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    Work and power (A-level only) — AQA A-Level Physics

    Teaching note checked against the official specification.

    Work is the energy transferred when a force moves an object through a distance, calculated as W = Fs cos(theta), while power is the rate of doing work or transferring energy, P = W/t. In AQA A-Level Physics, you must apply these definitions to mechanical systems, resolve forces, interpret gradients and areas on force-displacement and power-time graphs, and use the work-energy principle to link work done to changes in kinetic and potential energy.

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    Work and power (A-level only) explained

    For rotation, work done by a constant torque T through an angular displacement θ is W = Tθ, and power is P = Tω, where ω is angular velocity.

    Read the full explanation

    These equations mirror linear W = Fs and P = Fv. In rotating machinery, frictional torque opposes motion, so the net torque is the applied torque minus the frictional torque. This reduces the useful work output and power delivered. Students should calculate work and power using the correct angular quantities, convert between degrees and radians where needed, and account for frictional torque by subtracting it from the driving torque. They should also interpret efficiency as the ratio of useful output to total input, recognising that friction causes energy dissipation as thermal energy.

    Read the Work and power (A-level only) study guideFull revision notes for AQA A-Level Physics

    Your focus

    1. Use W = Tθ and P = Tω to solve problems involving constant torque and angular displacement or angular velocity.
    2. Account for frictional torque when determining net torque, useful work and useful power in rotating machinery.
    3. Explain how friction reduces the efficiency of rotating machinery and leads to energy transfer by heating.

    Which command word is this?

    You must use the given data and appropriate equations to find a numerical answer. Show all working, including the equation, substitution, and final answer with correct units and significant figures. No marks are awarded for the correct answer without working in 'show that' questions, but in 'calculate' questions, full marks can be obtained with correct working even if the final answer is slightly wrong due to rounding.

    Work and power (A-level only) exam tips

    Topic Overview

    This topic covers the definitions of work and power and their application to mechanical systems. You will learn that work is the energy transferred when a force moves an object through a distance, and that power is the rate of doing work or transferring energy. These concepts are fundamental to understanding energy conservation, efficiency, and the motion of vehicles and machines.

    In AQA A-Level Physics, you must be able to calculate work done by a force at an angle, use the work-energy principle to relate net work to changes in kinetic energy, and apply P = Fv to problems involving constant power. This topic links closely to Newton's laws, energy conservation, and momentum, and is essential for analysing real-world systems such as lifts, pumps, and accelerating vehicles.

