Work done

What you'll learn

How physicists describe energy stores and energy transfers in a system.
What work done means, and how to calculate it using force and distance.
How work links to gravitational, kinetic, elastic, thermal and electrical energy changes.
How to choose the right equation and avoid common exam traps.

Start point: systems, stores and transfers

Before “work done” makes sense, you need the energy language.

A system is the object or group of objects you are focusing on. For example, the system could be a falling ball, a kettle and water, or a motor lifting a weight.

An energy store is a way of accounting for where energy is. Common GCSE stores include:

Kinetic store — energy of a moving object.
Gravitational store — energy due to position in a gravitational field.
Elastic store — energy in a stretched or compressed object.
Thermal store — energy linked to temperature.
Chemical store — energy stored in fuels, food and batteries.

An energy transfer is a process that moves energy from one store to another. Energy can be transferred mechanically by forces, electrically by a current, by heating, or by radiation.

Definition

Closed system

A closed system is a system where no energy is transferred into it or out of it. Inside the system, energy can move between stores, but the total amount of energy stays the same.

Key Idea

Conservation of energy

Energy is not created or destroyed. In a closed system there is no net change to the total energy — energy is only redistributed between stores.

Example

Following energy transfers in a falling ball

A ball is dropped from rest and then hits the ground.

At the top, the ball has energy in its gravitational store because it is above the ground. Its kinetic store is very small because it starts from rest.
As the ball falls, energy is transferred from the gravitational store to the kinetic store, so the ball speeds up.
If air resistance is included, some energy is transferred to the thermal stores of the ball and the surrounding air. When the ball hits the ground, energy is also transferred to thermal and sound stores.

Common Mistake

Energy is not used up

Do not write that energy is “lost” or “used up” as if it has vanished. A better GCSE answer is: energy is transferred to less useful stores, often the thermal store of the surroundings.

Work done by a force

In physics, work done has a very specific meaning. It is not just “effort”.

Definition

Work done

Work done is the energy transferred when a force causes an object to move through a distance in the direction of the force.

For a force and distance in the same direction:

W = F d

where:

$W$ is the work done in joules, J
$F$ is the force in newtons, N
$d$ is the distance moved in metres, m

One joule is the same as one newton metre:

m1\ \text{J} = 1\ \text{N m}

For OCR Gateway J249, you should learn to recall or select the relevant energy equations, including this one. Even if an equation sheet is supplied for your exam series, you still need to choose the correct equation and use the units properly.

A block being pulled horizontally, showing force, distance and energy store changes

Example

Calculating work done by a pulling force

A student pulls a crate with a force of 75 N for a distance of 6.0 m. The force is in the direction of motion.

The force and distance are in the same direction, so use $W = F d$ .
Substitute the values:
$75\ \text{N} \times 6.0\ \text{m} = 450\ \text{J}$
The pulling force transfers 450 J of energy. If the crate moves at constant speed on a rough floor, most of this energy is transferred to thermal stores rather than increasing the kinetic store.

Common Mistake

Force without movement

If you hold a heavy bag still, you exert an upward force, but you do no work on the bag in the GCSE mechanical sense because the bag has not moved in the direction of the force.

Work done and mechanical energy stores

When work is done by forces, energy is transferred mechanically. The store that changes depends on the situation.

Lifting an object: gravitational potential energy

An object raised above ground level gains energy in its gravitational store. This is often called gravitational potential energy.

ΔEp=mgh\Delta E_p = mgh

where:

$ΔEp\Delta E_p$ is the change in gravitational potential energy in joules, J
$m$ is mass in kilograms, kg
$g$ is gravitational field strength in newtons per kilogram, N/kg
$h$ is height change in metres, m

On Earth, use the value of $g$ given in the question. If not given, GCSE questions often use about 10 N/kg.

Example

Lifting a box

A 8.0 kg box is lifted vertically by 1.5 m. Take $10\ \text{N/kg}$ .

