-3 from memory, then stop.

Question 1 / CHECK FOR UNDERSTANDING

A constant 12 N force acts along a movement of 3.0 m. Use W=Fs to calculate its work done.

Question 2 / CHECK FOR UNDERSTANDING

Convert a movement of 40 cm into metres.

Question 3 / CHECK FOR UNDERSTANDING

State the SI unit of work done. Complete: one newton metre is equal to one ____.

Try again A · Retry if directed

Use this support first: Use page 2: divide centimetres by 100, multiply force by distance along it, and label transferred energy in joules.

(a) Convert 20 cm to metres. (b) Find work by an 8.0 N force along 2.0 m. (c) State the unit and complete 1 N m = __ J.

Use the named support first, then answer independently. This replaces part of practice time.

025–15 min10 minutes

Read the named-force model

Read page 2 and follow the complete models.

YOUR SELECTED READING

Energy Calculations and Power

For a constant force acting along a displacement, work done = force × distance moved along the force: W = Fs. F is in newtons and s in metres, giving joules. This is also the energy transferred mechanically.

Work done by a force - selected constant-force reading

Section 4 (#concept-4), first three sentences through mechanical energy transfer.

Open the selected revision reading →

Worked model / WORKED MODEL

A constant 20 N pull acts along 1.5 m: calculate work. A different test transfers 18 J by a constant force along 30 cm: find that force. Also explain the work by vertical gravity when a trolley moves horizontally at unchanged height.

Pulling work=20×1.5=30 J. In the separate test convert 30 cm to 0.30 m, then F=W/s=18/0.30=60 N. For horizontal motion at fixed height, distance along vertical gravity is zero, so gravity does zero work. Select the named force; total path length does not apply to every force.

0315–20 min5 minutes

Check work when forces balance

Answer question 4 with reasons; use B only if directed.

Question 4 / CHECK FOR UNDERSTANDING

A trolley moves 2.0 m horizontally at constant speed. Its weight is 100 N vertically down. A 10 N horizontal pull is balanced by 10 N friction. Which pair is correct? Choose a letter and explain both work values.

Gravity 200 J; pull 20 J
Gravity 0 J; pull 0 J
Gravity 0 J; pull 20 J
Gravity 20 J; pull 200 J

Try again B · Retry if directed

Use this support first: Use page 2’s named-force rule. Match each force direction to the displacement; a zero resultant does not mean every force does zero work.

A different trolley moves 3.0 m horizontally at constant speed. Its weight is 80 N vertically down, pull is 6.0 N horizontally forward and friction is 6.0 N opposite. Find work by gravity and by the pull; justify both.

Use the named support first, then answer independently. This replaces part of practice time.

0420–45 min25 minutes

Apply work and energy to supplied tests

Complete 5-8; B replaces 5-6.
At about 30 minutes answer 9 independently and stop.
Continue 10-15; C replaces 10-11.

Questions 5-8

Complete questions 5-8 with the supplied independent-test contexts and lift givens.

The later questions concern separate theatre trolley/lift tests. Use the force named in each question; no experiment is required.

Complete 5-8, or 7-8 only if you completed B. Each question is a separate test. For the brake, calculate the amount transferred.

5. A worker pushes horizontally with a constant force of 30 N while a trolley moves 5.0 m horizontally. Calculate work done by this force.

6. A brake exerts a constant 50 N force opposite the trolley motion over 0.40 m. Calculate the amount of energy transferred by the brake.

7. A constant 24 N force does 96 J of work while acting parallel to trolley motion. Calculate the distance moved.

8. A force parallel to the motion transfers 5.0 J over 25 cm. Calculate the force.

Download these questions with writing space in the pupil handout (PDF) ↓

Question 9 / CHECK FOR UNDERSTANDING

A constant horizontal force does 45 J of work along 75 cm of motion. Find the force. A separate vertical weight of 90 N acts during that same horizontal movement: state its work and explain.

Questions 10-12

Complete questions 10-12 with the supplied independent-test contexts and lift givens.

