Full chapter

Energy

All 4 topics and the revision summary on one page.

Energy stores and transfers

Explain an energy change by naming the objects, the stores that change and the pathway that transfers energy.

A battery-powered motor raises a load. The battery's chemical store decreases, and the gravitational store of the load and Earth increases. Energy reaches the motor electrically, then reaches the rising load mechanically through the lifting force.

Energy is measured in joules, J. A store describes energy associated with the state of an object or system. A transfer pathway describes how energy is transferred. An energy account can include more than one object: the load and Earth form a useful system when considering gravitational energy.

A battery supplies a motor that lifts a load

This simplified steady lift ignores transfers to internal stores. The arrows label energy-transfer pathways.

The battery's chemical store decreases. An electric current transfers energy to the motor, and the lifting force transfers energy mechanically as the load rises. The final store belongs to the load-Earth system; a real motor can also increase internal stores.

Recognise the store from the physical situation

Kinetic: Associated with motion. A moving trolley has kinetic energy. If it slows down, this store decreases.
Gravitational potential: Associated with the positions of masses in a gravitational field. Lifting a load near Earth increases the gravitational store of the load-Earth system. Choose a reference height when assigning a value.
Chemical: Associated with the chemical state of substances. A battery, food or fuel can supply energy when chemical reactions change the substances.
Elastic potential: Associated with an object being stretched or compressed. Stretching a spring increases its elastic store. It can transfer energy as it returns towards its original shape.
Nuclear: Associated with the state of atomic nuclei. Changes in nuclei can release energy from nuclear stores, as in nuclear fuel undergoing fission.
Internal: Associated with the motion and interactions of the particles within a substance. A warmer metal block generally has a greater internal energy than the same block at a lower temperature in the same state.

Gravitational, chemical and elastic stores are examples of potential energy stores. One object may have several stores: a moving, warm trolley has kinetic energy as well as internal energy. Choose the stores whose changes matter to the situation.

Four ways energy is transferred

Mechanically, by a force acting over a distance. A string pulls a load upwards as it moves upwards. The force transfers energy to the load-Earth gravitational store. A force without the relevant movement does not automatically transfer energy mechanically.
Electrically, by an electric current. A battery supplies energy to a motor through a circuit. In an electric heater, an electrical transfer increases internal energy. Describe the current as the pathway, rather than inventing an extra store called "electricity".
By heating, because of a temperature difference. A hot metal block in contact with a cooler block transfers energy to it. The hotter block's internal store decreases and the cooler block's increases. Heat describes a transfer, not a substance stored inside an object.
By waves, both electromagnetic and mechanical. Sunlight is an electromagnetic wave: an absorbing surface can gain internal energy from it. Sound is a mechanical wave travelling through a medium such as air: it can make a receiving membrane vibrate. The energy travels with the disturbance; the medium does not need to travel bodily from source to receiver.

Some descriptions overlap: energy transferred by infrared waves from a hotter object to a cooler one is also a transfer by heating. Count that amount once in the energy account.

Include the surroundings

When a sliding trolley slows because of friction, its kinetic store decreases while internal stores of the trolley, track and surroundings increase. Saying that the energy is "lost" without identifying its destination leaves the account incomplete.

Energy cannot be created or destroyed. It can be transferred between stores. If no energy enters or leaves the system being considered, its total energy stays constant. If energy crosses the boundary, include that transfer when comparing the starting and final stores.

Keep stores and pathways distinct. Chemical and internal describe stores. Mechanically, electrically, heating and waves describe transfers. A complete explanation names what changes as well as how the transfer happens.

Optional check A loudspeaker makes a nearby thin membrane vibrate without touching it. Which description identifies the transfer between them?

Kinetic and gravitational energy

Kinetic energy depends on mass and speed. A change in gravitational potential energy near Earth depends on mass and vertical height change.

Use mass in kg, speed in m/s and height in m. Gravitational field strength g is measured in N/kg. Review unit conversions before substituting.

Kinetic energy: E_k = 1/2 mv²m = mass in kg; v = speed in m/s. E_k is measured in J. Square the speed, not the whole expression.