Calculate the weight of the box:
$8.0\ \text{kg} \times 10\ \text{N/kg} = 80\ \text{N}$
The force lifts the box through 1.5 m, so the work done is:
$80\ \text{N} \times 1.5\ \text{m} = 120\ \text{J}$
Ignoring losses, this 120 J is the increase in the box’s gravitational store, so $J\Delta E_p = 120\ \text{J}$ .

Speeding up an object: kinetic energy

A moving object has energy in its kinetic store. The faster it moves, the more kinetic energy it has.

Ek=12mv2E_k = \frac{1}{2}mv^2

where:

$E_k$ is kinetic energy in joules, J
$m$ is mass in kilograms, kg
$v$ is speed in metres per second, m/s

A useful thing to notice is the squared speed, $v^2$ . If speed doubles, kinetic energy becomes four times bigger, not just twice as big.

Stretching a spring: elastic potential energy

A stretched or compressed spring stores energy in its elastic store.

Ee=12ke2E_e = \frac{1}{2}ke^2

where:

$E_e$ is elastic potential energy in joules, J
$k$ is the spring constant in newtons per metre, N/m
$e$ is the extension in metres, m

Common Mistake

Stretching a spring too far

The spring energy equation is used when the spring behaves elastically, meaning it returns to its original length when the force is removed. If it is stretched beyond its elastic limit, this simple model no longer works.

Example

Calculating energy stored in a spring

A spring has spring constant 200 N/m and is stretched by 0.30 m.

The situation involves a stretched spring, so use $Ee=12ke2E_e = \frac{1}{2}ke^2$ .
Substitute the values, remembering to square the extension:
$m)2E_e = \frac{1}{2} \times 200\ \text{N/m} \times \left(0.30\ \text{m}\right)^2$
Calculate the energy stored:
$JE_e = 100 \times 0.090 = 9.0\ \text{J}$

Tip

Choosing a mechanical equation

Ask what is changing: height means use $ΔEp=mgh\Delta E_p = mgh$ , speed means use $Ek=12mv2E_k = \frac{1}{2}mv^2$ , and stretch or compression means use $Ee=12ke2E_e = \frac{1}{2}ke^2$ .

Heating and specific heat capacity

Energy can be transferred by heating, increasing the thermal store of an object. This may cause a temperature increase.

The specific heat capacity of a material is the energy needed to raise the temperature of 1 kg of the material by 1 °C.

ΔE=mcΔθ\Delta E = mc\Delta\theta

where:

$ΔE\Delta E$ is energy transferred in joules, J
$m$ is mass in kilograms, kg
$c$ is specific heat capacity in joules per kilogram per degree Celsius, J/kg °C
$Δθ\Delta\theta$ is temperature change in °C

Example

Heating water

A 2.0 kg sample of water is heated from 20 °C to 35 °C. The specific heat capacity of water is 4200 J/kg °C.

Find the temperature change:
$∘C\Delta\theta = 35\ ^\circ\text{C} - 20\ ^\circ\text{C} = 15\ ^\circ\text{C}$
Use $ΔE=mcΔθ\Delta E = mc\Delta\theta$ :
$∘C\Delta E = 2.0\ \text{kg} \times 4200\ \text{J/kg}^\circ\text{C} \times 15\ ^\circ\text{C}$
Calculate the energy transferred:
$J\Delta E = 126000\ \text{J}$

Common Mistake

Temperature is not energy

Temperature tells you how hot something is. Energy transferred by heating depends on mass, material and temperature change, so two objects at the same temperature can have different amounts of energy in their thermal stores.

Work done when a current flows

When an electric current flows through a component, energy is transferred electrically. For example:

A lamp transfers energy to light and thermal stores.
A motor transfers energy to kinetic and gravitational stores.
A heater transfers energy mainly to thermal stores.

Power is the rate of energy transfer. If you know power and time:

ΔE=Pt\Delta E = Pt

where $P$ is power in watts, W, and $t$ is time in seconds, s.

You may also see:

ΔE=IVt\Delta E = IVt

because electrical power can be calculated using $P = I V$ .

Kilowatt-hours

Electricity meters in homes measure energy in kilowatt-hours, kWh. This is a unit of energy, not power.

