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7.1 Work done

7.1.1 Conservation of energy in a closed system

Systems and boundaries

Definition

System

The object or group of objects being studied, separated from its surroundings by a boundary.

Definition

Closed system

A system in which no energy is transferred into or out of the system, so its total energy stays constant.

  1. In physics, a system is the object or group of objects you choose to study, such as a falling ball and the Earth, or a kettle and the water inside it.
  2. Everything outside the system is called the surroundings, and the imaginary line that separates the two is the boundary of the system.
  3. Energy can cross the boundary in either direction, for example when a mug of hot tea transfers energy to the cooler air in the room.
  4. A closed system is one where no energy crosses the boundary, so no energy enters from the surroundings and none escapes to them.
  5. Very few single objects form a closed system on their own, because almost every object exchanges some energy with the air, the ground or whatever is supporting it.
  6. You can still make a system closed by drawing the boundary wide enough to include every object that the energy reaches, such as a ball, the Earth and the air around them.
  7. The whole Universe can be treated as the largest closed system, because there is nothing outside it for energy to be transferred to.

Conservation of energy

Definition

Law of conservation of energy

The law stating that energy cannot be created or destroyed, only transferred from one store to another.

  1. The law of conservation of energy states that energy cannot be created or destroyed.
  2. Energy can only be transferred, either from one energy store to another or from one place to another.
  3. For example, a moving object holds energy in its kinetic store, and an object raised above the ground holds energy in its gravitational potential energy store.
  4. Whenever a system changes, the energy in at least one store goes down while the energy in one or more other stores goes up.
  5. In a closed system the total decrease is always exactly equal to the total increase, because no energy has left the system and no new energy has appeared.
  6. This means there is no net change in the total energy of a closed system, however many transfers happen inside it.
  7. A net change is the overall change once every increase and every decrease has been added together, so a rise of 30 J30\ \text{J}30 J in one store and a fall of 30 J30\ \text{J}30 J in another give a net change of zero.
  8. Individual stores can change by large amounts, so conservation of energy never means that each separate store stays the same.
  9. It is the sum of all the stores in the closed system that stays constant from the start of a process to the end.
Example

A swinging pendulum

  • A pendulum bob is pulled to one side and released, and the system is the bob, the string, the Earth and the surrounding air.
  • At the highest point of the swing the bob is momentarily still, so its kinetic store is empty and its gravitational potential energy store is at its largest.
  • As the bob swings down, its gravitational potential energy store decreases and its kinetic store increases by almost the same amount.
  • At the lowest point the bob moves fastest, so its kinetic store is at its largest and its gravitational potential energy store is at its smallest.
  • As the bob rises on the other side, energy moves back from the kinetic store to the gravitational potential energy store.
  • Air resistance and friction at the pivot transfer a small amount of energy to the thermal stores of the air and the pivot on every swing, so each swing is slightly lower than the one before.
  • The air and the pivot are inside the boundary, so the total energy of the system stays exactly the same from the first swing to the last.
Example

Keeping track of the totals

  • A toy car rolls down a slope, and the system is the car, the slope, the Earth and the air.
  • The gravitational potential energy store of the car decreases by 5.0 J5.0\ \text{J}5.0 J.
  • The kinetic store of the car increases by 3.5 J3.5\ \text{J}3.5 J.
  • Friction and air resistance increase the thermal stores of the car, the slope and the air by 1.2 J1.2\ \text{J}1.2 J.
  • The rattling sound of the wheels carries 0.3 J0.3\ \text{J}0.3 J into the air, where it also ends up in a thermal store.
3.5+1.2+0.3=5.0 J 3.5 + 1.2 + 0.3 = 5.0\ \text{J} 3.5+1.2+0.3=5.0 J
  • The increases add up to exactly the decrease, so the net change in the total energy of the closed system is zero.

Choosing the boundary

  1. Whether a system counts as closed depends entirely on where you draw its boundary.
  2. If the boundary surrounds only a ball rolling across a carpet, the total energy of the ball falls as it slows down.
  3. This happens because friction and air resistance transfer energy out of the ball to the thermal stores of the carpet and the air.
  4. The ball on its own is therefore not a closed system, even though no energy has been destroyed.
  5. If the boundary is widened to include the carpet and the air, every joule that leaves the ball stays inside the system.
  6. The wider system is closed, so its total energy stays constant while the energy moves from the kinetic store of the ball to the thermal stores of its surroundings.
  7. In the end, much of the energy is spread thinly through the thermal stores of the surroundings, but every joule of it still exists.
  8. Always state the system you are describing, because the claim that total energy stays constant is only true for a closed system.
Common Mistake
  • Do not say that energy is used up, lost or destroyed, because energy is only ever transferred between stores.
  • Do not write heat energy or movement energy, because the correct names are the thermal store and the kinetic store.
  • Do not claim that every store stays the same in a closed system, because only the total of all the stores stays constant.
  • Do not apply conservation of energy to a single object that is clearly transferring energy to its surroundings, unless you have widened the boundary to include those surroundings.
Exam technique

Describing conservation of energy

  • Start by naming the objects in the system, such as the ball, the Earth and the air, so it is clear which boundary you are using.
  • Name the store that decreases and every store that increases, using the full store names.
  • Link the changes to the law: the decrease in one store equals the total increase in the others, so there is no net change in the total energy.
  • Use precise phrases such as no net change or the total energy stays the same, because vague phrases such as the energy balances out often fail to earn credit.
Self review
  • What is meant by a system in physics?
  • State what makes a system a closed system.
  • State the law of conservation of energy.
  • Explain what is meant by no net change in the total energy of a closed system.
  • A ball rolling across a carpet slows down. Explain why the ball on its own is not a closed system.
  • Explain why it is wrong to say that energy is used up.

7.1.2 Energy stores and transfers

Stores of motion and position

Definition

Energy store

A way in which energy is held by an object or system, such as a kinetic, gravitational, elastic, thermal or chemical store.

Definition

Kinetic energy

The energy held in the kinetic store of an object because it is moving.

Definition

Gravitational potential energy

The energy held in the gravitational store of an object because of its height in a gravitational field.

Definition

Elastic potential energy

The energy held in the elastic store of an object that has been stretched, compressed or bent elastically.

  1. An energy store describes where energy is held, and the amount of energy in each store is measured in joules, J\text{J}J.
  2. Describing a change to a system means saying which stores decrease and which stores increase, from the start of the process to the end.
  3. The kinetic store holds energy because an object is moving, so it increases when the object speeds up and decreases when it slows down.
  4. A heavier object holds more energy in its kinetic store than a lighter one moving at the same speed, so a loaded lorry at 20 m/s20\ \text{m/s}20 m/s holds far more than a cyclist at 20 m/s20\ \text{m/s}20 m/s.
  5. The gravitational potential energy store holds energy because of an object's height in a gravitational field, so it increases when the object is raised and decreases when it falls.
  6. This store belongs to the object and the Earth together, because it depends on the gravitational attraction between them.
  7. The elastic potential energy store holds energy when an object is stretched, squashed or bent and can return to its original shape, such as a stretched bungee cord or a compressed spring.
  8. These three stores change whenever an object speeds up, slows down, rises, falls, stretches or springs back.

