Full chapter
Energy
All 4 topics and the revision summary on one page.
01
Energy stores and transfers
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A ball is thrown straight upwards. At the top of its flight, which energy store has gained energy compared with the moment it left the hand?
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The gravitational potential store, while the kinetic store has fallen.
As the ball rises, its kinetic store empties into the gravitational potential store of the ball and Earth. At the top it is momentarily at rest, so nearly all the energy is in the gravitational store.
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.
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.
Check your understanding
A loudspeaker makes a nearby thin membrane vibrate without touching it. Which description identifies the transfer between them?
Show the answer
Energy is transferred by sound, a mechanical wave travelling through the air.
The sound wave transfers energy through the air and can make the membrane move. Sound describes a transfer by waves, rather than an extra store inside the membrane.
02
Kinetic and gravitational energy
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A car speeds up from 10 m/s to 20 m/s. What happens to its kinetic energy?
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It becomes four times as large, because it depends on speed squared.
Kinetic energy is 1/2 m v2, so doubling the speed squares to a factor of four. The 20 m/s car has four times the kinetic 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.
Worked example
A moving 2.0 kg object
At a speed of 4.0 m/s:
Ek = 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 Ek = 0.5 x 2.0 x 64 = 64 J. Doubling the speed makes the kinetic energy four times as large.
Try it step by step
Work it out in steps: Speed in m/s, then Speed squared in m2/s2, then Kinetic energy in J.
Show the answer
Problem 1.
Speed in m/s: v = 54 x 1000 / 3600 = 15 m/s
Speed squared in m2/s2: v2 = 15 x 15 = 225 m2/s2
Kinetic energy in J: Ek = 1/2 m v2 = 0.5 x 800 x 225 = 90000 J, which is 90 kJ (2 sf)
Problem 2.
Speed in m/s: v = 72 x 1000 / 3600 = 20 m/s
Speed squared in m2/s2: v2 = 20 x 20 = 400 m2/s2
Kinetic energy in J: Ek = 0.5 x 1200 x 400 = 240000 J, which is 240 kJ (2 sf)
Problem 3.
Speed in m/s: v = 36 x 1000 / 3600 = 10 m/s
Speed squared in m2/s2: v2 = 10 x 10 = 100 m2/s2
Kinetic energy in J: Ek = 0.5 x 60 x 100 = 3000 J (2 sf)
- 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
Raising a mass increases the gravitational store of the mass-Earth system; lowering it decreases the store. The numerical value Ep = 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 Ep = 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 Ep / (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 = 2Ek / m, then take the square root. Do not treat the value of v2 as the speed.
Try it step by step
Work it out in steps: Weight mg in N, then Height h = Ep / (mg) in m.
Show the answer
Problem 1.
Weight mg in N: mg = 3.0 x 10 = 30 N
Height h = Ep / (mg) in m: h = 450 / 30 = 15 m (2 sf)
Problem 2.
Weight mg in N: mg = 8.0 x 10 = 80 N
Height h = Ep / (mg) in m: h = 960 / 80 = 12 m (2 sf)
Problem 3.
Weight mg in N: mg = 5.0 x 10 = 50 N
Height h = Ep / (mg) in m: h = 275 / 50 = 5.5 m (2 sf)
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.
Check your understanding
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?
Show the answer
4.0 J
Use the vertical rise, 0.80 m: increase = mgh = 0.50 x 10 x 0.80 = 4.0 J. The route taken does not change this store increase.
03
Work and conservation
Make a guess. You are not marked.
Two balls of equal mass are released from rest at the same height. One falls straight down. The other slides down a smooth curved track to the same level. Air resistance and friction are negligible. Which ball is moving faster when it reaches the lower level?
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Both reach the same speed, because they drop the same height.
Each ball loses the same gravitational potential energy, mgh, and with no friction all of it becomes kinetic energy. Same mass, same kinetic energy, same speed.
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.
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:
= 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.
Ep: gravitational energy of the object and Earth
Ek: 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.
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: Ek = 0 because it starts from rest. Ep = 0.50 x 10 x 1.8 = 9.0 J.
- At zero height: Ep = 0, so conservation gives Ek = 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.
Try it step by step
Work it out in steps: Gravitational potential energy lost in J, then Kinetic energy gained in J, then Speed in m/s.
Show the answer
Problem 1.
