Work, Energy and Power

Work done: transferring energy with a force

Work is done whenever a force moves its point of application, and doing work is one of the ways of transferring energy from one store to another. For a constant force, the work done is the product of the force and the displacement of the object in the direction of the force: W = F s. Work is a form of energy transfer, so it is measured in the same unit as energy, the joule (J).

One joule is the work done when a force of 1 newton moves its point of application 1 metre in the direction of the force, so 1 J = 1 N m. A force of 25 N that pulls a box 4.0 m along a floor does 25 x 4.0 = 100 J of work on the box. Two things follow from the equation. If the object does not move, s = 0 and no work is done, however large the force and however tired your arms become. And if the force is at right angles to the motion, it transfers no energy either, because there is no displacement along its line of action.

When the force is not along the direction of motion you must use its component in that direction: W = F s cos theta, where theta is the angle between the force and the displacement. A girl pulling a sledge with a rope at 60 degrees to the horizontal does work with the horizontal component of the tension only; the upward component has no displacement along its own line and transfers nothing.

• W = F s, where the displacement s is measured along the line of the force.
• If the force is at an angle theta to the motion, W = F s cos theta - only the component along the motion does work.
• 1 J = 1 N m; the joule is the unit of both work and energy.
• If the object does not move, or the force is at right angles to the motion, no work is done.

Deriving delta E p = m g delta h

Close to the Earth's surface the gravitational field is uniform: the field strength g = 9.81 m s-2 has the same magnitude and direction everywhere over the sort of heights you meet in the laboratory. The weight of an object of mass m is therefore m g at every height, and to lift the object at a steady speed you must pull upwards with a force equal to its weight, F = m g, because the resultant force on it is zero.

Lifting the object through a change in height delta h moves the point of application of that force along its own line, so the work done is W = F s = m g delta h. The lifting force is external to the object and the object ends up higher than it started, so this work is transferred to its gravitational potential store: delta E p = m g delta h.

Only the change in height matters, not the path taken. Carrying the object up a slope, or up in a lift, or along a winding staircase transfers exactly the same energy to the gravitational store for the same vertical rise, and moving it sideways at a constant height transfers none at all. For a 2.5 kg mass lifted through 1.8 m, delta E p = 2.5 x 9.81 x 1.8 = 44 J.

Deriving E k = 1/2 m v^2

Start with an object of mass m that is at rest and is accelerated by a constant resultant force F through a distance s, reaching a speed v. The equations of motion link speed, acceleration and distance without mentioning time: v^2 = u^2 + 2 a s, and with u = 0 this becomes v^2 = 2 a s.

Rearranging gives the acceleration in terms of things you can measure: a = v^2 / (2 s). Newton's second law then gives the force that produced it, F = m a, so F = m v^2 / (2 s). The work done by that force is W = F s = (m v^2 / 2 s) x s = 1/2 m v^2. All of this work goes into the kinetic store of the object, which is why E k = 1/2 m v^2.

The equation is proportional to the square of the speed, and that is the single most examinable feature of it. A 0.50 kg ball moving at 12 m s-1 has E k = 0.5 x 0.50 x 144 = 36 J; the same ball at 24 m s-1 has 144 J, four times as much. The derivation works for any object, because the resultant force is the force that changes the speed, whatever the object's shape or composition.

The principle of conservation of energy

The principle of conservation of energy states that energy cannot be created or destroyed; it can only be transferred from one store to another. The total energy of a closed system stays constant, which is why any energy account you write always balances: whatever leaves one store appears in another. Name the stores precisely in an answer - kinetic, gravitational potential, elastic potential, thermal, chemical, nuclear, electrostatic and magnetic are the ones you will use most.

Free fall is the cleanest example. A stone of mass m dropped from rest through a height h loses m g h from its gravitational potential store and gains 1/2 m v^2 in its kinetic store, so m g h = 1/2 m v^2 and v = sqrt(2 g h). The mass cancels, so every object dropped from the same height arrives at the same speed. For h = 5.0 m, v = sqrt(2 x 9.81 x 5.0) = 9.9 m s-1, whether the object is a coin or a cricket ball.