J1\ \text{kWh} = 1000\ \text{W} \times 3600\ \text{s} = 3.6 \times 10^6\ \text{J}

Example

Calculating energy used by an appliance

A 2.0 kW heater is switched on for 3.0 hours.

The power is in kW and the time is in hours, so calculate energy directly in kWh:
$h\Delta E = Pt = 2.0\ \text{kW} \times 3.0\ \text{h}$
Work out the energy transferred:
$kWh\Delta E = 6.0\ \text{kWh}$
Convert to joules if needed:
$J6.0\ \text{kWh} = 6.0 \times 3.6 \times 10^6\ \text{J} = 2.16 \times 10^7\ \text{J}$

Common Mistake

Joules and kilowatt-hours

Use watts with seconds to get joules, or kilowatts with hours to get kilowatt-hours. Mixing kW with seconds or W with hours is a very common source of wrong answers.

The big picture

Work done is one way energy is transferred. When a force moves an object, energy shifts between stores. In a real situation, some energy is often transferred to the thermal store of the surroundings due to friction or air resistance.

That does not break conservation of energy. It just means the useful energy transfer is not the whole story.

Exam technique

In the exam

Identify the process first: lifting, speeding up, stretching, heating, or electrical transfer.
Choose the matching equation and check units before substituting: kg, m, s, N, J, W, kW and h must be used carefully.
Describe energy changes using stores and transfers: say where energy starts, where useful energy goes, and where unwanted energy is transferred.

Self review

Check yourself

What has to happen for a force to do work on an object?
Why is it better to say energy is “transferred to thermal stores” rather than “lost”?
Which equation would you use for a stretched spring, and what must be true about the spring?

Work done

What you'll learn

How physicists describe energy stores and energy transfers in a system.
What work done means, and how to calculate it using force and distance.
How work links to gravitational, kinetic, elastic, thermal and electrical energy changes.
How to choose the right equation and avoid common exam traps.

Start point: systems, stores and transfers

Before “work done” makes sense, you need the energy language.

A system is the object or group of objects you are focusing on. For example, the system could be a falling ball, a kettle and water, or a motor lifting a weight.

An energy store is a way of accounting for where energy is. Common GCSE stores include:

Kinetic store — energy of a moving object.
Gravitational store — energy due to position in a gravitational field.
Elastic store — energy in a stretched or compressed object.
Thermal store — energy linked to temperature.
Chemical store — energy stored in fuels, food and batteries.

An energy transfer is a process that moves energy from one store to another. Energy can be transferred mechanically by forces, electrically by a current, by heating, or by radiation.

Definition

Closed system

A closed system is a system where no energy is transferred into it or out of it. Inside the system, energy can move between stores, but the total amount of energy stays the same.

Key Idea

Conservation of energy

Energy is not created or destroyed. In a closed system there is no net change to the total energy — energy is only redistributed between stores.

Example

Following energy transfers in a falling ball

A ball is dropped from rest and then hits the ground.

At the top, the ball has energy in its gravitational store because it is above the ground. Its kinetic store is very small because it starts from rest.
As the ball falls, energy is transferred from the gravitational store to the kinetic store, so the ball speeds up.
If air resistance is included, some energy is transferred to the thermal stores of the ball and the surrounding air. When the ball hits the ground, energy is also transferred to thermal and sound stores.

Common Mistake

Energy is not used up

Do not write that energy is “lost” or “used up” as if it has vanished. A better GCSE answer is: energy is transferred to less useful stores, often the thermal store of the surroundings.

Work done by a force

In physics, work done has a very specific meaning. It is not just “effort”.

Definition

Work done

Work done is the energy transferred when a force causes an object to move through a distance in the direction of the force.

For a force and distance in the same direction:

W = F d

where:

$W$ is the work done in joules, J
$F$ is the force in newtons, N
$d$ is the distance moved in metres, m

One joule is the same as one newton metre:

m1\ \text{J} = 1\ \text{N m}

A block being pulled horizontally, showing force, distance and energy store changes

Example

Calculating work done by a pulling force

A student pulls a crate with a force of 75 N for a distance of 6.0 m. The force is in the direction of motion.