Stores of particles and fields

Definition

Thermal energy store

The energy store of an object that increases when its temperature rises.

Definition

Chemical energy store

The energy store held in fuels, foods and batteries that is transferred when chemical reactions take place.

Definition

Nuclear energy store

The energy store held in the nuclei of atoms that is transferred when nuclei split apart or join together.

Definition

Magnetic energy store

The energy store that changes when magnets or magnetic materials that attract or repel each other move closer together or further apart.

Definition

Electrostatic energy store

The energy store that changes when electric charges that attract or repel each other move closer together or further apart.

  1. The thermal store of an object increases when its temperature rises, because its particles vibrate or move faster.
  2. A large object at a low temperature can hold more energy in its thermal store than a small object at a high temperature, so a swimming pool at 25 ∘C25\ ^\circ\text{C}25 ∘C holds far more than a mug of tea at 80 ∘C80\ ^\circ\text{C}80 ∘C.
  3. The chemical store is found in fuels such as petrol and natural gas, in food and in batteries, and it decreases when a chemical reaction such as burning or respiration takes place.
  4. The nuclear store is held in the nuclei of atoms and decreases when nuclei split apart in a nuclear power station or join together in the Sun.
  5. The magnetic store changes when magnets or magnetic materials that attract or repel each other move closer together or further apart, as when two fridge magnets snap together.
  6. The electrostatic store changes when electric charges that attract or repel each other move closer together or further apart, as when a charged balloon is pulled towards a wall.
  7. Electricity, light and sound are not stores, because they are ways of transferring energy from one store to another.

Projecting an object upwards

  1. When you throw a ball straight up, your muscles transfer energy from the chemical store in your body to the kinetic store of the ball.
  2. As the ball rises it slows down, so its kinetic store decreases while its gravitational potential energy store increases.
  3. At the top of its flight the ball is momentarily still, so its kinetic store is empty and its gravitational potential energy store is at its largest.
  4. As the ball falls back down, its gravitational potential energy store decreases and its kinetic store increases, so it speeds up.
  5. Air resistance acts on the ball throughout its flight and increases the thermal stores of the ball and the air, so the ball returns to your hand slightly slower than it left.
  6. An object pushed up a slope behaves in the same way, with its kinetic store decreasing and its gravitational potential energy store increasing as it climbs.
  7. Friction between the object and the slope also increases the thermal stores of both surfaces, so a trolley pushed up a ramp stops lower down than it would on a perfectly smooth ramp.

A ball held high and then released, showing its gravitational potential energy store decreasing and its kinetic store increasing as it falls back towards the ground.

Example

A child on a playground slide

  • At the top of the slide the child is still, so the child's kinetic store is empty and the gravitational potential energy store is at its largest.
  • As the child slides down, the gravitational potential energy store decreases.
  • Most of that energy goes into the child's kinetic store, so the child speeds up.
  • Friction between the child's clothes and the slide increases the thermal stores of the child and the slide, which both become slightly warm.
  • On the flat run-out at the bottom, friction slows the child down, so the kinetic store decreases to zero.
  • By the end, all the energy that left the gravitational potential energy store is in the thermal stores of the child, the slide and the surrounding air.

Hitting an obstacle

  1. A moving object that hits an obstacle, such as a car hitting a crash barrier or a hammer hitting a nail, loses most or all of its kinetic store in a fraction of a second.
  2. During the impact the object and the obstacle are squashed or bent, and the energy is transferred to their thermal stores and carried away to the surroundings as sound.
  3. If the change of shape is elastic, as when a tennis ball hits a wall, some energy is held briefly in the elastic potential energy store and then returned to the kinetic store as the ball bounces away.
  4. If the change of shape is plastic, as in the crumple zone of a car, the shape change is permanent and the energy ends up in thermal stores rather than being returned.
  5. When a hammer drives a nail into wood, the kinetic store of the hammer decreases and the thermal stores of the nail, the hammer head and the wood increase, which is why the nail feels warm afterwards.
  6. The sound produced by any impact also ends up in the thermal store of the surrounding air once the sound waves have died away.

Accelerating with a constant force

  1. When a constant resultant force acts on an object in its direction of motion, the object speeds up, so its kinetic store increases steadily.
  2. The energy always comes from the store that supplies the force, such as the chemical store of a car's petrol, a cyclist's muscles or the battery in an electric scooter.
  3. An electric car pulling away from traffic lights transfers energy from the chemical store of its battery to its kinetic store.
  4. Not all of that energy reaches the kinetic store, because the motor, the tyres and the air all warm up, so their thermal stores increase too.
  5. A rocket at launch shows several changes at once: the chemical store of its fuel decreases while its kinetic store and gravitational potential energy store both increase.

Slowing a vehicle down

  1. When a driver brakes, the brake pads press against discs fixed to the wheels, and friction between them acts against the motion.
  2. The kinetic store of the car decreases as it slows down.
  3. The thermal stores of the brake pads and discs increase, and the discs can glow red after repeated hard stops from high speed.
  4. Energy then spreads from the hot brakes to the thermal store of the surrounding air, and squealing brakes or tyres also carry some energy away as sound.
  5. Air resistance and friction in the moving parts act on a vehicle all the time, so a car that coasts with its engine off also slows down and its kinetic store decreases.
  6. A cyclist who stops pedalling on a flat road shows the same change more slowly, with the kinetic store going to thermal stores in the air, the tyres and the bearings.

Boiling water in a kettle

  1. An electric kettle is plugged into the mains, so energy is transferred electrically from the mains supply to the kettle's heating element.
  2. The thermal store of the element increases, and because the element is hotter than the water, energy is transferred by heating to the water around it.
  3. The thermal store of the water increases and its temperature rises steadily from room temperature towards 100 ∘C100\ ^\circ\text{C}100 ∘C.
  4. Once the water reaches 100 ∘C100\ ^\circ\text{C}100 ∘C, its temperature stops rising and the energy supplied turns some of the liquid water into steam.
  5. The plastic body of the kettle, its lid and the air in the kitchen also warm up, so their thermal stores increase as well.
  6. Steam escaping from the spout carries energy into the room, and the rumbling sound of the boiling water carries a little more away.
  7. In every one of these situations, the decreases and increases balance, so the total energy is unchanged once all the stores are included.
Common Mistake
  • Do not write heat energy, movement energy or height energy, because these are not the names of energy stores.
  • Do not describe an electrical energy store, because electricity is a way of transferring energy, not a store.
  • Do not stop at the useful store, because a full description also includes the thermal stores of the surroundings that increase.
  • Do not say that a stationary object at the top of its flight has no energy, because its gravitational potential energy store is at its largest there.
Exam technique

Describing every change

  • Work through the event in time order, describing what happens to each store at the start, during the process and at the end.
  • Pair every store with the word increases or decreases, such as the gravitational potential energy store of the ball decreases.
  • Give the reason for each change, such as the kinetic store increases because the ball speeds up.
  • Finish by naming the thermal stores of the surroundings that increase because of friction, air resistance or sound, because an answer that leaves them out describes only part of the change.
Self review
  • Name the energy store that increases when an object speeds up.
  • Name two places where energy is held in a chemical store.
  • Describe the changes in energy stores when a ball is thrown upwards and caught again.
  • Describe the changes in energy stores when a car brakes to a stop.
  • Describe the changes in energy stores when a moving hammer hits a nail.
  • Describe the changes in energy stores when water is brought to the boil in an electric kettle.