Worked example

A moving 2.0 kg object

At a speed of 4.0 m/s:

E_k = 0.5 x 2.0 x 4.0² = 0.5 x 2.0 x 16 = 16 J.

At 8.0 m/s, the same mass has E_k = 0.5 x 2.0 x 64 = 64 J. Doubling the speed makes the kinetic energy four times as large.

At fixed speed, doubling the mass doubles kinetic energy.
At fixed mass, doubling speed multiplies kinetic energy by four; halving speed makes it one quarter.
The direction of motion does not change this calculation. Use speed, and give energy in J without a spatial direction.

Use a vertical height change

Gravitational potential energy change = mghm = mass in kg; g = gravitational field strength in N/kg; h = vertical height change in m. The result is in J. Near Earth, take g as approximately constant over the height considered.

Raising a mass increases the gravitational store of the mass-Earth system; lowering it decreases the store. The numerical value E_p = mgh is measured relative to a chosen zero-height level. Changing that reference changes the assigned value, but not the energy change between the same two positions.

Worked example

Raise a load through 1.5 m

A total mass of 2.0 kg is raised vertically by 1.5 m. Use g = 10 N/kg.

Increase in E_p = 2.0 x 10 x 1.5 = 30 J.

Its weight is mg = 20 N. Lifting it to the same height along a longer ramp still increases the gravitational store by 30 J. Extra transfers due to friction would change the input required, not this gravitational change.

At fixed m and g, doubling the vertical rise doubles the gravitational store increase. At fixed rise and g, doubling the mass also doubles the increase. If the same load descends by 1.5 m, the store decreases by 30 J.

Rearrange for the quantity asked for

If the height change is unknown, h = change in E_p / (mg). For the 2.0 kg load, a 30 J increase at g = 10 N/kg gives h = 30 / (2.0 x 10) = 1.5 m.

If speed is unknown, first rearrange to v² = 2E_k / m, then take the square root. Do not treat the value of v² as the speed.

Height is not route length. Identify the starting and final vertical levels before using mgh. For kinetic energy, identify the speed at the instant being considered and square it.

Optional check A 0.50 kg load is taken up a ramp of length 2.0 m. Its final position is 0.80 m vertically above its starting position. Use g = 10 N/kg. What is its increase in gravitational potential energy?

Work and conservation

Work is a mechanical energy transfer. Conservation lets you account for that energy across all the stores and transfers that matter.

Kinetic energy is 1/2 mv²; a gravitational change is mgh. These describe stores. The work done by a force describes a transfer.

Work done = force x distance moved in the direction of the force
W = FdW = work in J; F = force in N; d = distance in m along the force. Apply this relationship to a constant force and the movement along its direction.

Work done may be written W or E, since it is energy transferred. Here W means work in joules. In W = mg, W instead means weight in newtons; after a power value, such as 60 W, W is the unit watt. Read the quantity and unit together.

Worked example

A 12 N pull through 3.0 m

A constant force of 12 N acts on a body while it moves 3.0 m along the force's direction.

Work done by the force = 12 x 3.0 = 36 J.

This tells us the mechanical transfer by that force. To say that the kinetic store increases by 36 J, we must also establish that no other transfer or store change takes part. For example, friction could transfer some of the input to internal stores.

A force can act without doing work on the object

Pushing a stationary wall: the wall does not move, so the work done on it by the push is zero.
Supporting a load during purely horizontal motion: the upward support is perpendicular to the movement. There is no distance moved in that force's direction, so its work on the load is zero.

These statements concern the specified force and object. They do not mean that a person's body transfers no energy internally while pushing or supporting.

For the 2.0 kg load lifted at constant speed through 1.5 m, take g = 10 N/kg and neglect losses. The lifting force balances its 20 N weight, so the lift does 20 x 1.5 = 30 J of work. That equals the 30 J gravitational store increase; the speed is unchanged, so kinetic energy is unchanged.

Conservation is an account of the whole change

The principle of conservation of energy states that energy cannot be created or destroyed. For an isolated system, the total energy remains constant. More generally:

Initial energy + energy transferred in
= final energy + energy transferred outChoose the objects included in the system and the same start and end points for every term. Do not count a transfer twice.

