Ch 5 · 4.5 Energy, Work and Power · A Level Mathematics 9709
Topic 5 of 5 in Mechanics — A Level Mathematics 9709 — 68 lessons.
Introduction
Energy methods are a shortcut: a child sliding down a curved water slide has a path far too complicated for suvat, but you can find the speed at the bottom from heights alone. Power tells you how quickly work is done, which sets the top speed of a car or a cyclist. (Syllabus section 4.5, Paper 4.)
Key ideas — Work done by a force
A constant force F whose point of application moves a distance d in a direction at angle θ to the force does work
🧑🏫 Teacher's explanation
Imagine you are pulling a heavy suitcase on wheels across a flat floor. If you pull the handle at an angle, some of your effort is trying to lift the suitcase up, while some is actually pulling it forward. Only the part of your pull that is horizontal—the part aligned with the floor—is actually doing the 'work' of moving the suitcase from A to B.
📖 Definitions
🔍 Worked example — Easy
A constant force of 50 N is applied to a crate, acting at an angle of 60^° to the horizontal. Calculate the work done when the crate moves 4 m horizontally.
🔍 Worked example — Standard
A car engine provides a constant driving force of 2000 N. The car moves at a constant velocity of 25 m/s. Calculate the power output of the engine.
🔍 Worked example — Exam-style
A block is pulled up a smooth plane inclined at 30^° to the horizontal by a force of 100 N acting parallel to the plane. The block moves 5 m up the plane. Calculate the work done by the force and the change in gravitational potential energy (assume g = 9.8 m/s^2 and mass m = 10 kg).
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — Kinetic and potential energy
KE = tfrac12mv^2 GPE = mgh
🧑🏫 Teacher's explanation
Imagine you are pushing a heavy shopping trolley. If you push it hard, it gains speed; that speed represents 'stored' energy of motion. If you lift that trolley up a flight of stairs, you are fighting gravity to store energy based on how high you lift it. In mechanics, we call these Kinetic Energy (KE) and Gravitational Potential Energy (GPE).\n\nThink of energy like a bank account. Kinetic energy is like 'cash in hand'—it is immediately available to do something (like moving). Gravitational potential energy is like 'money in a savings account'—it is stored due to your position. When you move from a high shelf to the floor, you 'spend' your GPE, and that energy is converted into KE (speed) as you fall. The total amount of energy stays the same unless an external force (like friction) 'taxes' it by turning it into heat.
📖 Definitions
🔍 Worked example — Easy
A ball of mass 0.5 kg is moving at 4 m s^-1. Calculate its kinetic energy.
🔍 Worked example — Standard
A crate of mass 20 kg is lifted vertically through a height of 3 m. Calculate the increase in its gravitational potential energy. (Assume g = 9.8 m s^-2)
⚠️ Common errors
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🎯 Exam question
Key ideas — The work–energy principle
The change in the total mechanical energy of a body equals the work done on it by forces other than its weight:
🧑🏫 Teacher's explanation
Imagine you are pushing a heavy shopping trolley up a ramp. You are doing work to move it, which increases its speed (Kinetic Energy) and its height (Gravitational Potential Energy). If the floor is slippery (no friction), all the effort you put in goes directly into that speed and height. If the floor is carpeted (resistance), some of your effort is 'wasted' overcoming friction, meaning less goes into speed and height.
📖 Definitions
🔍 Worked example — Easy
A block of mass 2 kg is pushed up a smooth (frictionless) ramp from a height of 0.5 m to a height of 1.5 m. Calculate the increase in its kinetic energy if its velocity increases from 1 m/s to 3 m/s.
🔍 Worked example — Standard
A car of mass 1200 kg accelerates from rest up a hill. A constant driving force of 5000 N acts on the car. The car moves 100 m up the hill, gaining 20 m in vertical height. Calculate the work done against resistance if the car's final velocity is 15 m/s.
🔍 Worked example — Exam-style
A particle of mass 0.5 kg slides down a smooth curved track from a height of 2 m to a height of 0.5 m. Calculate its velocity at the bottom of the track. (Assume g = 9.81 m/s^2)
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — Power
Power is the rate of doing work, measured in watts (1 W = 1 J s⁻¹):
🧑🏫 Teacher's explanation
Imagine you are climbing a staircase. If you run up the stairs, you reach the top faster than if you walk. In both cases, you have done the same amount of 'work' against gravity to lift your body mass, but because you did it in less time, your 'power' output was much higher. Power is essentially how 'fast' you can perform work.
📖 Definitions
🔍 Worked example — Easy
A constant force of 50 N moves an object through a distance of 10 m in 5 s. Calculate the average power exerted.