7.1.3 Energy changes by heating, forces and current

Three ways to change a system

  1. The energy stored in a system changes only when energy is transferred into it, out of it or between its parts.
  2. There are three main ways this happens: by heating, by work done by forces and by work done when a current flows.
  3. Each of these is a transfer pathway, which describes how energy moves rather than where it is held.
  4. A pathway always links a store that decreases to one or more stores that increase, so naming the pathway completes the description of a change.

Changes caused by heating

Definition

Heating

The transfer of energy from a hotter object to a cooler one because of the temperature difference between them.

Definition

Internal energy

The total energy stored by the particles of a system in their kinetic and potential energy stores.

Definition

Specific heat capacity

The energy needed to raise the temperature of 1 kg1\ \text{kg}1 kg of a substance by 1 ∘C1\ ^\circ\text{C}1 ∘C.

  1. Heating transfers energy from a hotter object to a cooler one whenever there is a temperature difference between them.
  2. Energy keeps flowing from hot to cold until both objects reach the same temperature, so a radiator warms a room and an ice pack cools a sprained ankle.
  3. Energy transferred to an object by heating increases its internal energy, which is the total energy held by its particles.
  4. If the object does not change state, its particles move or vibrate faster, so its temperature rises and its thermal store increases.
  5. The size of the temperature rise depends on three things: the energy transferred, the mass of the object and the material it is made from.
  6. For the same energy, a larger mass gives a smaller temperature rise, because the energy is shared among more particles.
  7. The specific heat capacity of a material is the energy needed to raise the temperature of 1 kg1\ \text{kg}1 kg of it by 1 ∘C1\ ^\circ\text{C}1 ∘C, measured in J/kg ∘C\text{J/kg}\ ^\circ\text{C}J/kg ∘C.
  8. A material with a high specific heat capacity, such as water at about 4200 J/kg ∘C4200\ \text{J/kg}\ ^\circ\text{C}4200 J/kg ∘C, needs a lot of energy for each degree of temperature rise, so it warms up and cools down slowly.
  9. A material with a low specific heat capacity, such as copper at about 385 J/kg ∘C385\ \text{J/kg}\ ^\circ\text{C}385 J/kg ∘C, needs far less energy for each degree, so the same energy gives it a much larger temperature rise.
  10. This is why a metal saucepan on a hob is hot to the touch within seconds, while the water inside takes several minutes to boil.
  11. It is also why water is used in central heating systems and hot water bottles, because each kilogram carries a large amount of energy for a small temperature change.
  12. Heating does not always raise the temperature, because at a melting point or boiling point the energy transferred changes the state of the substance instead.
  13. Objects can also lose internal energy by heating, so a cup of tea left on a table cools as energy is transferred from it to the cooler air and table.
Example

Heating water and aluminium

  • An electric heater transfers 9000 J9000\ \text{J}9000 J to 1 kg1\ \text{kg}1 kg of water, and an identical heater transfers 9000 J9000\ \text{J}9000 J to 1 kg1\ \text{kg}1 kg of aluminium.
  • Each kilogram of water needs about 4200 J4200\ \text{J}4200 J for every 1 ∘C1\ ^\circ\text{C}1 ∘C rise, while each kilogram of aluminium needs about 900 J900\ \text{J}900 J.
rise for water=90004200≈2.1 ∘C \text{rise for water} = \frac{9000}{4200} \approx 2.1\ ^\circ\text{C} rise for water=42009000​≈2.1 ∘C rise for aluminium=9000900=10 ∘C \text{rise for aluminium} = \frac{9000}{900} = 10\ ^\circ\text{C} rise for aluminium=9009000​=10 ∘C
  • The aluminium becomes almost five times hotter, because its specific heat capacity is almost five times smaller.
  • Both thermal stores increased by the same 9000 J9000\ \text{J}9000 J, so the temperature rise tells you how the energy is shared, not how much energy was transferred.

Changes caused by forces

Definition

Work done

The energy transferred when a force moves an object through a distance in the direction of the force.

  1. A force does work when it moves an object through a distance along the line of action of the force.
  2. The work done by the force equals the energy it transfers, so 1 J1\ \text{J}1 J of work done means 1 J1\ \text{J}1 J of energy transferred.
  3. Energy transferred by a force in this way is said to be transferred mechanically.
  4. The store that increases depends on what the force does to the object.
  5. A force that makes an object speed up increases its kinetic store, as when a footballer kicks a ball.
  6. A force that lifts an object increases its gravitational potential energy store, as when a crane raises a steel beam.
  7. A force that stretches, squashes or bends an elastic object increases its elastic potential energy store, as when an archer draws back a bowstring.
  8. A force such as friction or air resistance that acts against the motion decreases the kinetic store and increases the thermal stores of the surfaces or the air.
  9. Rubbing your hands together on a cold day is a simple example, because the work done against friction makes your palms warm.
  10. The energy for the work always comes from another store, such as the chemical store of your muscles, of a fuel or of a battery.
  11. If a force acts but the object does not move, as when you push hard against a brick wall, no work is done on the wall and no energy is transferred to it.

Changes caused by a current

Definition

Electrical work

The energy transferred when charge flows through a component because of the potential difference across it.