For question 10, a 24 kg load rises vertically by 2.5 m, as in Figure 1. The upward cable force is 240 N. Its GPE gain is 600 J (given). No GPE equation is needed. Questions 11 and 12 are separate tests.

Use the diagram in the pupil handout, page 6. Figure 1: the supplied lift. The cable force acts upwards along the 2.5 m rise.

10. In the supplied lift, calculate the work done by the 240 N upward cable force through 2.5 m. Compare it with the given 600 J GPE gain.

11. A trolley of weight 240 N moves 5.0 m horizontally at the same height. A trainee says its weight does 1200 J of work. Explain why gravity does no work in this movement.

12. In a separate lift the motor receives 750 J electrically. The load starts and finishes at rest and gains 600 J of GPE. All other transferred energy warms the hoist and surroundings. Calculate this amount and correct “the missing energy was destroyed”.

Download these questions with writing space in the pupil handout (PDF) ↓

Question 13

A trolley moves 4.0 m horizontally at constant speed. A pulling force of 18 N acts along the motion and friction is 18 N opposite. Calculate work done by the pull.

Question 14

For the same trolley, explain why zero resultant force does not mean zero work by the pull. Name the receiving energy stores when friction warms the trolley and surface.

Question 15

Express your pulling-work answer from question 13 in newton metres. Explain why the numerical value is unchanged from the value in joules.

Try again C · Retry if directed

Use this support first: Use page 2: convert cm to m, rearrange F=W/s, then check the direction of each separate force.

In another horizontal test a constant force does 32 J over 40 cm along the force. Find the force. State work by vertical gravity during the same movement and explain.

Use the named support first, then answer independently. This replaces part of practice time.

0545–55 min10 minutes

Correct and explain the physics

Correct practice and explain an error in 16.

Question 16

Correct practice in a different colour and explain one error using the named force, distance along that force or units. If all answers were correct, explain how pulling can transfer energy while a trolley’s kinetic energy stays constant.

0655–60 min5 minutes

Apply direction and energy independently

Cover earlier work and complete 17 independently.
Hand in all pages; D is next lesson only.

Question 17 / CHECK FOR UNDERSTANDING

A trolley moves 1.5 m horizontally at constant speed. A pull of 18 N acts along the motion; friction is 18 N opposite and weight is 90 N vertically down. (a) Find work by the pull. (b) State work by gravity and explain. (c) Explain where the pull’s transferred energy goes despite zero resultant force. (d) In a separate test a pull does 36 J over 60 cm along its direction. Find that force.

Try again D · Follow-up for next lesson

Use this support first: Next lesson revisit page 2’s equations, units and named-force rules. Cover the models before the new independent attempt.

Next lesson: a trolley moves 2.0 m horizontally at constant speed. Pull is 12 N forward, friction 12 N opposite and weight 80 N vertically down. Find pulling work, explain gravity’s work and identify receiving stores. In a separate test, 20 J is transferred by a force acting along 25 cm: find that force.

This follow-up is outside today’s core lesson.

THE RESOURCES, IN THE RIGHT PLACE

Print the handout. Keep the source links.

PUPIL MATERIALS

Ready-to-print pupil handout

Pupil booklet (PDF) - 8 pages with reading, figures and writing space. Reading, numbered questions, diagrams and writing space are included. Print one booklet per pupil; no separate worksheet is needed.

Download pupil handout (PDF) ↓

MAINTAINED WORKSHEET

AQA GCSE Physics Kinetic Energy, GPE and Work Done Worksheet

Q4(a-g), pages 4-5: 7 parts, 12 marks.

Use the 7 embedded work/energy parts. Supply omitted Q3(a) lift data and 600 J GPE gain as a given; no mgh calculation.

Open the selected resource →

TARGETED REVISION

Energy Calculations and Power

Section 4 (#concept-4), first three sentences through mechanical energy transfer.

Use page 2’s supplied constant-force reading, worked/rearranged model and direction/friction rules. Do not assign spring work or the GPE worked example.