If no energy enters or leaves the system and only kinetic and gravitational stores change, their total remains constant. These are model assumptions, not a rule that air resistance or friction must be absent from every real situation.

Account for the initial 9.0 J throughout the fall

A 0.50 kg object is released from rest at height 1.8 m; g = 10 N/kg. First, take the object and Earth as the system, with negligible air resistance and no other store changes.

E_p: gravitational energy of the object and Earth

E_k: kinetic energy of the object

Released from rest: h = 1.8 m

During the fall: h = 0.80 m

At the chosen zero height: h = 0 m

The zero-height level sets the gravitational reference. The last bar is before any impact.

Separate case: resistance increases internal stores

Start with the same 9.0 J. Now include the object, Earth and surrounding air in the account. At zero height, internal stores have increased by 2.0 J.

2.0 J is the increase in internal energy of the object and surroundings.

A 0.50 kg body is released from rest at a height of 1.8 m, with g = 10 N/kg. The first three bars show a fall with negligible air resistance: kinetic and gravitational stores total 9.0 J. The separate resistance case accounts for 7.0 J kinetic and a 2.0 J internal increase. All heights use the same chosen zero level.

Worked example

Find the speed during a fall

A 0.50 kg body is released from rest 1.8 m above the chosen zero-height level. Use g = 10 N/kg. Neglect air resistance and all other store changes, and consider the body just before it reaches that level, before any impact.

Initial account: E_k = 0 because it starts from rest. E_p = 0.50 x 10 x 1.8 = 9.0 J.
At zero height: E_p = 0, so conservation gives E_k = 9.0 J.
Use the kinetic-energy equation: 9.0 = 0.5 x 0.50 x v², so v² = 36 and v = 6.0 m/s.

At a height of 0.80 m, the remaining gravitational energy is 0.50 x 10 x 0.80 = 4.0 J. The kinetic energy is 9.0 - 4.0 = 5.0 J. Then v² = (2 x 5.0) / 0.50 = 20, giving v = about 4.5 m/s.

Include a transfer to internal stores

Now keep the same initial conditions, but suppose 2.0 J increases internal stores of the body and surrounding air by the time the body reaches zero height. Its final kinetic energy is 9.0 - 2.0 = 7.0 J.

The speed follows from v² = (2 x 7.0) / 0.50 = 28, giving about 5.3 m/s. It is smaller than 6.0 m/s because less energy remains in the kinetic store.

The final account is 7.0 J kinetic + 2.0 J internal increase = 9.0 J. The total has not decreased. Expanding the system to include the surroundings explains the destination of energy that has left the body's mechanical stores.

State the conditions before equating two stores. Starting from rest, negligible air resistance and an unchanged set of other stores justify E_p lost = E_k gained here. If a condition changes, change the account.

Optional check A 0.50 kg body is released from rest 1.8 m above a chosen zero-height level. Use g = 10 N/kg. By the time it reaches that level, 2.0 J has increased internal stores. No other energy changes occur. What is its final kinetic energy?

Power and measurements

Power is the rate of energy transfer. The same energy transfer can happen quickly or slowly.

Energy and work are measured in J. Time is measured in s. A power calculation needs both the amount transferred and the time taken for that transfer.

Average power = energy transferred / time taken
P = E / tP = average power in watts, W; E = energy transferred in J; t = elapsed time in s. 1 W = 1 J/s.

Same energy transfer, different times

Each motor transfers 240 J. The energy bars use one common scale; the time bars use a separate common scale.

Motor A: 240 J in 8.0 s

Average power = 240 J / 8.0 s = 30 W.

Motor B: 240 J in 4.0 s

Average power = 240 J / 4.0 s = 60 W.

Transferring the same energy in half the time gives twice the average power.

Both motors transfer 240 J. Motor A takes 8.0 s, giving 30 W; motor B takes 4.0 s, giving 60 W. The transferred energy is equal, while the time and average power differ.

Worked comparison

The same transfer in half the time

Motor A: P = 240 / 8.0 = 30 W.
Motor B: P = 240 / 4.0 = 60 W.

B transfers the energy twice as quickly. It does not transfer twice as much energy in this comparison: both amounts are 240 J.