  1. An electric current is a flow of charge, and in a metal wire the charge is carried by electrons.
  2. A cell, a battery or the mains supply provides a potential difference that drives the charge around the circuit.
  3. As the charge flows through a component it transfers energy to that component, which is called work done by a current or energy transferred electrically.
  4. In a battery-powered circuit the energy comes from the chemical store of the battery, which decreases while the current flows.
  5. In a resistor or a heating element, the moving electrons collide with the vibrating ions of the metal, so the thermal store of the component increases.
  6. In a motor, the current produces a turning effect, so energy is transferred mechanically to the kinetic store of the motor and whatever it drives.
  7. In a filament lamp, the filament becomes hot enough to glow, so its thermal store increases and energy is carried to the surroundings by light and infrared radiation.
  8. The more charge that flows and the larger the potential difference driving it, the more energy is transferred to the component.
  9. In every electrical device some energy goes into thermal stores that you did not want, which is why phone chargers, laptops and hairdryer motors feel warm while they run.
Example

A motor lifting a load

  • A battery-powered motor winds up a string and lifts a small mass from the floor to a bench.
  • The chemical store of the battery decreases as the current flows.
  • Energy is transferred electrically from the battery to the motor.
  • The motor exerts a force on the string, so energy is transferred mechanically to the mass.
  • The gravitational potential energy store of the mass increases as it rises.
  • The thermal stores of the motor, the wires and the surrounding air also increase, because the current heats the motor coils and friction acts in its bearings.
Common Mistake
  • Do not call heating a store, because heating is a pathway and it is the thermal store that increases.
  • Do not confuse temperature with energy, because temperature in ∘C^\circ\text{C}∘C tells you how hot an object is, while energy is measured in joules.
  • Do not write that energy is stored in a wire or in a current, because a current only transfers energy from one store to another.
  • Do not describe a force as a type of energy, because a force transfers energy only when it moves an object through a distance.
Exam technique

Naming the pathway

  • Name the store that decreases, then the pathway, then the store that increases, so the chain of transfer is complete.
  • Use the exact pathway words: by heating, mechanically for work done by a force, and electrically for work done by a current.
  • When a question mentions specific heat capacity, compare materials by the energy needed per kilogram for each degree, rather than saying one material holds heat better.
  • For a device with more than one output, describe the unwanted increase in thermal stores as well as the useful change.
Self review
  • State what is meant by heating.
  • Define specific heat capacity.
  • Explain why a metal block warms up more than the same mass of water when both receive the same energy.
  • State when a force does work on an object.
  • Describe the energy changes when a battery drives a current through a resistor.

7.1.4 Quantifying energy transfers in a system

Energy on a common scale

Definition

Joule

The unit of energy, equal to the work done when a force of one newton moves an object one metre along the line of action of the force.

  1. Every energy store and every energy transfer is measured in the same unit, the joule, J\text{J}J.
  2. This common scale lets you compare transfers by completely different pathways, such as the work done by a crane and the energy needed to boil a kettle.
  3. Larger amounts are written in kilojoules, where 1 kJ=1000 J1\ \text{kJ} = 1000\ \text{J}1 kJ=1000 J, and megajoules, where 1 MJ=1 000 000 J1\ \text{MJ} = 1\,000\,000\ \text{J}1 MJ=1000000 J.
  4. Calculating each change in joules shows the redistribution of energy in a system, which means how much energy went to each store.
  5. Because energy is conserved, the energy put into a system always equals the total of all the energy that ends up in its other stores.
  6. This can be written as total input energy=useful output energy+energy transferred to other stores\text{total input energy} = \text{useful output energy} + \text{energy transferred to other stores}total input energy=useful output energy+energy transferred to other stores.
  7. In a Sankey diagram like this one, the width of each arrow is proportional to the energy it carries, so the arrows leaving always add up to the arrow entering.

A Sankey diagram where the input arrow splits into a useful output arrow and an arrow of energy transferred to other stores, with the two output widths adding up to the input width.

Mechanical work

Definition

Work done

The energy transferred when a force moves an object through a distance in the direction of the force.

  1. Energy transferred by a force is calculated using W=F sW = F\,sW=Fs, which you must learn.
  2. In this equation WWW is the work done in joules, J\text{J}J, FFF is the force in newtons, N\text{N}N, and sss is the distance moved along the line of action of the force in metres, m\text{m}m.
  3. Rearranging gives F=WsF = \dfrac{W}{s}F=sW​ to find a force, and s=WFs = \dfrac{W}{F}s=FW​ to find a distance.
  4. One newton-metre, N m\text{N m}N m, of work is exactly the same amount of energy as one joule, so 1 N m=1 J1\ \text{N m} = 1\ \text{J}1 N m=1 J.
  5. The distance must be measured along the direction of the force, so when an object is lifted the distance is the vertical height gained, not the length of a slope.
  6. When a force lifts an object at a steady speed, the force equals the object's weight, so the work done equals the weight multiplied by the height gained.
  7. When friction brings a moving object to rest, the work done against friction equals the energy transferred from its kinetic store to thermal stores.
Example

Work done by a braking force

  • A cyclist brakes, and the brakes provide a constant friction force of 150 N150\ \text{N}150 N while the bicycle travels 12 m12\ \text{m}12 m.
W=F s W = F\,s W=Fs W=150×12 W = 150 \times 12 W=150×12 W=1800 J W = 1800\ \text{J} W=1800 J
  • The kinetic store of the cyclist and bicycle decreases by 1800 J1800\ \text{J}1800 J, and the thermal stores of the brakes and the surroundings increase by the same amount.

Electrical work

Definition

Kilowatt-hour

The energy transferred by a device with a power of one kilowatt when it runs for one hour.

  1. The energy transferred when charge flows through a component is calculated using E=Q VE = Q\,VE=QV, which you must learn.
  2. Here EEE is the energy transferred in joules, J\text{J}J, QQQ is the charge that flows in coulombs, C\text{C}C, and VVV is the potential difference in volts, V\text{V}V.
  3. The energy transferred by an appliance in a given time is calculated using E=P tE = P\,tE=Pt, which you must also learn.
  4. Here PPP is the power in watts, W\text{W}W, and ttt is the time in seconds, s\text{s}s, so a 60 W60\ \text{W}60 W lamp transfers 60 J60\ \text{J}60 J every second.
  5. Convert minutes and hours into seconds before using E=P tE = P\,tE=Pt to find an answer in joules, using 1 min=60 s1\ \text{min} = 60\ \text{s}1 min=60 s and 1 h=3600 s1\ \text{h} = 3600\ \text{s}1 h=3600 s.
  6. Homes use a much larger unit of energy, the kilowatt-hour, kW h\text{kW h}kW h, because household appliances transfer millions of joules every day.
  7. To find the energy in kilowatt-hours, use the same equation E=P tE = P\,tE=Pt with the power in kilowatts and the time in hours.
  8. One kilowatt-hour equals 1000 W×3600 s=3 600 000 J1000\ \text{W} \times 3600\ \text{s} = 3\,600\,000\ \text{J}1000 W×3600 s=3600000 J, which is 3.6 MJ3.6\ \text{MJ}3.6 MJ.
  9. An electricity meter records the energy supplied to a home in kilowatt-hours, and each kilowatt-hour is called a unit on the electricity bill.
  10. The cost of running an appliance is the energy it uses in kilowatt-hours multiplied by the price of one unit.
Example

Energy used by a fan heater

  • A 2.5 kW2.5\ \text{kW}2.5 kW fan heater runs for 3.03.03.0 hours, and electricity costs 28p28\text{p}28p per unit.
E=P t E = P\,t E=Pt E=2.5×3.0=7.5 kW h E = 2.5 \times 3.0 = 7.5\ \text{kW h} E=2.5×3.0=7.5 kW h cost=7.5×28=210p=£2.10 \text{cost} = 7.5 \times 28 = 210\text{p} = \pounds 2.10 cost=7.5×28=210p=£2.10
  • In joules, the same energy is 7.5×3 600 000=2.7×107 J7.5 \times 3\,600\,000 = 2.7 \times 10^{7}\ \text{J}7.5×3600000=2.7×107 J.
Example

Energy transferred by a flow of charge

  • A charge of 40 C40\ \text{C}40 C flows through a lamp connected to a 12 V12\ \text{V}12 V supply.
E=Q V E = Q\,V E=QV E=40×12=480 J E = 40 \times 12 = 480\ \text{J} E=40×12=480 J
  • The supply transfers 480 J480\ \text{J}480 J to the lamp, which increases the thermal store of the filament before light and infrared radiation carry the energy to the surroundings.