Open the selected resource →

FOR THE COVERING TEACHERGuide, answers and AfL decisions

START HERE

Quick subject briefing

Before this lesson: Consolidation after work done, force directions, balanced forces, energy stores and cm-to-m conversion have been taught. Both courses/tiers use the same route. The omitted GPE result is supplied; no mgh, spring or trigonometric calculation is required.

Work is energy transferred by a force through displacement. For a constant force along motion use W=Fs, naming which force is involved. If movement is perpendicular to a force, the distance along it is zero and so is its work. Zero resultant force does not mean each force does zero work: a pull can transfer energy while friction increases thermal stores and kinetic energy stays constant. A braking energy amount is a positive magnitude, not a kinetic-energy gain.

Words and symbols you will need

Work W, joule J: Energy transferred mechanically by a force. W here is the quantity work, not watts.
Force F, newton N: The named push/pull, rather than automatically the resultant.
Distance s, metre m: Movement along the force. Convert cm to m by dividing by 100.
Resultant / balanced: The overall force after combining forces. Zero here means unchanged velocity.
Thermal store: Energy retained in warmer objects/surroundings; energy is not destroyed.

Help pupils choose a method

Name the force and compare its direction with the movement.
Choose the distance along that force and convert it to metres.
Use W=Fs, F=W/s or s=W/F; state J, N or m for the unknown.
Explain receiving stores separately from any change in speed.

Use the supplied model — pupil page 2

Pupil page 2’s 20 N pull does 30 J. The separate unknown-force model converts 30 cm to 0.30 m and gives 60 N. Vertical gravity does zero work on horizontal motion. These are different tests, not one combined energy account.

Worked answer: Pulling work=20×1.5=30 J. In the separate test convert 30 cm to 0.30 m, then F=W/s=18/0.30=60 N. For horizontal motion at fixed height, distance along vertical gravity is zero, so gravity does zero work. Select the named force; total path length does not apply to every force.

If pupils are still stuck

Use the matching Say/Ask/Expect coaching, then the named fresh retry in its replacement time. If still insecure, help label the named force, direction and units before selecting the equation. Record the exact remaining difficulty for the usual teacher. Keep D for next lesson; do not extend today’s lesson.

KEEP THIS OPEN DURING THE LESSON

Timed teacher run sheet

Print the complete 8-page pupil booklet, one per pupil, stapled in order. Double-sided printing is fine.
Read teacher page 1, then use the timed run sheet on page 2. Keep the 10-page teacher guide separate from pupil copies.
Pupils need pens, a ruler and a calculator. Reading, data and figures are supplied. No login, pupil devices or practical apparatus are required.

0-5 min

Pupil page / Q
1 / 1-3

Read aloud / check: Start from memory; check work, conversion and unit.

Next: If insecure A uses first 2 reading minutes.

Answers: 1, 2, 3.

Open check 1 and its coaching →

5-15 min

Pupil page / Q
2 / reading

Read aloud / check: Read the named-force rules and complete models.

Next: Check A if used; distinguish work W from watts.

15-20 min

Pupil page / Q
3 / 4

Read aloud / check: Commit to both work values and reasons; stop.

Next: If insecure B replaces 5-6.

Answers: 4.

Open check 2 and its coaching →

20-30 min

Pupil page / Q
4 / 5-8

Read aloud / check: Calculate work, distance and force for separate tests.

Next: If B used, complete 7-8 only; leave 5-6 blank.

Answers: 5, 6, 7, 8.

30-33 min

Pupil page / Q
5 / 9

Read aloud / check: Find force independently, then explain gravity’s work.

Next: If insecure C replaces 10-11.

Answers: 9.

Open check 3 and its coaching →

33-45 min

Pupil page / Q
6-7 / 10-15

Read aloud / check: Use the supplied GPE gain and complete the constant-speed application.

Next: If C used leave 10-11 blank; complete 12-15.

Answers: 10, 11, 12, 13, 14, 15.

