Ch 2 · 5 Work, Energy and Power · A Level Physics 9702
Topic 2 of 3 in Forces, Energy and Materials — A Level Physics 9702 — 64 lessons.
Introduction
Energy is the most useful idea in physics because it is always conserved. Work is how forces transfer energy, power is how fast they do it, and efficiency tells you how much is useful. (Syllabus topic 5: sections 5.1–5.2.)
Key ideas — 5.1 (a) Work
Work done by a force is
🧑🏫 Teacher's explanation
Imagine you are pushing a heavy shopping trolley. If you push it forward and it moves forward, you are doing work. If you push it forward but it stays stuck and doesn't move, you might be sweating, but in physics terms, you have done zero work because there is no displacement.
📖 Definitions
🔍 Worked example — Easy
A constant force of 50 N is applied to a crate, causing it to move 4 m in the direction of the force. Calculate the work done.
🔍 Worked example — Standard
A person pulls a sled with a force of 120 N at an angle of 30^° to the horizontal. If the sled moves 10 m horizontally, calculate the work done by the person.
🔍 Worked example — Exam-style
A car engine provides a constant driving force of 2500 N. The car travels at a constant velocity of 20 m/s. Calculate the work done by the engine in 15 s.
⚠️ Common errors
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🎯 Exam question
Key ideas — 5.1 (b) Conservation of energy
Principle of conservation of energy: energy cannot be created or destroyed; it can only be transferred from one form to another. The total energy of a closed system is constant.
🧑🏫 Teacher's explanation
Imagine you are playing a game of billiards. When the white cue ball hits a red ball, the cue ball slows down and the red ball speeds up. The energy didn't vanish; it just moved from one ball to the other. This is the core of conservation of energy: energy is like money in a bank account—you can move it from savings to checking, or spend it on different things, but the total amount remains the same unless you add or remove funds from the system.
📖 Definitions
🔍 Worked example — Easy
A 2.0 kg mass is dropped from a height of 5.0 m above the ground. Calculate its kinetic energy just before it hits the ground, assuming no air resistance.
🔍 Worked example — Exam-style
A pendulum of mass 0.50 kg is released from a height of 0.20 m above its lowest point. Calculate the velocity of the mass at the lowest point, ignoring air resistance.
⚠️ Common errors
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🎯 Exam question
Key ideas — 5.1 (c) Efficiency
efficiency = useful energy output/total energy input × 100\%
🧑🏫 Teacher's explanation
Imagine you are using a mobile phone to play a game. You charge the battery with 100 J of electrical energy. However, the screen and processor only use 70 J to actually run the game; the remaining 30 J is lost as heat, making the phone warm in your hand. Efficiency is simply a measure of how much of that 'input' actually does the job you intended it to do.
📖 Definitions
🔍 Worked example — Easy
An electric motor is used to lift a 5.0 kg mass through a height of 2.0 m. If the motor consumes 150 J of electrical energy to perform this task, calculate the efficiency of the motor. (Use g = 9.81 m/s^2)
🔍 Worked example — Standard
A car engine produces 45 kW of useful power to move the vehicle, but the total power input from fuel is 120 kW. Calculate the efficiency of the engine.
🔍 Worked example — Exam-style
An electric hoist lifts a crate of mass 200 kg to a height of 10 m in 5.0 s. The total electrical energy supplied to the hoist is 2500 J. Calculate the efficiency of the hoist. (Use g = 9.81 m/s^2)
⚠️ Common errors
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🎯 Exam question
Key ideas — 5.1 (d) Power
Power is the work done per unit time (rate of energy transfer):
🧑🏫 Teacher's explanation
Think of power as the 'speed' of energy transfer. If you carry a heavy box up a flight of stairs, you do the same amount of work whether you walk slowly or run quickly. However, the person who runs will be more 'powerful' because they transfer that energy in less time. Power tells us how fast a process is happening, not just how much total energy is used.
📖 Definitions
🔍 Worked example — Easy
A crane lifts a load of mass 500 kg through a vertical height of 20 m in 25 s. Calculate the average power output of the crane. (Assume g = 9.81 m/s^2)
🔍 Worked example — Standard
A car travels at a constant velocity of 30 m/s. The total resistive force acting on the car (air resistance and friction) is 850 N. Calculate the power required by the engine to maintain this constant velocity.
⚠️ Common errors
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🎯 Exam question
Key ideas — 5.2 (a) Gravitational potential energy
Derivation. Lifting a mass m at constant speed through a height Δ h in a uniform gravitational field needs a force equal to its weight, F = mg. The work done is
🧑🏫 Teacher's explanation
Imagine you are lifting a heavy box up a flight of stairs. You are exerting a force upwards to counteract the weight of the box, and you are moving that box a certain vertical distance. The energy you 'spend' to do this isn't lost; it is simply being stored in the box due to its new position in the Earth's gravitational field. This 'stored' energy is what we call Gravitational Potential Energy (GPE).
📖 Definitions
🔍 Worked example — Easy
A 2.5 kg mass is lifted vertically through a height of 1.2 m at a constant speed. Calculate the work done by the lifting force. (Use g = 9.81 m/s^2)
🔍 Worked example — Standard
An object of mass 0.45 kg is raised from the ground to a height of 5.0 m. Calculate the change in its gravitational potential energy. (Use g = 9.81 m/s^2)
⚠️ Common errors
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🎯 Exam question
Key ideas — 5.2 (b) Kinetic energy
Derivation. A resultant force F accelerates a mass m from rest through displacement s. Work done W = Fs = mas. From v^2 = u^2 + 2as with u = 0: as = tfrac12 v^2. So
🧑🏫 Teacher's explanation
Kinetic energy is the energy an object possesses due to its motion. If an object is moving, it has the capacity to do work on something else when it hits it or moves against a force. Think of a bowling ball rolling down a lane: a heavy, fast-moving ball has much more 'oomph' to knock down pins than a light, slow-moving ball. This 'oomph' is the kinetic energy.
📖 Definitions
🔍 Worked example — Standard
A car of mass 1200 kg increases its speed from 10 m s^-1 to 20 m s^-1. Calculate the increase in its kinetic energy.
🔍 Worked example — Exam-style
A constant resultant force of 50 N acts on a 2.0 kg mass, accelerating it from rest over a distance of 10 m. Calculate the final velocity of the mass.
⚠️ Common errors
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🎯 Exam question
Worked example
Example 1: work at an angle. A child pulls a sledge 50 m with a force of 30 N along a rope at 25° to the horizontal. Work done = 30 × 50 × cos 25° = 1360 J.
🔍 Worked example — Standard
A 0.20 kg ball is dropped from a height of 5.0 m. Calculate its speed just before it hits the ground, ignoring air resistance.
🔍 Worked example — Exam-style
A car travels at a constant velocity of 25 m s^-1 against a total resistive force of 800 N. Calculate the useful power output of the engine.
⚠️ Common errors
Common mistakes
Practice
Answers: (1) P = W/t = FΔ s/Δ t = Fv. (2) W = 600 × 9.81 × 15 = 8.83 × 10^4 J; P = 8.83 × 10^4 / 20 = 4.4 kW. (3) tfrac12 × 1200 × (20^2 - 10^2) = 1.8 × 10^5 J. (4) F = P/v = 250/10 = 25 N. (5) 0.65 × 2000 × 60 = 7.8 × 10^4 J.