The force and distance are in the same direction, so use $W = F d$ .
Substitute the values:
$75\ \text{N} \times 6.0\ \text{m} = 450\ \text{J}$
The pulling force transfers 450 J of energy. If the crate moves at constant speed on a rough floor, most of this energy is transferred to thermal stores rather than increasing the kinetic store.

Common Mistake

Force without movement

If you hold a heavy bag still, you exert an upward force, but you do no work on the bag in the GCSE mechanical sense because the bag has not moved in the direction of the force.

Work done and mechanical energy stores

When work is done by forces, energy is transferred mechanically. The store that changes depends on the situation.

Lifting an object: gravitational potential energy

An object raised above ground level gains energy in its gravitational store. This is often called gravitational potential energy.

ΔEp=mgh\Delta E_p = mgh

where:

$ΔEp\Delta E_p$ is the change in gravitational potential energy in joules, J
$m$ is mass in kilograms, kg
$g$ is gravitational field strength in newtons per kilogram, N/kg
$h$ is height change in metres, m

On Earth, use the value of $g$ given in the question. If not given, GCSE questions often use about 10 N/kg.

Example

Lifting a box

A 8.0 kg box is lifted vertically by 1.5 m. Take $10\ \text{N/kg}$ .

Calculate the weight of the box:
$8.0\ \text{kg} \times 10\ \text{N/kg} = 80\ \text{N}$
The force lifts the box through 1.5 m, so the work done is:
$80\ \text{N} \times 1.5\ \text{m} = 120\ \text{J}$
Ignoring losses, this 120 J is the increase in the box’s gravitational store, so $J\Delta E_p = 120\ \text{J}$ .

Speeding up an object: kinetic energy

A moving object has energy in its kinetic store. The faster it moves, the more kinetic energy it has.

Ek=12mv2E_k = \frac{1}{2}mv^2

where:

$E_k$ is kinetic energy in joules, J
$m$ is mass in kilograms, kg
$v$ is speed in metres per second, m/s

A useful thing to notice is the squared speed, $v^2$ . If speed doubles, kinetic energy becomes four times bigger, not just twice as big.

Stretching a spring: elastic potential energy

A stretched or compressed spring stores energy in its elastic store.

Ee=12ke2E_e = \frac{1}{2}ke^2

where:

$E_e$ is elastic potential energy in joules, J
$k$ is the spring constant in newtons per metre, N/m
$e$ is the extension in metres, m

Common Mistake

Stretching a spring too far

Example

Calculating energy stored in a spring

A spring has spring constant 200 N/m and is stretched by 0.30 m.

The situation involves a stretched spring, so use $Ee=12ke2E_e = \frac{1}{2}ke^2$ .
Substitute the values, remembering to square the extension:
$m)2E_e = \frac{1}{2} \times 200\ \text{N/m} \times \left(0.30\ \text{m}\right)^2$
Calculate the energy stored:
$JE_e = 100 \times 0.090 = 9.0\ \text{J}$

Tip

Choosing a mechanical equation

Ask what is changing: height means use $ΔEp=mgh\Delta E_p = mgh$ , speed means use $Ek=12mv2E_k = \frac{1}{2}mv^2$ , and stretch or compression means use $Ee=12ke2E_e = \frac{1}{2}ke^2$ .

Heating and specific heat capacity

Energy can be transferred by heating, increasing the thermal store of an object. This may cause a temperature increase.

The specific heat capacity of a material is the energy needed to raise the temperature of 1 kg of the material by 1 °C.

ΔE=mcΔθ\Delta E = mc\Delta\theta

where:

$ΔE\Delta E$ is energy transferred in joules, J
$m$ is mass in kilograms, kg
$c$ is specific heat capacity in joules per kilogram per degree Celsius, J/kg °C
$Δθ\Delta\theta$ is temperature change in °C

Example

Heating water

A 2.0 kg sample of water is heated from 20 °C to 35 °C. The specific heat capacity of water is 4200 J/kg °C.