45-55 min

Pupil page / Q
8 / 16

Read aloud / check: Share practice keys; correct and explain a physical reason.

Next: Keep independent exit answers hidden.

Answers: 16.

55-60 min

Pupil page / Q
8 / 17

Read aloud / check: Cover earlier work and complete the exit independently.

Next: Collect all booklets. D is next lesson only.

Answers: 17.

Open check 4 and its coaching →

Estimated 60 minutes. Practice: 10 minutes, 3-minute mid-check and 12 minutes. A uses reading time; B replaces 5-6; C replaces 10-11; D is next lesson only. Timings not classroom-trialled.

Check entry, hinge and mid-practice at the marked pauses. At 45 minutes share only practice answers. Keep the independent exit and next-lesson answers hidden until submission.

Download teacher guide & answers ↓

If several pupils are stuck

Estimated 60 minutes. Practice: 10 minutes, 3-minute mid-check and 12 minutes. A uses reading time; B replaces 5-6; C replaces 10-11; D is next lesson only. Timings not classroom-trialled.

ENTRY CHECK

Check 1: Question 1, Question 2, Question 3

Look for: Q1 36 J; Q2 0.40 m; Q3 joule and 1 N m=1 J.

If secure: Continue to page 2 reading/model.

If: Divides force by distance or gives watts/newtons for work.Open coaching

Likely issue: The equation or difference between force, work and power is insecure.

Say: Work is the energy a named force transfers through movement. For a constant force along the movement multiply newtons by metres; the result is joules, not watts.
Ask: For question 1, what is 12×3 and which unit measures transferred energy?
Expect: 36 J. One newton metre is one joule.

Next: Use page 2’s equation/units, then complete Try again A on page 1 in the first 2 reading minutes.

Pupil page 2, W=Fs and units rows.

If: Uses 40 m or 0.04 m for question 2.Open coaching

Likely issue: The centimetre conversion is wrong.

Say: There are 100 centimetres in one metre. Divide the centimetre number by 100 before using the work equation.
Ask: In question 2, what is 40/100, and is 40 cm less than a metre?
Expect: 0.40 m; yes.

Next: Use page 2’s equation/units, then complete Try again A on page 1 in the first 2 reading minutes.

Pupil page 2, conversions row.

HINGE CHECK

Check 2: Question 4

Look for: C with gravity 0 J from perpendicular displacement and pull 20 J despite balanced forces.

If secure: Continue to questions 5-8.

If: A: gravity does 200 J.Open coaching

Likely issue: Horizontal path length has been used with vertical weight.

Say: Work uses distance along the force. The trolley does not rise or fall, so the distance along vertical gravity is zero.
Ask: For question 4, how much vertical displacement does the trolley have?
Expect: Zero, so gravity does 0 J.

Next: Use page 2’s named-force and direction rules, then complete Try again B on page 3 instead of 5-6; complete 7-8.

Pupil page 2, perpendicular-force rule.

If: B: the pull does zero work because forces balance.Open coaching

Likely issue: Resultant force has replaced the named force in the work equation.

Say: The pull is still 10 N through 2 m. Balanced forces keep the speed unchanged; they do not make each force’s energy transfer zero.
Ask: For question 4, which force is named for pulling work and what is 10×2?
Expect: The 10 N pull; work is 20 J.

Next: Use page 2’s named-force and direction rules, then complete Try again B on page 3 instead of 5-6; complete 7-8.

Pupil page 2, balanced-forces rule.

If: D, or C without an explanation.Open coaching

Likely issue: The force values may be swapped or the correct letter guessed.

Say: Label each force and its direction before calculating. Weight is vertical; the pull is the smaller horizontal force. Use neither force in place of the other.
Ask: In question 4, pair the vertical weight and horizontal pull with their distances along the force.
Expect: Weight: 0 m, giving 0 J. Pull: 2 m, giving 20 J.

Next: Use page 2’s named-force and direction rules, then complete Try again B on page 3 instead of 5-6; complete 7-8.

Pupil page 2, named-force model.