Thermal processes

Definition

Specific heat capacity

The energy needed to raise the temperature of 1 kg1\ \text{kg}1 kg of a substance by 1 ∘C1\ ^\circ\text{C}1 ∘C.

Definition

Specific latent heat

The energy needed to change the state of 1 kg1\ \text{kg}1 kg of a substance without any change in its temperature.

  1. The energy needed to change the temperature of a material without changing its state is calculated using ΔE=m c Δθ\Delta E = m\,c\,\Delta\thetaΔE=mcΔθ.
  2. Here ΔE\Delta EΔE is the change in thermal energy in joules, mmm is the mass in kilograms, ccc is the specific heat capacity in J/kg ∘C\text{J/kg}\ ^\circ\text{C}J/kg ∘C and Δθ\Delta\thetaΔθ is the temperature change in ∘C^\circ\text{C}∘C.
  3. This is a select and apply equation, so you do not need to memorise it, but you must recognise when to use it.
  4. The temperature change is always the final temperature minus the starting temperature, never a single thermometer reading.
  5. The energy needed to change the state of a material without changing its temperature is calculated using E=m LE = m\,LE=mL, which is also a select and apply equation.
  6. Here LLL is the specific latent heat in J/kg\text{J/kg}J/kg, the energy needed to change the state of 1 kg1\ \text{kg}1 kg of the substance.
  7. When an electric heater warms a substance, you can set the energy supplied, E=P tE = P\,tE=Pt, equal to m c Δθm\,c\,\Delta\thetamcΔθ if you assume that no energy escapes to the surroundings.
Example

Heating water in a kettle

  • A 2000 W2000\ \text{W}2000 W kettle heats 1.5 kg1.5\ \text{kg}1.5 kg of water from 20 ∘C20\ ^\circ\text{C}20 ∘C to 100 ∘C100\ ^\circ\text{C}100 ∘C, and the specific heat capacity of water is 4200 J/kg ∘C4200\ \text{J/kg}\ ^\circ\text{C}4200 J/kg ∘C.
  • The temperature change is 100−20=80 ∘C100 - 20 = 80\ ^\circ\text{C}100−20=80 ∘C.
ΔE=m c Δθ \Delta E = m\,c\,\Delta\theta ΔE=mcΔθ ΔE=1.5×4200×80=504 000 J \Delta E = 1.5 \times 4200 \times 80 = 504\,000\ \text{J} ΔE=1.5×4200×80=504000 J
  • The kettle supplies 2000 J2000\ \text{J}2000 J every second, so the shortest possible heating time comes from rearranging E=P tE = P\,tE=Pt.
t=EP=504 0002000=252 s t = \frac{E}{P} = \frac{504\,000}{2000} = 252\ \text{s} t=PE​=2000504000​=252 s
  • The real kettle takes longer than 252 s252\ \text{s}252 s, because some energy also increases the thermal stores of the kettle body and the kitchen air.

Showing the redistribution

Definition

Joulemeter

A meter connected between a power supply and an appliance that records the energy transferred to the appliance in joules.

  1. A full redistribution calculation finds the energy input, then each output in joules, and then checks that the outputs add up to the input.
  2. Any energy not accounted for by the useful output must have gone to other stores, usually the thermal stores of the device and its surroundings.
  3. Subtracting the useful output from the input gives the energy transferred to these other stores.
  4. The electrical energy supplied to a device can be measured directly with a joulemeter, which counts the joules passing from the supply to the device.
Example

A motor lifting a crate

  • A motor is supplied with 1500 J1500\ \text{J}1500 J while it lifts a crate of weight 200 N200\ \text{N}200 N through a height of 5.0 m5.0\ \text{m}5.0 m.
W=F s=200×5.0=1000 J W = F\,s = 200 \times 5.0 = 1000\ \text{J} W=Fs=200×5.0=1000 J
  • This 1000 J1000\ \text{J}1000 J goes into the gravitational potential energy store of the crate.
energy to other stores=1500−1000=500 J \text{energy to other stores} = 1500 - 1000 = 500\ \text{J} energy to other stores=1500−1000=500 J
  • The 1500 J1500\ \text{J}1500 J input has been redistributed as 1000 J1000\ \text{J}1000 J in the crate's gravitational potential energy store and 500 J500\ \text{J}500 J in the thermal stores of the motor, the cable and the surrounding air.
Practical

Measuring the energy used by appliances

  • Aim: to use a joulemeter to measure and compare the energy transferred to different low-voltage electrical appliances in the same time.
  • Apparatus: joulemeter, low-voltage power supply set to 12 V12\ \text{V}12 V, connecting leads, a 12 V12\ \text{V}12 V filament lamp in a holder, a 12 V12\ \text{V}12 V immersion heater in a beaker of water, a small 12 V12\ \text{V}12 V motor clamped to the bench, a stopwatch, and an ammeter and voltmeter for a cross-check.
  • Variables: the appliance is the independent variable, the energy recorded by the joulemeter is the dependent variable, and the supply potential difference and the running time are controlled.
  • Method:
    • Connect the power supply to the input terminals of the joulemeter and the appliance to its output terminals, so that every joule reaching the appliance passes through the meter.
    • With the supply switched off, reset the joulemeter to zero, or write down its starting reading if it cannot be reset.
    • Switch the supply on and start the stopwatch at the same moment.
    • Run the appliance for exactly 300 s300\ \text{s}300 s, then switch off and record the final joulemeter reading.
    • Subtract the starting reading from the final reading to find the energy transferred, if the meter was not reset.
    • Repeat the run three times for each appliance, letting the lamp and heater cool between runs, and calculate a mean after leaving out any anomalous reading.
    • For one run, place an ammeter in series and a voltmeter in parallel with the appliance and calculate E=V I tE = V\,I\,tE=VIt as a check on the joulemeter value.
  • Results: the appliances record very different energies in the same 300 s300\ \text{s}300 s, so a heater drawing 3.0 A3.0\ \text{A}3.0 A at 12 V12\ \text{V}12 V should record about 12×3.0×300=10 800 J12 \times 3.0 \times 300 = 10\,800\ \text{J}12×3.0×300=10800 J, while a small lamp records only a fraction of that.
  • Maths: divide the mean energy by the running time in seconds to find the energy transferred each second in watts, and compare it with the value from V×IV \times IV×I.
  • Watch out: a heater that is not fully immersed overheats and gives a reading that cannot be compared, a motor that stalls draws a much larger current, and the joulemeter only tells you the energy supplied, not where it goes afterwards.
  • Safety: never switch on an immersion heater unless it is fully under water, keep the beaker away from the power supply and leads, let the lamp and heater cool before handling them, and keep fingers clear of the spinning motor shaft.
Common Mistake
  • Do not leave the time in minutes or hours when you want an answer in joules, because E=P tE = P\,tE=Pt only gives joules when PPP is in watts and ttt is in seconds.
  • Do not mix units in a kilowatt-hour calculation, because the power must be in kilowatts and the time in hours.
  • Do not use the length of a slope as sss when calculating the work done in lifting, because only the vertical height is along the line of action of the weight.
  • Do not confuse energy in joules with power in watts, because power is the energy transferred each second.
Exam technique

