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Energy

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Energy is conserved: it is transferred between stores rather than created or destroyed. Useful models include kinetic, gravitational potential, elastic, chemical, magnetic, electrostatic, nuclear and thermal stores. Mechanical work transfers energy when a force moves an object.

For a moving object, \(E_{k}\) = \(\dfrac{1}{2}mv^{2}\). Near Earth's surface, a change in gravitational potential energy is Δ\(E_{p}\) = mgΔh. Power is the rate of energy transfer, P = \(\dfrac{E}{t}\). Efficiency compares useful output with total input: efficiency = useful output / total input. In real processes some energy is dissipated, often to thermal stores of the surroundings, making it less available for the intended purpose.

Energy stores and transfer pathways

A useful starting point is to separate an energy store from a transfer pathway. Stores describe where energy is held: kinetic, thermal, chemical, gravitational potential, elastic, magnetic, electrostatic and nuclear. Pathways describe how energy moves between stores: mechanically when a force does work, electrically when charge moves through a potential difference, by heating because of a temperature difference, or by radiation such as light and infrared. In an exam answer, avoid saying that an object 'contains movement energy' or that energy is 'used up'. Name the initial store, the transfer pathway and the final store.

Energy is conserved in a closed system. This means the total amount of energy does not disappear, even when a device is inefficient. Inefficiency means that some of the transferred energy ends up in less useful, more widely spread thermal stores in the surroundings. Sankey diagrams represent this idea visually: the total width entering equals the total width leaving, and the useful and wasted branches together equal the input.

Work, power and common calculations

Work done is an energy transfer caused by a force moving an object through a distance. When force and motion are in the same direction, W = Fs. Lifting an object increases its gravitational potential energy, Δ\(E_{p}\) = mgΔh. Changing speed changes kinetic energy, \(E_{k}\) = \(\dfrac{1}{2}mv^{2}\). Stretching an elastic object in its linear region stores elastic energy, \(E_{e}\) = \(\dfrac{1}{2}kx^{2}\).

Power describes how quickly energy is transferred: P = \(\dfrac{E}{t}\), or equivalently P = \(\dfrac{W}{t}\). A 1000 W device transfers 1000 J each second. Always check prefixes: 1 kW = 1000 W and 1 kWh is a unit of energy, not power. For domestic-energy questions, energy transferred can be calculated using E = Pt, with units chosen consistently.

Efficiency and exam reasoning

Efficiency = useful output / total input. It may be written as a decimal or multiplied by 100 for a percentage. A machine cannot have an efficiency above 1 or above 100%. If a question gives power rather than energy, the same ratio can be used with useful and total power because both are energy-per-second quantities.

For explanation questions, distinguish conservation from usefulness. Friction may reduce the useful kinetic-energy transfer, but the 'lost' energy has usually been transferred to thermal stores of the object and surroundings. Good answers identify the mechanism causing dissipation and describe why the energy becomes less useful for the intended purpose.

Choosing an equation

Before substituting numbers, ask what is changing. Height change suggests gravitational potential energy; speed change suggests kinetic energy; force through distance suggests work; energy per time suggests power. Write the equation first, convert units, substitute, then give the final unit. A quick estimate is valuable: doubling speed quadruples kinetic energy because speed is squared, while doubling height only doubles gravitational potential-energy change.

Worked example

Worked example 1: A 1.5 kW kettle transfers 135 kJ of energy. Convert 1.5 kW to 1500 W and 135 kJ to 135000 J. From P = \(\dfrac{E}{t}\), t = \(\dfrac{135000}{1500}\) = 90 s.

Worked example 2: A 950 kg car travelling at 18 m/s has kinetic energy \(E_{k}\) = \(\dfrac{1}{2}\) × 950 × 18² = 153900 J. If an average braking force of 3800 N removes this energy, Fd = 153900 gives d = 40.5 m.

Worked example 3: A 600 N load is lifted 2.0 m in 4.0 s. Work done = 600 × 2.0 = 1200 J and power = \(\dfrac{1200}{4.0}\) = 300 W.

Misconception check

Energy is not 'used up'. When a device is inefficient, energy has still been conserved; a larger fraction has been transferred to unwanted stores and dispersed into the surroundings.

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State the equation linking power, energy transferred and time.

P = \(\dfrac{E}{t}\), with power in watts when energy is in joules and time is in seconds.

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