    Key Concepts
    • →Work done by a constant force is defined as W = Fs cos(theta), where F is the magnitude of the force, s is the displacement, and theta is the angle between the force and the displacement. Only the component of force in the direction of displacement does work.
    • →Power is the rate of doing work or transferring energy: P = W/t. The instantaneous power delivered by a force moving an object at velocity v is P = Fv, where F is the component of force in the direction of motion.
    • →The work-energy principle states that the net work done on an object equals its change in kinetic energy: W_net = delta E_k = 0.5mv^2 - 0.5mu^2. This is a powerful tool for solving problems involving forces, distances, and speeds.
    • →Efficiency is the ratio of useful output energy or power to total input energy or power, often expressed as a percentage: efficiency = (useful power output / total power input) x 100%.
    • →For a vehicle moving at constant speed, the driving force equals the total resistive force, and the power required is P = Fv. At maximum speed, the resultant force is zero, so the driving force equals the resistive force.
    Marking Points
    • Recalls and applies W = Tθ for work done by a constant torque through an angular displacement.
    • Recalls and applies P = Tω for mechanical power in rotational motion.
    • Converts angular displacement between degrees and radians correctly when using W = Tθ.
    • Explains that frictional torque opposes rotation and must be subtracted from the applied torque to find the net torque.
    • Calculates useful work or power by using the net torque after accounting for friction.
    • Relates energy lost to friction to heating of components and reduced efficiency of the machine.
    Examiner Tips
    • 💡Write down the equation W = Tθ or P = Tω before substituting values, and check that θ is in radians.
    • 💡When friction is present, calculate the net torque first, then use it in the work or power equation.
    • 💡State clearly whether you are calculating total input work or useful output work, and show how friction affects the result.
    • 💡Always show your working clearly, including the formula, substitution, and final answer with correct units. In calculation questions, method marks are awarded for correct use of equations even if the final answer is wrong due to arithmetic errors.
    • 💡When a force acts at an angle, draw a diagram and resolve the force into components. State explicitly that you are using the component of force in the direction of motion. This demonstrates understanding and secures full marks.
    • 💡For questions involving power and speed, remember that P = Fv applies when the force and velocity are in the same direction. If the force is at an angle, use P = Fv cos(theta). At maximum speed, set resultant force to zero and equate driving force to resistive force.
    Common Mistakes
    • Using degrees directly in W = Tθ without converting to radians; the correction is to convert θ to radians before substitution.
    • Ignoring frictional torque and using the applied torque as the net torque; the correction is to subtract the frictional torque to find the net torque.
    • Confusing torque with force or angular velocity with linear velocity; the correction is to use the rotational definitions T, θ and ω consistently.
    • Students often think that if an object is moving, work is being done on it. In fact, work is only done when a force causes a displacement. An object moving at constant velocity on a frictionless surface has no net work done on it.
    • Many students confuse power and energy, believing that a more powerful machine does more work. Power is the rate of doing work; a more powerful machine can do the same work in less time, or more work in the same time, but the total energy transferred depends on both power and time.
    • Students frequently forget that work is a scalar quantity and can be negative. When the force opposes the displacement (e.g., friction), the work done by that force is negative, representing energy transferred away from the object.
    Revision Plan
    1. 1Start by revising the definitions of work and power, and learn the equations W = Fs cos(theta) and P = W/t = Fv. Make flashcards with the equations and their conditions of use.
    2. 2Practice resolving forces at angles and calculating work done by each force. Use past paper questions to become familiar with the style of AQA questions, focusing on showing clear method.
    3. 3Work through problems involving the work-energy principle, including those with friction and changes in height. Always draw a diagram and label all forces and displacements.
    4. 4Study vehicle motion problems where power is constant, and understand how driving force varies with speed. Practice calculating maximum speed and acceleration at a given speed.
    5. 5Complete a timed past paper section on work and power, then review your answers against the mark scheme. Identify any recurring mistakes and revisit those concepts.
    Exam Question Types
    • 📋Calculation of work done by a force at an angle: Resolve the force into components, use W = Fs cos(theta), and ensure the correct component is used. Watch for forces that do no work (perpendicular to motion).
    • 📋Application of the work-energy principle: Calculate net work done and equate to change in kinetic energy. Often involves friction and multiple forces; draw a free-body diagram and consider the sign of each work term.
    • 📋Power and efficiency calculations: Use P = W/t or P = Fv, and efficiency = useful output / total input. Be prepared to convert between power, energy, and time, and to calculate mass flow rates in pump problems.
    • 📋Graphical analysis: Interpret force-displacement graphs (area under graph = work done) and power-time graphs (area under graph = energy transferred). Calculate gradients where appropriate.
    Command Word Expectations (AQA)
    Calculate

    You must use the given data and appropriate equations to find a numerical answer. Show all working, including the equation, substitution, and final answer with correct units and significant figures. No marks are awarded for the correct answer without working in 'show that' questions, but in 'calculate' questions, full marks can be obtained with correct working even if the final answer is slightly wrong due to rounding.

    Explain

    You must provide a reasoned justification for a physical situation, using correct terminology and linking cause and effect. For example, explaining why a car's acceleration decreases as it speeds up at constant power requires stating that P = Fv, so as v increases, F decreases, hence resultant force and acceleration decrease. Simply stating 'because power is constant' is insufficient.

    Determine

    You must use given data or a graph to find a value, often requiring reading from a graph or using a derived equation. For example, determining the work done from a force-displacement graph requires calculating the area under the graph. You should state the method and show the calculation.