Setting out energy calculations

  • Write the equation first, in symbols or in words, before substituting any numbers.
  • Show the substituted values on their own line, because a correct substitution earns credit even when the final answer is wrong.
  • Convert grams to kilograms, minutes to seconds and kilojoules to joules before substituting, and write the converted value down.
  • Give the final answer with its unit, such as J\text{J}J, kJ\text{kJ}kJ or kW h\text{kW h}kW h.
  • If you got an earlier part of a question wrong, carry your own value forward into later parts, because an error carried forward can still earn the method marks.
  • In a redistribution question, finish by stating where the remaining energy went rather than leaving a bare number.
Self review
  • State the equation that links work done, force and distance.
  • Calculate the energy transferred when 25 C25\ \text{C}25 C of charge flows through a potential difference of 6.0 V6.0\ \text{V}6.0 V.
  • How many joules are there in one kilowatt-hour?
  • Calculate the energy in kilowatt-hours used by a 1.2 kW1.2\ \text{kW}1.2 kW iron in 303030 minutes.
  • State what a joulemeter measures.
  • A motor is supplied with 800 J800\ \text{J}800 J and the load it lifts gains 600 J600\ \text{J}600 J of gravitational potential energy. How much energy is transferred to other stores?

7.1.5 Kinetic, elastic and gravitational potential energy

Energy of a moving object

Definition

Kinetic energy

The energy held in the kinetic store of an object because it is moving.

  1. Every moving object has energy in its kinetic store, and the amount depends on its mass and its speed.
  2. The kinetic energy is calculated using Ek=12 m v2E_k = \dfrac{1}{2}\,m\,v^2Ek​=21​mv2, which you must learn.
  3. Here EkE_kEk​ is the kinetic energy in joules, J\text{J}J, mmm is the mass in kilograms, kg\text{kg}kg, and vvv is the speed in metres per second, m/s\text{m/s}m/s.
  4. Kinetic energy is directly proportional to mass, so doubling the mass at the same speed doubles the kinetic energy.
  5. Kinetic energy is proportional to the square of the speed, so doubling the speed makes the kinetic energy four times as large.
  6. This is why a car travelling at twice the speed needs roughly four times the braking distance, because the brakes must remove four times as much energy.
  7. Square the speed before multiplying, because squaring the whole of m vm\,vmv is one of the commonest slips in this calculation.
  8. To find a speed from a known kinetic energy, rearrange the equation to v=2 Ekmv = \sqrt{\dfrac{2\,E_k}{m}}v=m2Ek​​​.
Example

Kinetic energy of a trolley

  • A trolley of mass 0.80 kg0.80\ \text{kg}0.80 kg moves at 1.5 m/s1.5\ \text{m/s}1.5 m/s.
Ek=12 m v2 E_k = \frac{1}{2}\,m\,v^2 Ek​=21​mv2 Ek=0.5×0.80×1.52 E_k = 0.5 \times 0.80 \times 1.5^2 Ek​=0.5×0.80×1.52 Ek=0.5×0.80×2.25=0.90 J E_k = 0.5 \times 0.80 \times 2.25 = 0.90\ \text{J} Ek​=0.5×0.80×2.25=0.90 J
  • The kinetic store of the trolley holds 0.90 J0.90\ \text{J}0.90 J.
Example

Speed from kinetic energy

  • A football of mass 0.45 kg0.45\ \text{kg}0.45 kg has 45 J45\ \text{J}45 J of kinetic energy just after it is kicked.
v=2 Ekm v = \sqrt{\frac{2\,E_k}{m}} v=m2Ek​​​ v=2×450.45=200 v = \sqrt{\frac{2 \times 45}{0.45}} = \sqrt{200} v=0.452×45​​=200​ v=14 m/s v = 14\ \text{m/s} v=14 m/s
  • The ball leaves the boot at about 14 m/s14\ \text{m/s}14 m/s, given to two significant figures.
Practical

Investigating kinetic energy with light gates

  • Aim: to measure the kinetic energy of a trolley at the foot of a ramp and find how it changes with the release height.
  • Apparatus: dynamics trolley, ramp about 1 m1\ \text{m}1 m long propped on blocks, light gate connected to a data logger, a card of measured length fixed upright on the trolley, metre rule, top-pan balance, and a box of padding at the end of the runway.
  • Variables: the release height is the independent variable, the speed and kinetic energy at the foot of the ramp are the dependent variables, and the trolley, the ramp surface and the card length are controlled.
  • Method:
    • Measure the mass of the trolley with its card on a top-pan balance and convert it into kilograms.
    • Measure the length of the card with a ruler, for example 10.0 cm10.0\ \text{cm}10.0 cm, and convert it to 0.100 m0.100\ \text{m}0.100 m.
    • Clamp the light gate at the foot of the ramp at a height where the card breaks the beam cleanly before any other part of the trolley does.
    • Mark a release line on the ramp and measure its vertical height above the bench with the metre rule.
    • Hold the trolley with its back wheels on the line and let go without pushing.
    • Record the time the card takes to pass through the beam, which the data logger shows to 0.001 s0.001\ \text{s}0.001 s.
    • Release the trolley three times from the same line, then repeat from at least four other heights.
  • Results: the trolley passes through the gate faster when it is released from higher up, and its kinetic energy rises roughly in proportion to the release height.
  • Maths: the speed through the gate is v=card lengthtimev = \dfrac{\text{card length}}{\text{time}}v=timecard length​, and the kinetic energy is then Ek=12 m v2E_k = \dfrac{1}{2}\,m\,v^2Ek​=21​mv2 using the mean time for each height.
  • Watch out: a push at release adds energy, a card that is not upright gives the wrong length, and friction means the kinetic energy at the bottom is always slightly less than the gravitational potential energy lost.
  • Safety: keep the padding at the end of the runway so the trolley cannot run off the bench, and keep feet out of the way in case it falls.

