Work, energy and power — AQA A-Level Physics
Teaching note checked against the official specification.
Work, energy and power are central concepts in AQA A-Level Physics that describe how forces transfer energy and how quickly that transfer occurs. The work-energy principle states that the net work done on an object equals its change in kinetic energy, while power measures the rate of energy transfer, and conservation of energy underpins all mechanical analysis.
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Work, energy and power explained
The area under a force–displacement graph represents the work done by the force.
Read the full explanation
Work is the energy transferred when a force moves its point of application through a displacement in the direction of the force. For a constant force, work W = Fs cos θ, and the area of a rectangle on an F–s graph gives F × s. For a variable force, the area under the curve must be found by splitting the graph into strips, counting squares or integrating F with respect to s. For example, a spring obeying Hooke's law has F = kx, so the area under its straight-line graph is ½kx², the elastic potential energy stored. The area is a signed quantity: area below the displacement axis represents negative work, meaning energy is transferred away from the object. Always check the axes and units before calculating area.
Your focus
- Interpret the area under a force–displacement graph as the work done by the force.
- Calculate work done from the area under a force–displacement graph for constant and variable forces.
- Explain the significance of positive and negative areas in terms of energy transfer.
Which command word is this?
You must use the given data and appropriate equations to find a numerical answer. Marks are awarded for correct substitution, accurate arithmetic, and the correct unit. No marks for just writing the equation.
Work, energy and power exam tips
Topic Overview
This topic covers the fundamental principles of work, energy, and power, which are essential for understanding mechanics in A-Level Physics. You will learn how to calculate work done by forces, apply the work-energy principle to solve problems involving kinetic and potential energy, and analyse power as the rate of energy transfer. These concepts are crucial for explaining real-world systems, from simple machines to vehicle motion and energy efficiency.
Work, energy, and power form the backbone of classical mechanics and appear throughout the AQA specification, linking to topics such as momentum, circular motion, and materials. Mastery of these ideas enables you to solve complex problems involving forces, motion, and energy conservation, and provides a foundation for further study in fields like engineering and astrophysics. Understanding energy transfers and efficiency is also vital for addressing global challenges like sustainable energy use.
Key Concepts
- →Work done by a constant force is defined as the product of the force and the displacement in the direction of the force: W = F s cosθ, where θ is the angle between force and displacement.
- →The work-energy principle states that the net work done on an object equals its change in kinetic energy: W_net = ΔEk = 0.5 m v^2 - 0.5 m u^2.
- →Gravitational potential energy near Earth's surface is given by ΔEp = m g Δh, and elastic potential energy in a stretched spring is Ep = 0.5 k x^2.
- →Power is the rate of doing work or transferring energy: P = W / t = F v, and efficiency is the ratio of useful output energy to total input energy, often expressed as a percentage.
- →The principle of conservation of energy states that energy cannot be created or destroyed, only transferred between stores; in closed systems, total energy remains constant.
Marking Points
- States that the area under a force–displacement graph equals the work done by the force.
- Explains that work done equals energy transferred, so the area also represents the energy transferred to or from the object.
- For a constant force, calculates area as a rectangle: area = force × displacement.
- For a variable force, finds area by counting squares, using the trapezium rule or integrating the force–displacement relationship.
- Interprets area below the displacement axis as negative work done, indicating energy transferred away from the object.
- Applies the result to a spring obeying Hooke's law, where the triangular area gives elastic potential energy ½kx².
Examiner Tips
- 💡Always label the axes and state the unit of area, for example N m or J, before or after calculating the area.
- 💡When a graph is not a simple shape, count squares within the curve and multiply by the area represented by one square, showing your working clearly.
- 💡If the force–displacement graph is a straight line through the origin, use the triangle area formula ½ × base × height rather than counting squares.
- 💡Check whether the question asks for work done on the object or by the object, and use the sign of the area to support your answer.
- 💡Always show your working clearly, including the equation used, substitution of values with units, and the final answer with appropriate significant figures and units. Marks are awarded for correct method even if the final answer is wrong due to a calculation error.
- 💡When asked to explain energy transfers, use precise terminology such as 'energy transferred from the chemical store of the fuel to the kinetic store of the car' rather than vague statements like 'energy is lost'. Specify the stores involved and whether the transfer is useful or wasteful.
- 💡In questions involving power, remember that power can be calculated as P = F v for an object moving at constant speed against a resistive force. This is often quicker than calculating work done and time separately.
Common Mistakes
- Confusing the area under a force–displacement graph with the gradient; the gradient gives stiffness or force per unit displacement, not work done. Correct by recalling that area = work done and gradient = force constant for a spring.
- Using the maximum force instead of the average force for a variable force; for a linear variation the average force is half the maximum, but for a non-linear graph the area must be found by squares or integration.
- Ignoring the sign of the area when the force acts opposite to the displacement; area below the axis represents negative work done, not a magnitude to be made positive.