    How Students Lose Marks (Examiner Pitfalls)
    Pitfall: Students often calculate work done using the full force rather than the component of force in the direction of displacement, especially when the force acts at an angle to the motion.
    ❌ Weak Answer (Loses Marks):A student calculates W = Fs using the full 50 N force when the force acts at 30 degrees to the horizontal, giving W = 50 x 10 = 500 J.
    Example improved answer:The work done is the product of the component of force in the direction of displacement and the displacement: W = Fs cos(theta) = 50 x 10 x cos(30) = 433 J (to 3 s.f.). Only the component of force parallel to the displacement transfers energy.
    Examiner Tip: Always resolve the force into components parallel and perpendicular to the displacement. Only the parallel component does work. If the force is perpendicular to motion, no work is done by that force.
    Pitfall: Confusing power with work or energy, and failing to recognise that a constant power source produces a decreasing force as speed increases, which is essential for explaining vehicle motion.
    ❌ Weak Answer (Loses Marks):A student states that a car engine of constant power 40 kW produces a constant driving force of 2000 N at all speeds, and calculates acceleration using F = ma without considering the variation of force with speed.
    Example improved answer:Power is the rate of doing work, P = Fv, so for constant power the driving force is inversely proportional to speed: F = P/v. At 20 m/s, F = 40000/20 = 2000 N; at 40 m/s, F = 1000 N. The resultant force decreases as speed increases, so acceleration decreases, explaining why a car reaches a maximum speed when driving force equals resistive force.
    Examiner Tip: Use P = Fv for problems involving constant power and variable speed. Remember that at maximum speed, resultant force is zero, so driving force equals total resistive force.
    Step-by-Step Worked Solutions

    Question: A crate of mass 25 kg is pulled 8.0 m along a rough horizontal floor by a rope inclined at 25 degrees to the horizontal. The tension in the rope is 120 N and the frictional force is 45 N. Calculate (a) the work done by the tension, (b) the work done against friction, (c) the net work done on the crate, and (d) the final speed of the crate assuming it starts from rest.

    1. 1.Step 1: Identify given facts: m = 25 kg, s = 8.0 m, T = 120 N at 25 degrees, f = 45 N, u = 0 m/s.
    2. 2.Step 2: Calculate work done by tension using W = Ts cos(theta) = 120 x 8.0 x cos(25) = 120 x 8.0 x 0.9063 = 870 J (to 2 s.f.).
    3. 3.Step 3: Calculate work done against friction: W_f = f x s = 45 x 8.0 = 360 J. This is energy transferred away from the crate.
    4. 4.Step 4: Net work done = work by tension - work against friction = 870 - 360 = 510 J.
    5. 5.Step 5: Apply work-energy principle: net work done = change in kinetic energy = 0.5mv^2 - 0. So 510 = 0.5 x 25 x v^2 = 12.5v^2.
    6. 6.Step 6: Solve for v: v^2 = 510/12.5 = 40.8, so v = 6.4 m/s (to 2 s.f.).
    Final Answer: (a) Work done by tension = 870 J; (b) Work done against friction = 360 J; (c) Net work done = 510 J; (d) Final speed = 6.4 m/s.

    Question: A pump is used to raise water from a well 15 m deep and eject it at a speed of 8.0 m/s. The pump has a power output of 2.5 kW and operates at 70% efficiency. Calculate the mass of water raised per minute.