Energy of a stretched spring

Definition

Elastic potential energy

The energy held in the elastic store of an object that has been stretched, compressed or bent elastically.

Definition

Extension

The increase in the length of an object when it is stretched, equal to its stretched length minus its original length.

Definition

Spring constant

The force needed to produce each metre of extension in a spring, measured in newtons per metre.

  1. A spring that has been stretched or compressed holds energy in its elastic potential energy store.
  2. The energy stored is calculated using Ee=12 k x2E_e = \dfrac{1}{2}\,k\,x^2Ee​=21​kx2, which is a select and apply equation that you need to recognise rather than memorise.
  3. Here EeE_eEe​ is the elastic potential energy in joules, J\text{J}J, kkk is the spring constant in newtons per metre, N/m\text{N/m}N/m, and xxx is the extension in metres, m\text{m}m.
  4. The equation only applies while the spring has a linear force-extension relationship, because beyond that point each extra newton stretches the spring by a larger amount.
  5. The extension is the increase in length, so subtract the original length from the stretched length before substituting.
  6. Extensions are usually measured in centimetres or millimetres, so convert them to metres first, for example 5.0 cm=0.050 m5.0\ \text{cm} = 0.050\ \text{m}5.0 cm=0.050 m.
  7. Because the extension is squared, doubling the extension makes the stored energy four times as large.
  8. A stiffer spring with a larger spring constant stores more energy for the same extension.
  9. The energy stored in a spring equals the work done in stretching it, which is why both quantities are measured in joules.
Example

Energy stored in a spring

  • A spring with a spring constant of 40 N/m40\ \text{N/m}40 N/m is stretched from a length of 12.0 cm12.0\ \text{cm}12.0 cm to a length of 27.0 cm27.0\ \text{cm}27.0 cm.
  • The extension is 27.0−12.0=15.0 cm27.0 - 12.0 = 15.0\ \text{cm}27.0−12.0=15.0 cm, which is 0.150 m0.150\ \text{m}0.150 m.
Ee=12 k x2 E_e = \frac{1}{2}\,k\,x^2 Ee​=21​kx2 Ee=0.5×40×0.1502 E_e = 0.5 \times 40 \times 0.150^2 Ee​=0.5×40×0.1502 Ee=0.5×40×0.0225=0.45 J E_e = 0.5 \times 40 \times 0.0225 = 0.45\ \text{J} Ee​=0.5×40×0.0225=0.45 J
  • The elastic potential energy store of the spring holds 0.45 J0.45\ \text{J}0.45 J.

Energy of a raised object

Definition

Gravitational potential energy

The energy held in the gravitational store of an object because of its height in a gravitational field.

Definition

Gravitational field strength

The gravitational force acting on each kilogram of mass, which is 10 N/kg10\ \text{N/kg}10 N/kg near the Earth's surface.

  1. Lifting an object means doing work against its weight, so the object's gravitational potential energy store increases.
  2. The change in gravitational potential energy is calculated using Ep=m g hE_p = m\,g\,hEp​=mgh, which you must learn.
  3. Here EpE_pEp​ is the gravitational potential energy in joules, J\text{J}J, mmm is the mass in kilograms, kg\text{kg}kg, ggg is the gravitational field strength in newtons per kilogram, N/kg\text{N/kg}N/kg, and hhh is the height in metres, m\text{m}m.
  4. Near the Earth's surface g=10 N/kgg = 10\ \text{N/kg}g=10 N/kg, and a question gives you this value when you need it.
  5. The height is the vertical height gained, so a person walking up a long gentle ramp gains the same gravitational potential energy as one climbing a steep ladder to the same level.
  6. The equation gives the change in gravitational potential energy, so hhh is measured from whatever starting level the question uses, such as the floor or the ground.
  7. On the Moon, where ggg is only about 1.6 N/kg1.6\ \text{N/kg}1.6 N/kg, lifting the same mass to the same height stores far less energy.
  8. When an object falls with negligible air resistance, the gravitational potential energy it loses equals the kinetic energy it gains.
  9. Setting the two equal, m g h=12 m v2m\,g\,h = \dfrac{1}{2}\,m\,v^2mgh=21​mv2, lets you find the speed of a falling object from the height it has fallen.
Example

Energy of a lifted mass

  • A motor lifts a 2.0 kg2.0\ \text{kg}2.0 kg mass through a height of 1.5 m1.5\ \text{m}1.5 m, with g=10 N/kgg = 10\ \text{N/kg}g=10 N/kg.
Ep=m g h E_p = m\,g\,h Ep​=mgh Ep=2.0×10×1.5=30 J E_p = 2.0 \times 10 \times 1.5 = 30\ \text{J} Ep​=2.0×10×1.5=30 J
  • The gravitational potential energy store of the mass increases by 30 J30\ \text{J}30 J.
Example

Speed after a fall

  • An apple of mass 0.20 kg0.20\ \text{kg}0.20 kg falls 3.2 m3.2\ \text{m}3.2 m from a branch, with g=10 N/kgg = 10\ \text{N/kg}g=10 N/kg and negligible air resistance.
Ep=m g h=0.20×10×3.2=6.4 J E_p = m\,g\,h = 0.20 \times 10 \times 3.2 = 6.4\ \text{J} Ep​=mgh=0.20×10×3.2=6.4 J
  • All of this energy goes into the kinetic store, so Ek=6.4 JE_k = 6.4\ \text{J}Ek​=6.4 J just before the apple lands.
v=2 Ekm=2×6.40.20=64 v = \sqrt{\frac{2\,E_k}{m}} = \sqrt{\frac{2 \times 6.4}{0.20}} = \sqrt{64} v=m2Ek​​​=0.202×6.4​​=64​ v=8.0 m/s v = 8.0\ \text{m/s} v=8.0 m/s
  • The apple hits the ground at 8.0 m/s8.0\ \text{m/s}8.0 m/s.
Practical

Investigating gravitational potential energy with a motor

  • Aim: to compare the electrical energy supplied to a motor with the gravitational potential energy gained by the mass it lifts.
  • Apparatus: small electric motor clamped to the edge of a bench, string wound on the motor axle, 100 g100\ \text{g}100 g mass hanger with extra slotted masses, joulemeter, low-voltage power supply, metre rule clamped vertically, and a cushion on the floor.
  • Variables: the mass lifted is the independent variable, the energy supplied and the gravitational potential energy gained are the dependent variables, and the lifting height and supply potential difference are controlled.
  • Method:
    • Clamp the motor firmly to the bench and tie the string to the mass hanger so it rests on the floor directly below the axle with the string just taut.
    • Clamp the metre rule vertically beside the string and mark a fixed lifting height, such as 1.00 m1.00\ \text{m}1.00 m above the floor.
    • Connect the motor to the power supply through the joulemeter and reset the meter to zero.
    • Switch on and let the motor wind the mass upwards, switching off the moment the base of the hanger reaches the mark.
    • Record the joulemeter reading, lower the mass back to the floor and repeat twice more.
    • Add slotted masses and repeat the whole procedure for at least four different masses.
  • Results: the gravitational potential energy gained by the mass is always less than the energy recorded by the joulemeter, because the motor, the wires and the bearings warm up and the motor hums.
  • Maths: calculate Ep=m g hE_p = m\,g\,hEp​=mgh with the mass in kilograms, then subtract it from the mean joulemeter reading to find the energy transferred to other stores.
  • Watch out: switching off late lets the mass overshoot the mark, slack string means the timing of the lift is uneven, and a mass that swings gains kinetic energy that is not counted in EpE_pEp​.
  • Safety: place a cushion beneath the masses in case the string slips or snaps, and keep hands and feet out of the area below the hanging masses.