🔍 Worked example — Standard
A car engine provides a constant power of 24 kW to move a car. At a constant speed of 20 m/s, calculate the driving force acting on the car.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — Cars, hills and maximum speed
At constant power, combine F = P/v with Newton's second law:
🧑🏫 Teacher's explanation
Imagine you are riding a bicycle up a hill. You are pedaling with a constant effort (this is your power). At first, you accelerate, but as you go faster, the air resistance increases and the hill feels harder to fight. Eventually, you reach a point where you are pedaling as hard as you can, but you aren't going any faster—you've hit your 'top speed' for that specific effort. This happens because the force you provide is exactly balanced by the forces pushing you back.
📖 Definitions
🔍 Worked example — Easy
A car engine provides a constant power of 50 kW to move a car up a hill. At a certain point, the car is moving at 20 m/s and the total resistance (including gravity) is 2500 N. Calculate the acceleration of the car.
🔍 Worked example — Standard
A car of mass 1200 kg moves up a hill with an angle of inclination α = 5^°. The engine provides a constant power of 30 kW. If the total resistance is 1500 N, find the velocity of the car when it is accelerating at 0.5 m/s^2. (Use g = 9.8 m/s^2)
🔍 Worked example — Exam-style
A car of mass m is climbing a hill of angle α with constant power P. The resistance is R. Show that the maximum speed v reached by the car is given by v = P/R + mgsinα.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Worked examples — Example 1: work at an angle
A sledge is pulled 8 m along level ground by a rope with tension 50 N at 30^° to the horizontal. Work done = 50 × 8 × cos 30^° = 346 J.
🔍 Worked example — Easy
A crate is pulled 10 m across a smooth floor by a rope exerting a force of 40 N at an angle of 60^° to the horizontal. Calculate the work done.
🔍 Worked example — Standard
A car engine provides a constant driving force of 2000 N acting at an angle of 5^° below the horizontal. If the car travels 500 m on a level road, calculate the work done by the engine.
⚠️ Common errors
Worked examples — Example 2: a smooth curved slide
A child starts from rest at the top of a smooth curved slide 3 m high. GPE lost = KE gained: mg(3) = tfrac12mv^2 ⇒ v = √(60) = 7.75 m s⁻¹ (the mass cancels).
🔍 Worked example — Easy
A child of mass 40 kg slides down a smooth curved slide from a height of 3 m. Calculate the speed of the child at the bottom.
🔍 Worked example — Standard
A block of mass 2 kg is released from rest at the top of a smooth curved track of height h. If the block reaches a speed of 5 m s^-1 at the bottom, find the value of h. (Use g = 9.81 m s^-2)
⚠️ Common errors
Worked examples — Example 3: work against friction
A 4 kg particle slides from rest 10 m down a rough plane inclined at 25^° and reaches 6 m s⁻¹. Find the friction force.
🔍 Worked example — Easy
A 5 kg block slides down a smooth (frictionless) incline of height 3 m. Calculate its speed at the bottom.
🔍 Worked example — Exam-style
A car of mass 1200 kg climbs a hill of height 20 m over a distance of 250 m along the slope. The engine provides a constant force of 5000 N in the direction of motion. Calculate the work done against friction if the car's speed remains constant.
⚠️ Common errors
Worked examples — Example 4: a car on a hill
A car of mass 1200 kg has an engine working at 30 kW and resistance 800 N.
🔍 Worked example — Easy
A car of mass 1200 kg has an engine providing 30 kW of power. If the resistance is 800 N, calculate the maximum speed on level ground.
🔍 Worked example — Standard
A car of mass 1200 kg travels up a hill with an angle α where sinα = 1/20. The engine provides a driving force of 2000 N and resistance is 800 N. Calculate the acceleration at a speed of 15 m s^-1.
🔍 Worked example — Exam-style
A car of mass 1200 kg has an engine with 30 kW power. It is driving up a hill where sinα = 1/20 against a resistance of 800 N. Calculate its maximum speed up the hill.
⚠️ Common errors
Common mistakes
Practice
Answers: (1) 200 J; DeltaKE = 200 - 125 = 75 J. (2) tfrac12(0.2)(144) = 0.2(10)h ⇒ h = 7.2 m. (3) 80(10)(15) - tfrac12(80)(144 - 16) = 12000 - 5120 = 6880 J. (4) 300 × 10 × 12/60 = 600 W. (5) P = 600 × 40 = 24 kW; F = 24000/25 = 960 N, a = 960 - 600/900 = 0.4 m s⁻².