    1. 1.Step 1: Identify given facts: depth h = 15 m, ejection speed v = 8.0 m/s, useful power output P_out = 2500 W, efficiency = 70%.
    2. 2.Step 2: Calculate energy required per kilogram of water: potential energy gain = gh = 9.81 x 15 = 147.15 J/kg; kinetic energy gain = 0.5v^2 = 0.5 x 64 = 32 J/kg. Total energy per kg = 147.15 + 32 = 179.15 J/kg.
    3. 3.Step 3: Useful power output is 2500 W, so energy transferred per minute = 2500 x 60 = 150000 J.
    4. 4.Step 4: Mass of water raised per minute = total energy / energy per kg = 150000 / 179.15 = 837 kg (to 3 s.f.).
    5. 5.Step 5: Note: efficiency is given but not needed if we use the useful power output directly. If input power were given, we would multiply by efficiency to find useful power.
    Final Answer: The mass of water raised per minute is approximately 840 kg (to 2 s.f.) or 837 kg (to 3 s.f.).
    Active Recall Memory Test
    State the equation for work done by a force at an angle to the displacement, defining each symbol.
    Key Fact: W = Fs cos(theta), where W is work done in joules (J), F is the magnitude of the force in newtons (N), s is the displacement in metres (m), and theta is the angle between the force and the displacement.
    What is the relationship between power, force, and velocity for a constant force?
    Key Fact: P = Fv, where P is power in watts (W), F is the component of force in the direction of motion in newtons (N), and v is the velocity in metres per second (m/s).
    State the work-energy principle.
    Key Fact: The net work done on an object is equal to its change in kinetic energy: W_net = delta E_k = 0.5mv^2 - 0.5mu^2.
    How is efficiency defined in terms of power?
    Key Fact: Efficiency = (useful power output / total power input) x 100%. It is a ratio of useful energy or power transferred to total energy or power supplied.
    Frequently Asked Questions
    What is the difference between work and power in physics?
    Work is the energy transferred when a force moves an object through a distance, measured in joules (J). Power is the rate at which work is done or energy is transferred, measured in watts (W), where 1 W = 1 J/s. For example, a machine that does 1000 J of work in 10 seconds has a power of 100 W, while another that does the same work in 5 seconds has a power of 200 W. Power tells you how quickly energy is transferred, not how much energy is transferred.
    How do you calculate work done when the force is at an angle?
    You use the equation W = Fs cos(theta), where F is the magnitude of the force, s is the displacement, and theta is the angle between the force and the direction of displacement. Only the component of the force in the direction of motion does work. For example, if a force of 50 N is applied at 30 degrees to the horizontal to move an object 10 m, the work done is 50 x 10 x cos(30) = 433 J. If the force is perpendicular to the motion, cos(90) = 0, so no work is done.
    What is the work-energy principle and how is it used?
    The work-energy principle states that the net work done on an object equals its change in kinetic energy: W_net = 0.5mv^2 - 0.5mu^2. To use it, calculate the work done by each force, sum them to find the net work, and set this equal to the change in kinetic energy. This allows you to find final speed, distance travelled, or net force. For example, if a 2 kg object accelerates from rest to 6 m/s, the net work done is 0.5 x 2 x 36 = 36 J. If friction does -10 J of work, the applied force must do 46 J of work.
    Why does a car's acceleration decrease as it reaches top speed?
    For a car with constant engine power, P = Fv, so the driving force F is inversely proportional to speed v. As the car speeds up, the driving force decreases. At the same time, resistive forces (such as air resistance and friction) increase with speed. The resultant force, which equals driving force minus resistive force, therefore decreases, so acceleration decreases. Eventually, the driving force equals the total resistive force, the resultant force is zero, and the car reaches its maximum speed.
    How do you calculate the efficiency of a motor or machine?
    Efficiency is calculated as the ratio of useful output energy or power to total input energy or power, expressed as a percentage: efficiency = (useful power output / total power input) x 100%. For example, if a motor is supplied with 200 W of electrical power and delivers 150 W of mechanical power, its efficiency is (150/200) x 100% = 75%. The remaining 50 W is dissipated, usually as heat due to friction and resistance. Always ensure you use the useful output, not the total output, and that both powers are in the same units.
    What are the units of work and power, and how are they related?
    Work is measured in joules (J), where 1 J = 1 N m (one newton metre). Power is measured in watts (W), where 1 W = 1 J/s (one joule per second). The relationship is P = W/t, so power is the rate of doing work. For example, a 60 W light bulb transfers 60 J of energy every second. In mechanics, power can also be expressed as P = Fv, where force is in newtons and velocity in metres per second, giving units of N m/s, which is equivalent to J/s or W.