In real experiments some of the energy always ends up in the thermal store of the surroundings through air resistance or friction, and that energy is spread out so thinly that it is no longer useful. The energy is still there. The loss from one store equals the gain in the others plus the energy transferred to the surroundings, so an equation with only two stores is wrong as soon as a question mentions air resistance or friction.

Power: the rate of transferring energy

Power is the work done per unit time, P = W / t. Its unit is the watt (W), and one watt is one joule per second (1 W = 1 J s-1). Power tells you how fast energy is transferred, not how much energy is transferred. A machine that does 2000 J of work in 4.0 s has a power of 500 W; a machine that does the same 2000 J in 40 s has a power of only 50 W, even though both machines transfer the same energy.

P = F v follows from the two equations you already have. For an object moving at a constant speed v under a constant driving force F, the work done in a time t is W = F s and the distance travelled is s = v t. Substituting both into P = W / t gives P = F s / t = F v t / t = F v, so P = F v. Read it together with P = W / t: the first is for a moving object and a force, the second for a machine and a time interval.

P = F v explains why a car cannot keep accelerating on a flat road. The engine delivers at most a fixed maximum power, so as the speed rises the maximum force it can supply falls, while the resistive forces from air resistance and friction grow. When the two meet, the resultant force is zero and the car sits at its top speed. The same equation is why the engine needs a low gear to pull away: a large force is only available at a low speed. A crane lifting 250 kg through 12 m in 8.0 s does W = m g h = 250 x 9.81 x 12 = 2.9 x 10^4 J of useful work, so its useful output power is 2.9 x 10^4 / 8.0 = 3.7 x 10^3 W.

• P = W / t, measured in watts, where 1 W = 1 J s-1.
• P = F v for a constant force driving an object along at speed v.
• Average power uses the total time taken; the same equations give the instantaneous power if you use the force and speed at that instant.
• A power rating says how fast a machine can transfer energy, not how much energy it can transfer in total.

Efficiency and where the energy goes

The efficiency of a system is the ratio of the useful energy output from the system to the total energy input: efficiency = useful energy output / total energy input. The same ratio can be written with power, efficiency = useful power output / total power input, because both powers are measured over the same time interval. Efficiency has no unit: it is a fraction, and it is usually turned into a percentage by multiplying by 100.

Energy is always conserved, so a machine never really loses energy - it transfers part of the input to stores that are not useful for the job. Almost always that means the thermal store of the surroundings, heated by friction in the bearings, air resistance, or resistance in wires and windings. A car engine transfers only about a quarter of the chemical energy of its fuel to the kinetic store of the car; most of the rest leaves in the thermal store of the exhaust gases and the coolant.

A worked example. An electric motor is supplied with 2400 J of electrical energy and does 2040 J of useful work. Efficiency = 2040 / 2400 = 0.85, which is 85%, and the remaining 360 J is transferred to the thermal store of the motor and the air. Improvements come from attacking those losses directly: lubricating the bearings, streamlining a body shape, or using a more efficient lamp. Efficiency can never be greater than 1, because that would mean more energy leaving the system than entered it.

Choosing the right equation and checking the answer

These five relationships cover every question on this topic, and choosing between them is most of the skill. Work and both energy equations give an answer in joules; power always needs a time or a speed; efficiency is the only one that comes out as a bare number or a percentage.

Check the units before you substitute anything. Lengths in centimetres must become metres, masses in grams must become kilograms, and times in minutes must become seconds before they go into any of these equations. Carry the calculation through without rounding in the middle, then quote the final answer to two or three significant figures with its unit - an otherwise perfect answer with no unit loses a mark.

Two traps catch most candidates. The first is using the whole force when only its component along the motion does work. The second is treating energy as something that can be destroyed: whenever a question mentions friction or air resistance, the missing energy is in the thermal store of the surroundings, not gone.