AQA GCSE PHYSICS · PAPER 2 · FOUNDATION & HIGHER
Work done and energy transfer
Forces · Lesson 4 of 20
Teacher copy - includes the notes for whoever is teaching from it.
Warm-up
Answer each one, then check.
1. What is energy measured in?
Joules
2. What is a force measured in?
Newtons
3. What is distance measured in?
Metres
4. What is friction?
A force opposing motion
5. Name an energy store that increases when something warms up.
Thermal
Learning Objectives
1. Define work done.
2. Recall and apply W = Fs.
3. Convert between newton-metres and joules.
4. Describe the energy transfers when work is done, including against friction.
Work Done
work done = force × distance moved along the line of action of the force W = Fs
One joule of work is done when a force of one newton moves an object one metre. 1 J = 1 Nm.
Work Done
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Work done is energy transferred by a force. |
A person pushing a box with a force of 50 N for 12 m, and a brake showing work done against friction.
Calculating Work Done
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A force of 50 N pushes a box 12 m along the floor. Calculate the work done. |
1. Write the equation
W = Fs
2. Substitute
W = 50 × 12
3. Answer
W = 600 J
Answer: 600 J
Finding the Force
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A crate is moved 8.0 m by a force. The work done is 1200 J. Find the force. |
1. Rearrange
F = W ÷ s
2. Substitute
F = 1200 ÷ 8.0
3. Answer
F = 150 N
Answer: 150 N
Work Against Friction
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Explain why the brakes on a bicycle get warm when it slows down. |
1. Friction force
The brake pads rub on the wheel
2. Work done
Work is done against the frictional force
3. Energy transfer
Energy is transferred from the kinetic store to the thermal store of the brakes
Answer: The kinetic energy of the bicycle is transferred to the thermal energy of the brakes and wheel.
Key Ideas
Learn these links.
▸ Work is energy transferred. When a force moves an object, energy is transferred.
▸ Distance. Measured in the direction of the force.
▸ Friction. Work done against friction raises the temperature of the object.
▸ Units. Convert kJ to J before calculating.
Key Terms
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Work done Energy transferred when a force moves an object. |
Joule The unit of work and energy. |
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Newton-metre A joule: one newton moving an object one metre. |
Friction A force between surfaces that opposes motion. |
|
Displacement The distance moved in a particular direction. |
Dissipated Transferred to less useful stores. |
Your Task: Work It Out
10 minutes
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A student pushes a trolley with a force of 40 N for 6.0 m. How much work is done? What happens to the energy if the trolley is moving at a steady speed on a rough floor? 1. Calculate the work. 2. Describe the energy transfer. |
A good answer shows: W = 40 × 6.0 = 240 J. The energy is transferred to the thermal store of the trolley and floor by friction.
Note: Ask what would be different on a smooth floor.
Can I...?
☐ Define work done.
☐ Use W = Fs.
☐ Rearrange for F or s.
☐ Convert Nm to J.
☐ Describe energy transfer.
☐ Explain heating due to friction.
☐ Use the unit joule.
☐ Show the working.
Summary
✓ W = Fs.
✓ 1 J = 1 Nm.
✓ Work done against friction raises the temperature.
✓ Work is energy transferred.
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EXAM FOCUS A force of 50 N moves a box 12 m. Calculate the work done. (2 marks) Multiply force by distance and give J. |
Exam Practice: Work done and energy transfer
Answer all questions. Use the mark allocation as a guide to how much to write. · 14 minutes
▸ Question 1 · 2 marks · Calculate. A person pushes a box with a force of 50 N. The box moves 12 m. Calculate the work done. Use the equation: work done = force × distance
▸ Question 2 · 3 marks · Calculate. A force does 1200 J of work moving a crate 8.0 m. Calculate the force.
▸ Question 3 · 2 marks · Explain. A bicycle rider brakes and the brake pads become warm. Explain why.
▸ Question 4 · 2 marks · Describe. State what one joule of work means in terms of force and distance.
▸ Question 5 · 4 marks · Calculate. A crane lifts a 200 kg load through 15 m at constant speed. Gravitational field strength = 9.8 N/kg. Calculate the work done on the load by…
Question 1 · 2 marks · Calculate
|
“A person pushes a box with a force of 50 N. The box moves 12 m. Calculate the work done. Use the equation: work done = force × distance” |
HOW TO ANSWER IT Command word: Calculate. Worth 2 marks, so plan before writing.
Question 1 · mark scheme
2 marks available. Award a mark for each point made.
▸ Correct substitution. 1 mark
▸ 600 J. 1 mark
▸ Model answer. 50 × 12 = 600 J
Question 2 · 3 marks · Calculate
|
“A force does 1200 J of work moving a crate 8.0 m. Calculate the force.” |
HOW TO ANSWER IT Command word: Calculate. Worth 3 marks, so plan before writing.
Question 2 · mark scheme
3 marks available. Award a mark for each point made.
▸ Rearranges to F = W ÷ s. 1 mark
▸ Correct substitution. 1 mark
▸ 150 N. 1 mark
▸ Model answer. F = W ÷ s = 1200 ÷ 8.0 = 150 N
Question 3 · 2 marks · Explain
|
“A bicycle rider brakes and the brake pads become warm. Explain why.” |
HOW TO ANSWER IT Command word: Explain. Worth 2 marks, so plan before writing.
Question 3 · mark scheme
2 marks available. Award a mark for each point made.
▸ Work done against friction. 1 mark
▸ Kinetic to thermal energy store. 1 mark
▸ Model answer. Work is done against the friction between the pads and the wheel. The kinetic energy of the bicycle is transferred to the thermal energy store of the brakes.
Question 4 · 2 marks · Describe
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“State what one joule of work means in terms of force and distance.” |
HOW TO ANSWER IT Command word: Describe. Worth 2 marks, so plan before writing.
Question 4 · mark scheme
2 marks available. Award a mark for each point made.
▸ Force of 1 N. 1 mark
▸ Moves an object 1 m. 1 mark
▸ Model answer. One joule is the work done when a force of one newton moves an object a distance of one metre in the direction of the force.
Question 5 · 4 marks · Calculate
|
“A crane lifts a 200 kg load through 15 m at constant speed. Gravitational field strength = 9.8 N/kg. Calculate the work done on the load by the crane.” |
HOW TO ANSWER IT Command word: Calculate. Worth 4 marks, so plan before writing.
Question 5 · mark scheme
4 marks available. Award a mark for each point made.
▸ Force = weight = mg. 1 mark
▸ 1960 N. 1 mark
▸ Correct substitution into W = Fs. 1 mark
▸ 29 400 J. 1 mark
▸ Model answer. F = 200 × 9.8 = 1960 N; W = 1960 × 15 = 29 400 J