A-Level · Physics · AQA · Mark scheme decoded
AQA A-Level Physics: Work, Power, and Efficiency in Mechanics — mark scheme explained
The short answer
In this section, we will explore the concepts of work, power, and efficiency, particularly focusing on how these principles apply to mechanical systems and electric motors. Understanding these concepts is crucial for solving problems related to energy transfer and the performance of machines.
The question
A force of 10 N is applied to move an object 5 m along a horizontal surface. The angle between the force and the direction of displacement is 30°. Calculate the work done.
[Paraphrased for study — not reproduced from any exam paper.]
Mark scheme, decoded
How the examiner actually awards the marks on this topic.
Gradora's own decode of the marking approach — not the exam board's published mark scheme.
How marks are awarded
For calculation questions, marks are typically awarded for correct use of formulae, substitution of values, and final answers. For conceptual questions, marks are given for clear and accurate explanations. Always show your working and units.
What the command words demand
- Calculate
- Perform a numerical calculation to find a specific value.
- Determine
- Find a value by reasoning or calculation, often involving multiple steps.
- Explain
- Provide a clear and detailed account of how something works or why something happens.
- Identify
- Recognize and name key concepts, values, or errors in a given context.
- Describe
- Give a detailed account of the characteristics or features of something.
Model answer
A full-mark response to the question above, worked through step by step.
Timing: Allocate about 1-1.2 minutes per mark. For a 6-mark question, spend approximately 6-7 minutes.
- Identify the given values: F = 10 N, s = 5 m, θ = 30°.0 marks
- Use the formula for work done: W = F × s × cos(θ).1 mark
- Calculate cos(30°) = √3/2 ≈ 0.866.1 mark
- Substitute the values into the formula: W = 10 N × 5 m × 0.866.1 mark
- W = 43.3 J.1 mark
Final answer: 43.3 J
Work through every step correctly and you earn all 4 marks.
Another worked example
A motor lifts a mass of 2 kg to a height of 10 m in 5 seconds. Calculate the useful output power of the motor.
- Identify the given values: m = 2 kg, h = 10 m, t = 5 s, g = 9.81 m/s 2 .0 marks
- Calculate the work done by the motor: W = m × g × h = 2 kg × 9.81 m/s 2 × 10 m = 196.2 J.2 marks
- Use the formula for power: P = W / t.1 mark
- Substitute the values into the formula: P = 196.2 J / 5 s.1 mark
- P = 39.24 W.1 mark
Final answer: 39.24 W
Work through every step correctly and you earn all 5 marks.
Common mistakes
Forgetting to include the cosine term in the work done formula when the force is not applied in the direction of displacement.
Why it happens: Students often overlook the importance of the angle between the force and the direction of displacement, leading to incorrect calculations.
Fix: Always check if the force is applied at an angle and use the cosine term accordingly.
Confusing work done with power in problems involving time.
Why it happens: Students sometimes mix up the concepts of work and power, especially when time is involved in the problem.
Fix: Remember that work is energy transferred (W = F × s × cos(θ)), while power is the rate at which work is done (P = W / t or P = F × v).
Using incorrect units for force, displacement, and time.
Why it happens: Students may use inconsistent units, leading to incorrect answers.
Fix: Always ensure that all values are in the correct SI units (N for force, m for displacement, s for time).
Forgetting to convert efficiency into a percentage when required.
Why it happens: Students sometimes forget to multiply by 100% when calculating efficiency, leading to incorrect answers.
Fix: Always remember to express efficiency as a percentage by multiplying the result by 100%.
Incorrectly identifying random and systematic errors in experiments.
Why it happens: Students may not fully understand the difference between random and systematic errors, leading to incorrect identification and analysis.
Fix: Review the definitions of random and systematic errors and practice identifying them in different experimental scenarios.
Failing to account for the area under a force-displacement graph when calculating work done by a variable force.
Why it happens: Students may not realize that the area under the curve represents the work done, especially in problems involving variable forces.
Fix: Always remember that the area under a force-displacement graph gives the work done. Practice integrating or using geometric methods to find this area.
Where the marks go
The question types you’ll meet on this topic and the marks each one carries — so you know what to expect and where to focus.
| Question type | What you’re asked to do | Marks |
|---|---|---|
| Calculate Work Done | Find the work done by a force acting at an angle to the displacement. | 4 |
| Calculate Power | Find the useful output power of a motor lifting a mass through a height. | 5 |
| Calculate Motor Efficiency | Find efficiency by comparing useful power output to total electrical power input. | 7 |
| Calculate Displacement | Rearrange the work-done equation to find displacement using force, work, and angle. | 4 |
| Calculate Efficiency | Find useful energy output and input energy, then calculate the motor's percentage efficiency. | 6 |
| Total across these question types | 26 | |
Question types and mark tariffs are Gradora’s guidance based on how this topic is typically examined — not the board’s official paper structure.