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A-Level Energy stores and transfers, work, kinetic and potential energy, fields, power and efficiency

What the A-Level syllabus expects for Energy stores and transfers, work, kinetic and potential energy, fields, power and efficiency, and how to practise it.

What the syllabus expects

How it's examined

Questions on this topic most often ask you to find, determine, state. About 5% of the past-paper style questions in Rae's bank for this subject sit in this topic.

Worked examples

Example 1 (2 marks)

The apparatus shown is designed to launch a ball of mass 40 g straight upward. A spring with force constant 60 N m⁻¹ sits underneath a platform, and the ball rests on the platform at the spring's top. Take the spring and platform to have no significant mass, and ignore air resistance. (a) The spring is squashed down by a distance of 0.12 m. Find the force the compressed spring pushes on the ball with the moment it is let go.

Show the worked answer

At the instant of release the spring is compressed by x = 0.12 m, so by Hooke's law the force it exerts on the ball is: F = k x = 60 x 0.12 = 7.2 N.

Example 2 (2 marks)

(a) In place of a petrol engine, a prototype car uses a flywheel. The flywheel is spun up to a high speed before the journey begins. To move off, the flywheel is coupled to the drive wheels through a gear system, and the rotational energy it stores is used to drive the car forward. The flywheel has a moment of inertia of 80 kg m². Ahead of each journey it is spun up until its angular speed reaches 320 rad s⁻¹. (i) State the factors that determine the moment of inertia of the flywheel.

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The moment of inertia of a body about an axis depends on: 1. The total mass of the flywheel. 2. The distribution of that mass relative to the axis of rotation (i.e. how far the mass lies from the axis – its radius/shape and the position of the axis). Mass placed farther from the axis contributes more to the moment of inertia.

Example 3 (3 marks)

This petrol car holds 0.055 m³ of fuel (1.87 GJ) and delivers 25% of that as useful work. Given 0.086 GJ of useful energy covers each 100 km, obtain the car's driving range. [3]

Show the worked answer

Principle: efficiency tells you what fraction of the energy stored in the fuel actually ends up doing useful work. Range is then found by seeing how many times the useful energy available covers the energy needed for a fixed distance. Step 1 - find the useful energy from a full tank. useful energy = efficiency × total energy in the fuel useful energy = 0.25 × 1.87 GJ useful energy = 0.4675 GJ (the other 75 per cent, about 1.40 GJ, is wasted mostly as heat and sound) Step 2 - use the energy cost per 100 km. The car needs 0.086 GJ of useful energy for every 100 km travelled. So the number of 100 km stretches the tank can cover is number of stretches = useful energy / energy per stretch number of stretches = 0.4675 GJ / 0.086 GJ number of stretches = 5.44 (both energies are in GJ, so the units cancel and this is a pure number - no conversion to joules is needed) Step 3 - convert that into a distance by multiplying by 100 km. range = 5.44 × 100 km range = 544 km The car's range is about 544 km. (The tank volume 0.055 m³ is not needed here, because the energy it holds, 1.87 GJ, has already been given.)

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More A-Level H2 Physics topics

Physical quantities, units, measurement uncertainty and vector basics · Types of force, turning effects and conditions for equilibrium · Kinematics, uniformly accelerated motion, momentum and Newton's laws · Falling freely, the gravitational potential energy of a uniform field, and how air resistance changes the motion · Impulse and the conservation of momentum and energy · Kinematics of uniform circular motion and centripetal acceleration · all of A-Level H2 Physics