Find the temperature change:
$∘C\Delta\theta = 35\ ^\circ\text{C} - 20\ ^\circ\text{C} = 15\ ^\circ\text{C}$
Use $ΔE=mcΔθ\Delta E = mc\Delta\theta$ :
$∘C\Delta E = 2.0\ \text{kg} \times 4200\ \text{J/kg}^\circ\text{C} \times 15\ ^\circ\text{C}$
Calculate the energy transferred:
$J\Delta E = 126000\ \text{J}$

Common Mistake

Temperature is not energy

Work done when a current flows

When an electric current flows through a component, energy is transferred electrically. For example:

A lamp transfers energy to light and thermal stores.
A motor transfers energy to kinetic and gravitational stores.
A heater transfers energy mainly to thermal stores.

Power is the rate of energy transfer. If you know power and time:

ΔE=Pt\Delta E = Pt

where $P$ is power in watts, W, and $t$ is time in seconds, s.

You may also see:

ΔE=IVt\Delta E = IVt

because electrical power can be calculated using $P = I V$ .

Kilowatt-hours

Electricity meters in homes measure energy in kilowatt-hours, kWh. This is a unit of energy, not power.

J1\ \text{kWh} = 1000\ \text{W} \times 3600\ \text{s} = 3.6 \times 10^6\ \text{J}

Example

Calculating energy used by an appliance

A 2.0 kW heater is switched on for 3.0 hours.

The power is in kW and the time is in hours, so calculate energy directly in kWh:
$h\Delta E = Pt = 2.0\ \text{kW} \times 3.0\ \text{h}$
Work out the energy transferred:
$kWh\Delta E = 6.0\ \text{kWh}$
Convert to joules if needed:
$J6.0\ \text{kWh} = 6.0 \times 3.6 \times 10^6\ \text{J} = 2.16 \times 10^7\ \text{J}$

Common Mistake

Joules and kilowatt-hours

Use watts with seconds to get joules, or kilowatts with hours to get kilowatt-hours. Mixing kW with seconds or W with hours is a very common source of wrong answers.

The big picture

That does not break conservation of energy. It just means the useful energy transfer is not the whole story.

Exam technique

In the exam

Identify the process first: lifting, speeding up, stretching, heating, or electrical transfer.
Choose the matching equation and check units before substituting: kg, m, s, N, J, W, kW and h must be used carefully.
Describe energy changes using stores and transfers: say where energy starts, where useful energy goes, and where unwanted energy is transferred.

Self review

Check yourself

What has to happen for a force to do work on an object?
Why is it better to say energy is “transferred to thermal stores” rather than “lost”?
Which equation would you use for a stretched spring, and what must be true about the spring?

Work done

What you'll learn

Start point: systems, stores and transfers

Work done by a force

Work done and mechanical energy stores

Lifting an object: gravitational potential energy

Speeding up an object: kinetic energy

Stretching a spring: elastic potential energy

Heating and specific heat capacity

Work done when a current flows

Kilowatt-hours

The big picture

Recap questions

Work done

What you'll learn

Start point: systems, stores and transfers

Work done by a force

Work done and mechanical energy stores

Lifting an object: gravitational potential energy

Speeding up an object: kinetic energy

Stretching a spring: elastic potential energy

Heating and specific heat capacity

Work done when a current flows

Kilowatt-hours

The big picture

Recap questions

Work done

What you'll learn

Start point: systems, stores and transfers

Work done by a force

Work done and mechanical energy stores

Lifting an object: gravitational potential energy

Speeding up an object: kinetic energy

Stretching a spring: elastic potential energy

Heating and specific heat capacity

Work done when a current flows

Kilowatt-hours

The big picture

Recap questions

Work done Revision Guide

Work done

What you'll learn

Start point: systems, stores and transfers

Work done by a force

Work done and mechanical energy stores

Lifting an object: gravitational potential energy

Speeding up an object: kinetic energy

Stretching a spring: elastic potential energy

Heating and specific heat capacity

Work done when a current flows

Kilowatt-hours

The big picture

Recap questions

Work done Revision Guide