Energy changes of a bouncing ball

  1. A bouncing ball shows all three stores in this article changing in turn.
  2. Held above the floor, the ball has energy in its gravitational potential energy store.
  3. As it falls, its gravitational potential energy store decreases and its kinetic store increases.
  4. When it hits the floor the ball squashes, so its kinetic store decreases and its elastic potential energy store increases.
  5. As the ball regains its shape, its elastic potential energy store decreases and its kinetic store increases, so it leaves the floor moving upwards.
  6. As it rises, its kinetic store decreases and its gravitational potential energy store increases until it stops at the top of the bounce.
  7. The ball never returns to its starting height, because during each squash some energy goes to the thermal stores of the ball and the floor and some is carried away as sound.
  8. Since Ep=m g hE_p = m\,g\,hEp​=mgh and neither mmm nor ggg changes, the fraction of energy still in the gravitational store after one bounce equals rebound heightdrop height\dfrac{\text{rebound height}}{\text{drop height}}drop heightrebound height​.
  9. This fraction shows how efficient a ball is at bouncing, so a ball that keeps a larger fraction of its energy bounces higher.
Practical

Investigating bouncing balls

  • Aim: to describe the energy changes of a bouncing ball and compare the fraction of energy different balls keep after one bounce.
  • Apparatus: a selection of balls such as a tennis ball, a rubber ball, a table tennis ball and a golf ball, two metre rules clamped vertically one above the other, a clamp stand, a hard floor tile and a phone camera that films in slow motion.
  • Variables: the type of ball is the independent variable, the rebound height is the dependent variable, and the drop height and floor surface are controlled.
  • Method:
    • Clamp the metre rules vertically with the zero mark resting on the floor tile.
    • Hold the first ball so that its bottom is level with the 1.00 m1.00\ \text{m}1.00 m mark, then release it without throwing.
    • Film the ball against the scale, then play the clip back frame by frame to read the height of the bottom of the ball at the top of its first bounce.
    • Line the camera up level with the expected rebound height, so that parallax does not make the reading too high or too low.
    • Drop each ball five times from the same height and calculate a mean rebound height, leaving out any anomalous drop.
    • Repeat for every type of ball using the same tile and drop height.
  • Results: a lively rubber ball might rebound to about 0.70 m0.70\ \text{m}0.70 m from a 1.00 m1.00\ \text{m}1.00 m drop, keeping about 70%70\%70% of its energy in the gravitational store, while a soft ball rebounds far lower.
  • Maths: the fraction of energy kept is mean rebound heightdrop height\dfrac{\text{mean rebound height}}{\text{drop height}}drop heightmean rebound height​, and multiplying by 100100100 gives it as a percentage.
  • Watch out: judging the top of the bounce by eye gives large random errors, measuring to the top of the ball instead of the bottom shifts every reading, and a warm ball bounces higher than a cold one.
  • Safety: clear the floor so nobody trips on a rolling ball, and do not stand on chairs or desks to reach greater drop heights.
Common Mistake
  • Do not forget to square the speed in Ek=12 m v2E_k = \frac{1}{2}\,m\,v^2Ek​=21​mv2 or the extension in Ee=12 k x2E_e = \frac{1}{2}\,k\,x^2Ee​=21​kx2.
  • Do not substitute a mass in grams or an extension in centimetres, because both equations need SI units.
  • Do not use the original length of a spring as xxx, because xxx is the extension.
  • Do not say that a bouncing ball loses energy, because the energy is transferred to thermal stores and carried away by sound.
Exam technique

Calculating stored energy

  • Learn Ek=12 m v2E_k = \frac{1}{2}\,m\,v^2Ek​=21​mv2 and Ep=m g hE_p = m\,g\,hEp​=mgh, because you are expected to recall them, while Ee=12 k x2E_e = \frac{1}{2}\,k\,x^2Ee​=21​kx2 is one you select and apply.
  • Write the equation, substitute the values, then calculate, keeping each step on a separate line.
  • In a two-step problem such as the speed after a fall, find the energy first and then use it to find the speed.
  • Round only at the end, and give the answer to the same number of significant figures as the data, usually two or three.
  • In a practical question, describe repeat readings, a mean and how you dealt with anomalies, as well as the variables you controlled.
Self review
  • State the equation for kinetic energy and the unit of each quantity.
  • Calculate the kinetic energy of a 1200 kg1200\ \text{kg}1200 kg car moving at 15 m/s15\ \text{m/s}15 m/s.
  • Calculate the energy stored in a spring with a spring constant of 200 N/m200\ \text{N/m}200 N/m and an extension of 4.0 cm4.0\ \text{cm}4.0 cm.
  • Calculate the gravitational potential energy gained by a 60 kg60\ \text{kg}60 kg climber who climbs 12 m12\ \text{m}12 m, with g=10 N/kgg = 10\ \text{N/kg}g=10 N/kg.
  • Explain why a bouncing ball never returns to the height it was dropped from.

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A system is the object or group of objects you decide to study. Physicists track energy using stores such as kinetic, gravitational, elastic, thermal, and chemical.

Energy can be transferred mechanically by forces, electrically by a current, or by heating. In a closed system the total energy stays constant, so energy is redistributed rather than created or destroyed.

Work done is one special kind of energy transfer. It happens when a force makes an object move in the direction of the force.

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Question 1

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A student throws a ball vertically upwards. The ball rises, stops for an instant at its highest point and then falls back down.

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What characterizes a closed system in terms of energy?

7.1 Work done Revision Guide

  1. GCSE
  2. /Physics
  3. /7.1 Work done

Revision notes for OCR GCSE Physics 7.1 Work done: explanations and worked examples on 7.1.1 Conservation of energy in a closed system, 7.1.2 Energy stores and transfers, 7.1.3 Energy changes by heating, forces and current, 7.1.4 Quantifying energy transfers in a system, and 7.1.5 Kinetic, elastic and gravitational potential energy.

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