MID PRACTICE CHECK

Check 3: Question 9

Look for: 0.75 m; F=45/0.75=60 N; vertical weight does 0 J with direction reason.

If secure: Continue to 10-15, using the supplied lift givens.

If: Gives 0.60 N, 33.75 N or uses the unrearranged equation.Open coaching

Likely issue: Conversion or solving for force is insecure.

Say: Convert the distance before rearranging. Since W=Fs, divide the work by the distance in metres to recover F.
Ask: For question 9, what is 75 cm in metres, and then 45/0.75?
Expect: 0.75 m and 60 N.

Next: Use page 2’s conversion/rearrangement and direction model, then complete Try again C on page 5 instead of 10-11; complete 12-15.

Pupil page 2, conversions and rearranged model.

If: Gives nonzero work by the vertical weight.Open coaching

Likely issue: Force direction has not been matched to displacement.

Say: The 90 N weight is a separate vertical force. Horizontal motion contains zero displacement along that force, regardless of the weight’s size.
Ask: In question 9, is any rise or fall stated? What work follows for weight?
Expect: No rise/fall; gravity does 0 J.

Next: Use page 2’s conversion/rearrangement and direction model, then complete Try again C on page 5 instead of 10-11; complete 12-15.

Pupil page 2, perpendicular-force rule.

EXIT CHECK

Check 4: Question 17

Look for: Q17: pulling 27 J; gravity 0 J; friction warms named thermal stores with constant kinetic energy; separate force 60 N.

If secure: Record secure calculations, named-force reasoning and energy-transfer explanation.

If: Pulling work or separate force is wrong, especially after using 60 as metres.Open coaching

Likely issue: Equation choice, conversion or rearrangement needs follow-up.

Say: For work, multiply named force by distance along it. For an unknown force, convert distance to metres then divide work by that distance.
Ask: For question 17, what is 18×1.5, and what is 36 divided by 0.60?
Expect: 27 J and 60 N.

Next: Next lesson revisit the relevant page 2 rules, then complete Try again D independently on page 8.

Pupil page 2, equation/units and rearranged model.

If: Gravity does work, or claims no energy transfer/destruction because resultant is zero.Open coaching

Likely issue: Direction or individual energy transfers remain insecure.

Say: Gravity has no displacement along it here. The pull still transfers energy, balanced by frictional transfers to thermal stores; constant speed means unchanged kinetic energy.
Ask: For question 17, which force acts along the movement and which objects warm?
Expect: Pull/friction act horizontally; trolley/surface/surroundings gain thermal energy. Gravity does 0 J; energy is not destroyed.

Next: Next lesson revisit the relevant page 2 rules, then complete Try again D independently on page 8.

Pupil page 2, direction, balanced forces and friction reading.