Rearrange to E = Pt when power and time are given, or t = E / P when the transfer amount and power are given. For example, transferring 240 J at a constant 60 W takes 240 / 60 = 4.0 s. Convert minutes to seconds and kilowatts to watts before combining with joules.

If the rate changes during the interval, total energy divided by total elapsed time gives the average power. It does not show how the rate varied at each instant.

Determine useful lifting power from measurements

A motor steadily raises a load. Its intended output is the increase in the load-Earth gravitational store. Measure that store change and the time over the same part of the lift.

Measure total lifted mass: include the load and its hanger. Record mass in kg; for example, 500 g = 0.500 kg.
Mark two vertical levels: use a fixed point on the load to identify its start and end positions. Measure their vertical separation h, not the length of a sloping string.
Time the marked interval: start as the chosen point passes the lower mark and stop at the upper mark. Use the steady part of the lift so the load's kinetic energy does not change over the interval.
Calculate the output: use the supplied g to find mgh, then divide by the corresponding time.

Example measurements for a steady lift
Quantity	Value
Total lifted mass / kg	0.500
Vertical rise / m	0.80
Time for that rise / s	2.0
Supplied g / N/kg	10

The intended energy increase is mgh = 0.500 x 10 x 0.80 = 4.0 J. Average useful lifting power = 4.0 / 2.0 = 2.0 W.

This method determines the rate at which the gravitational store increases. It does not determine the motor's total electrical input power: energy can also increase internal stores. The input requires a separate measurement.

Link improvements to a specific measurement

Repeat timings for the same mass, rise and operating conditions to estimate random variation. A longer suitable steady interval can make reaction time a smaller fraction of the measured time.
Keep the ruler vertical and use the same point on the load at both marks. Overestimating h makes the calculated energy and output power too large.
Keep the timing interval matched to those height marks. Timing extra motion outside the measured rise makes t too large and the calculated power too small.
Including the hanger matters: omitting part of the lifted mass makes the calculated gravitational change too small.

Repeated readings do not repair an incorrect height reference or a systematically mismatched timing interval. Correct the method causing that error.

Keep the unit attached to the claim. J describes an energy amount. W describes an amount per second. A high power alone does not establish a large total transfer.

Optional check A motor steadily lifts a total mass of 2.0 kg through a vertical height of 1.5 m in 6.0 s. Use g = 10 N/kg. What can these measurements determine?

Revision summary

Choose the quantity before calculating

Kinetic energy: E_k = 1/2 mv². Use m in kg and speed v in m/s; the result is J. Doubling speed gives four times the energy at fixed mass.
Gravitational energy change: Change = mgh. h is vertical height change, not path length. Use an approximately constant g in N/kg near Earth and a stated zero level for E_p.
Mechanical transfer: Work W (also written E) = Fd, where d is the distance moved in the constant force's direction. N x m gives J. Distinguish work W in J from weight W in N and the unit watt W. No displacement in that direction means no work by that force on that object.
Rate of transfer: Average P = E / t; E = Pt; t = E / P. Use E in J and t in s for P in W. 1 W = 1 J/s.

Stores and transfer pathways

Stores include kinetic, gravitational potential, chemical, elastic potential, nuclear and internal. Transfers occur mechanically, electrically, by heating or by waves. Include both electromagnetic examples such as sunlight and mechanical examples such as sound.

Name the objects and the stores that change. Gravitational energy concerns the mass-Earth interaction; internal energy concerns particles. Heat, sound and electricity should not be added as extra stores in this description.

Close the energy account

Energy cannot be created or destroyed. Initial energy + transfers in = final energy + transfers out. With no transfer across the boundary, the total for the system stays constant.

Equate a gravitational decrease with a kinetic increase only when the other changes are negligible. If friction or air resistance increases internal stores, include that amount. A body released from rest starts with zero kinetic energy; a moving starting body does not.

Power from a steady lift

Measure total lifted mass, vertical rise and elapsed time for the same steady interval. mgh/t gives average useful gravitational output power. It does not measure total electrical input power. Match height and timing marks, include the hanger and distinguish repeated random variation from an incorrect reference.

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