Gravitational potential energy lost in J: Ep lost = mgh = 0.20 x 10 x 3.2 = 6.4 J
Kinetic energy gained in J: No other store changes, so Ek gained = 6.4 J
Speed in m/s: 6.4 = 0.5 x 0.20 x v2, so v2 = 64 and v = 8.0 m/s (2 sf)
Problem 2.
Gravitational potential energy lost in J: Ep lost = 2.0 x 10 x 1.25 = 25 J
Kinetic energy gained in J: Smooth slope, so Ek gained = 25 J
Speed in m/s: 25 = 0.5 x 2.0 x v2, so v2 = 25 and v = 5.0 m/s (2 sf)
Problem 3.
Gravitational potential energy lost in J: Ep lost = 0.50 x 10 x 0.45 = 2.25 J
Kinetic energy gained in J: Ek gained = 2.25 J
Speed in m/s: 2.25 = 0.5 x 0.50 x v2, so v2 = 9.0 and v = 3.0 m/s (2 sf)
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 Ep lost = Ek gained here. If a condition changes, change the account.
Check your understanding
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?
Show the answer
7.0 J
Initial energy = 0.50 x 10 x 1.8 = 9.0 J. The final account is 9.0 = Ek + 2.0, so Ek = 7.0 J. Total energy is conserved.
04
Power and measurements
Make a guess. You are not marked.
Two motors each lift the same 20 kg load through the same 2.0 m. Motor A takes 4 s. Motor B takes 8 s. Which motor transfers more energy to the gravitational store?
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Both transfer the same energy, because the load and height match.
The gain in gravitational potential energy is mgh, which is the same load through the same height. Motor A has twice the power, not twice the energy.
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.
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.
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.
Try it step by step
Work it out in steps: Energy transferred mgh in J, then Power P = E / t in W.
Show the answer
Problem 1.
Energy transferred mgh in J: E = mgh = 40 x 10 x 2.5 = 1000 J
Power P = E / t in W: P = 1000 / 5.0 = 200 W (2 sf)
Problem 2.
Energy transferred mgh in J: E = 60 x 10 x 3.0 = 1800 J
Power P = E / t in W: P = 1800 / 12 = 150 W (2 sf)
Problem 3.
Energy transferred mgh in J: E = 25 x 10 x 4.0 = 1000 J
Power P = E / t in W: P = 1000 / 8.0 = 125 W (3 sf)
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.
| 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.
Check your understanding
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?
Show the answer
Its useful lifting output power is 5.0 W.
The intended gravitational store increase is mgh = 30 J. Dividing by 6.0 s gives 5.0 J/s = 5.0 W of useful lifting output. These measurements do not measure the motor's electrical input.
Revision summary
Choose the quantity before calculating
- Kinetic energy
- Ek = 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 Ep.
- 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.
Back to energy stores and transfersPractise exam questions
Write your answer first, then tick the mark points it makes.
Exam-style question
A 0.50 kg ball is dropped from a height of 3.2 m. Air resistance is negligible and g = 10 N/kg. Describe the energy transfer as it falls, and calculate its speed just before it hits the ground. [3]
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Exam-style question
State what is meant by the work done by a force. A boy pushes a box 3.0 m along the floor with a horizontal force of 40 N. Calculate the work done by the boy. [3]
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Exam-style question
An electric motor lifts a 50 kg load at a steady speed through a height of 6.0 m in 12 s. Take g = 10 N/kg. Calculate the useful power output of the motor. [3]
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Exam-style question
Data-based question: A trolley released on a ramp [9]
A trolley of mass 0.50 kg is released from rest at different heights h on a ramp. A light gate at the bottom of the ramp measures its speed v. Take g = 10 N/kg.
| h / m | v / (m/s) |
|---|---|
| 0.20 | 1.80 |
| 0.40 | 2.55 |
| 0.60 | 3.12 |
| 0.80 | 3.60 |
- (a) Calculate the gravitational potential energy lost by the trolley when it is released from 0.80 m. [1]
- (b) Calculate its kinetic energy at the bottom of the ramp for this release. [2]
- (c) Calculate the energy transferred to other stores, and name the main store it goes to. [2]
- (d) A student predicts the speed at the bottom using mgh = (1/2)mv2. State one assumption in this prediction. [1]
- (e) The student says: "The ramp wastes the same amount of energy whatever height the trolley starts from." Use the data to evaluate this claim. [3]
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Quick recall
A loudspeaker makes a nearby thin membrane vibrate without touching it. Which description identifies the transfer between them?