Answers keyed to the pupil handout

Question 1 Pupil page 1: 12×3.0=36 J.
Question 2 Pupil page 1: 40/100=0.40 m.
Question 3 Pupil page 1: Joule (J). 1 N m=1 J. Work is an amount of transferred energy, not power in watts.
Question 4 Pupil page 3: C: gravity does 0 J because motion has no vertical displacement; the pull does 10×2=20 J. The zero resultant does not make work by each individual force zero.
Question 5 Pupil page 4: W=Fs=30×5.0=150 J.
Question 6 Pupil page 4: Energy transferred=50×0.40=20 J. This is the positive magnitude of braking work, not a gain in kinetic energy.
Question 7 Pupil page 4: s=W/F=96/24=4.0 m.
Question 8 Pupil page 4: 25 cm=0.25 m. F=W/s=5.0/0.25=20 N.
Question 9 Pupil page 5: 75 cm=0.75 m; F=45/0.75=60 N. The vertical weight does 0 J because the motion has no displacement along the weight.
Question 10 Pupil page 6: Cable work=240×2.5=600 J, equal to the supplied GPE gain. Use the cable force, not the resultant force. No mgh calculation is needed.
Question 11 Pupil page 6: Weight acts vertically and displacement is horizontal. Distance along the weight is zero, so work done by gravity is zero. Do not multiply weight by the horizontal path length.
Question 12 Pupil page 6: 750-600=150 J increases thermal stores of the hoist and surroundings. Energy is dissipated into less useful stores, not destroyed; the complete account conserves energy.
Question 13 Pupil page 7: W=18×4.0=72 J. Use the pulling force, not the zero resultant force.
Question 14 Pupil page 7: The pull still does 72 J of work while the opposing forces balance. Kinetic energy is unchanged; friction transfers energy to thermal stores of trolley/surface/surroundings. Energy is not destroyed.
Question 15 Pupil page 7: 72 N m=72 J because 1 N m=1 J. Accept consistent numerical follow-through from question 13.
Question 16 Pupil page 8: Accept a correction with a physical reason, not a copied answer. All-correct: a pull can do work while friction transfers energy to thermal stores; balanced forces mean unchanged velocity/kinetic energy, not no individual transfers.
Question 17 Pupil page 8: (a) 18×1.5=27 J. (b) 0 J: no vertical displacement along gravity. (c) Kinetic energy is unchanged; 27 J is transferred through friction to thermal stores of trolley/surface/surroundings, not destroyed. (d) 60 cm=0.60 m; F=36/0.60=60 N. Accept consistent follow-through for the amount in (c).
Try again A: (a) 0.20 m. (b) 8.0×2.0=16 J. (c) Joules; 1 N m=1 J.
Try again B: Gravity: 0 J, because no vertical displacement. Pull: 6.0×3.0=18 J despite zero resultant force. Use each named force and the distance along it.
Try again C: 40 cm=0.40 m; F=32/0.40=80 N. Gravity does 0 J: horizontal displacement has no component along vertical weight.
Try again D: Pulling work=12×2=24 J. Gravity does 0 J with no vertical displacement. Kinetic energy stays constant; friction transfers energy to thermal stores of trolley/surface/surroundings, not destruction. Separate force: 25 cm=0.25 m; F=20/0.25=80 N.

Worksheet selection and marking

Seven maintained source parts total 12 marks; original trolley application adds 4 marks (13:1, 14:2, 15:1). Accept equivalent physics and consistent numerical follow-through. Do not require signed negative work or trigonometry.

Selection: Q4(a-g), pages 4-5: 7 parts, 12 marks.

Open the maintained worked solutions →

Collect and follow up: Collect every named booklet with working/corrections. Record who needs support with conversions/rearrangement, force direction, or frictional stores and who needs Try again D next lesson.

A USEFUL NEXT STEP

Choose the support that matches the answer.

AQA GCSE Physics Kinetic Energy, GPE and Work Done Worksheet →
Use the 7 embedded work/energy parts. Supply omitted Q3(a) lift data and 600 J GPE gain as a given; no mgh calculation.
Energy Calculations and Power →
Use page 2’s supplied constant-force reading, worked/rearranged model and direction/friction rules. Do not assign spring work or the GPE worked example.

Work done by forces.

Know what pupils are working towards.

By the end, pupils should be able to…

Course and starting points

Useful prior knowledge

From recall to independent explanation.

Recall work, distance units and joules

Read the named-force model

Energy Calculations and Power

Work done by a force - selected constant-force reading

Check work when forces balance

Apply work and energy to supplied tests

Correct and explain the physics

Apply direction and energy independently

Print the handout. Keep the source links.

Ready-to-print pupil handout

AQA GCSE Physics Kinetic Energy, GPE and Work Done Worksheet

Energy Calculations and Power

Quick subject briefing

Words and symbols you will need

Help pupils choose a method

Use the supplied model — pupil page 2

If pupils are still stuck

Timed teacher run sheet

If several pupils are stuck

Check 1: Question 1, Question 2, Question 3

Check 2: Question 4

Check 3: Question 9

Check 4: Question 17

Answers keyed to the pupil handout

Worksheet selection and marking

Choose the support that matches the answer.

Specification and review