- Forgetting to convert units such as cm to m or N to kN before calculating area, which gives an incorrect value for work done.
- Students often think that if an object moves at constant speed, no work is being done. In reality, work may be done against resistive forces like friction, and energy is transferred to thermal stores even though kinetic energy remains constant.
- Many students confuse power and energy, believing that a more powerful machine always does more work. Power is the rate of energy transfer, so a high-power device transfers energy quickly but may do less total work over a shorter time.
- Students frequently forget that work done by a force perpendicular to displacement is zero, such as the normal force on a horizontal surface or centripetal force in circular motion, leading to incorrect energy analyses.
Revision Plan
- 1Start by reviewing the definitions of work, energy, and power, and ensure you can state and apply the relevant equations (W = F s cosθ, Ek = 0.5 m v^2, Ep = m g h, P = W/t = F v).
- 2Work through a variety of problems, beginning with simple calculations of work done by a single force, then progressing to multi-step problems involving the work-energy principle and conservation of energy.
- 3Practise exam-style questions from past AQA papers, focusing on structured questions that require explanations of energy transfers and efficiency calculations. Mark your answers using the mark scheme to identify gaps.
- 4Create a summary sheet of key equations, units, and common pitfalls, and use it for quick revision. Test yourself with active recall questions and flashcards.
- 5After one week, attempt a full past paper under timed conditions to build exam technique and identify any remaining weaknesses. Review any errors and revisit the relevant concepts.
Exam Question Types
- 📋Calculation questions: These require you to use equations for work, energy, or power to find a numerical answer. Always show your working and pay attention to unit conversions and significant figures.
- 📋Explanation questions: You may be asked to describe energy transfers in a given scenario, such as a rollercoaster or a braking car. Use precise language and refer to conservation of energy and efficiency.
- 📋Graph questions: You might be given a force-displacement graph and asked to calculate work done as the area under the graph, or a power-time graph to find energy transferred. Remember that area under a force-displacement graph equals work done.
- 📋Practical-based questions: These may involve analysing experimental data to determine work done or power, and evaluating sources of error. Focus on identifying the independent and dependent variables and using appropriate apparatus.
Command Word Expectations (AQA)
You must use the given data and appropriate equations to find a numerical answer. Marks are awarded for correct substitution, accurate arithmetic, and the correct unit. No marks for just writing the equation.
You must provide a reasoned account that links cause and effect, using correct physics terminology. Typically, you need to refer to energy transfers, forces, or conservation principles, and make the logical connection clear.
You must use given data or information to find a value, often requiring rearrangement of an equation or interpretation of a graph. Usually, the answer is unique and you should show your method.
How Students Lose Marks (Examiner Pitfalls)
Step-by-Step Worked Solutions
Question: A crate of mass 25 kg is pulled 8.0 m along a rough horizontal floor by a horizontal force of 120 N. The frictional force is 50 N. Calculate the work done by the applied force, the work done against friction, and the final speed of the crate starting from rest.
- 1.Step 1: Identify given facts: mass m = 25 kg, displacement s = 8.0 m, applied force F_app = 120 N, friction F_fric = 50 N, initial speed u = 0 m/s.
- 2.Step 2: Calculate work done by applied force: W_app = F_app * s = 120 * 8.0 = 960 J.
- 3.Step 3: Calculate work done against friction: W_fric = F_fric * s = 50 * 8.0 = 400 J (negative work done on crate).
- 4.Step 4: Net work done = W_app - W_fric = 960 - 400 = 560 J.
- 5.Step 5: Apply work-energy principle: Net work = change in kinetic energy = 0.5 * m * v^2 - 0.5 * m * u^2. Since u = 0, 560 = 0.5 * 25 * v^2.
- 6.Step 6: Solve for v: v^2 = (560 * 2) / 25 = 44.8, so v = sqrt(44.8) = 6.69 m/s.
Question: A pump lifts 200 kg of water per minute from a well 15 m deep and delivers it with a speed of 4.0 m/s. Calculate the minimum power output of the pump. Assume g = 9.81 m/s^2.
- 1.Step 1: Identify given facts: mass per minute = 200 kg, so mass per second = 200/60 = 3.333 kg/s, height h = 15 m, final speed v = 4.0 m/s, g = 9.81 m/s^2.
- 2.Step 2: Calculate gain in gravitational potential energy per second: ΔEp = mgh = 3.333 * 9.81 * 15 = 490.5 J/s.
- 3.Step 3: Calculate gain in kinetic energy per second: ΔEk = 0.5 * m * v^2 = 0.5 * 3.333 * (4.0)^2 = 26.67 J/s.
- 4.Step 4: Total energy transferred per second = ΔEp + ΔEk = 490.5 + 26.67 = 517.17 J/s.
- 5.Step 5: Power = energy per second = 517.17 W.