Work, energy and power

Welcome to the foundation of mechanics! Almost every physical process you will study in A-Level Physics—from circuits to quantum mechanics—comes back to energy.

What you'll learn in this topic:

How to calculate the work done by a force, even when it acts at an angle to the direction of motion.
How to find the work done when a force is constantly changing (variable forces).
How to link power, force, and velocity for moving objects.
How to calculate the efficiency of real-world systems, such as an electric motor lifting a mass.

1. Work Done (Energy Transferred)

In everyday language, "doing work" means putting in effort. In physics, the definition is much stricter. Work is only done when a force causes an object to move, and energy is transferred as a result.

Definition

Work Done

Work done is the energy transferred when a force moves an object through a distance. One Joule (1 J) of work is done when a force of 1 Newton moves an object by 1 metre in the direction of the force.

If you push a block in a straight line and the force is acting in the exact same direction as the movement, the equation is simply $W = F s$ , where $W$ is work done in Joules (J), $F$ is force in Newtons (N), and $s$ is displacement in metres (m).

However, forces usually act at an angle. Think about pulling a heavy suitcase on wheels—you pull diagonally upwards on the handle, but the suitcase moves horizontally across the floor. Only the horizontal component of your pulling force is actually doing work to move the suitcase forward. The vertical component is just lifting it slightly against gravity.

To account for this, we use the general equation:

\cos \theta

Where:

$F$ is the magnitude of the applied force.
$s$ is the displacement of the object.
$θ\theta$ is the angle between the force vector and the direction of displacement.

Block pulled at an angle

Common Mistake

Using the wrong angle

Always ensure $θ\theta$ is the angle between the force and the direction of motion. If an exam question gives you the angle between the force and the vertical (for an object moving horizontally), you must subtract it from 90 degrees before using the $\cos \theta$ formula, or resolve the force yourself using trigonometry.

Example

Pulling a sledge

A child pulls a sledge across a flat snowfield for a distance of 45 m. The tension in the pulling rope is 120 N, and the rope is held at an angle of 30 degrees to the horizontal. Calculate the work done on the sledge.

Identify the given values: $120\text{ N}$ , $45\text{ m}$ , and $θ=30∘\theta = 30^\circ$ .
State the formula for work done at an angle:

\cos \theta

Substitute the values into the equation:

\times 45 \times \cos(30^\circ)

Calculate the result:

\approx 120 \times 45 \times 0.866 \approx 4676\text{ J}

State the final answer with appropriate significant figures and units: The work done is 4700 J (to 2 s.f.).

2. Variable Forces

The equation $W = F s$ assumes the force remains perfectly constant. But in the real world, forces often change. For example, the more you stretch a spring, the harder it pulls back. When a force is variable, you cannot just plug a single value of $F$ into the formula.

Instead, we look at a force–displacement graph.

Force-displacement graph

Key Idea

Area under the graph

For any force-displacement graph, the area under the line or curve represents the total work done (energy transferred).

If the force increases uniformly (forming a straight diagonal line), you can calculate the area of the resulting triangle using $12×base×height\frac{1}{2} \times \text{base} \times \text{height}$ . If the force forms a curve, you may need to estimate the area by counting squares on the exam paper grid.

3. Power

If work done is how much energy is transferred, power is how fast that energy is transferred.

Definition

Power

Power is the rate of doing work, or the rate of energy transfer. It is measured in Watts (W). One Watt is equal to one Joule per second ( $s−11\text{ W} = 1\text{ J s}^{-1}$ ).

The fundamental equation for power is:

\frac{\Delta W}{\Delta t}

Where $ΔW\Delta W$ is the work done (or energy transferred) and $Δt\Delta t$ is the time taken.

Power for moving objects

There is an incredibly useful shortcut for calculating the power of a vehicle (like a car or a train) that is moving at a constant speed against resistive forces.

We know that $W = F s$ . If we substitute this into the power equation, we get:

\begin{aligned} P &= \frac{Fs}{t} \\ P &= F \left( \frac{s}{t} \right) \end{aligned}

Since velocity $\frac{s}{t}$ , we can replace that part of the equation to arrive at:

P = F v

Where $F$ is the driving force and $v$ is the velocity. This equation is essential for AQA exams. Note that for this formula to work, the force and the velocity must be in the same direction.

Example

Car engine power

A car is travelling at a constant velocity of 25 m s⁻¹ along a flat road. The total resistive forces acting on the car (air resistance and friction) are 800 N. Calculate the useful power output of the car's engine.

Recognise that because the car is moving at a constant velocity, the driving force from the engine must perfectly balance the resistive forces. Therefore, the driving force $800\text{ N}$ .
Identify the velocity: $25\text{ m s}^{-1}$ .
State the formula linking power, force, and velocity:

P = F v

Substitute the values:

\times 25

Calculate the final answer and include units:

20000\text{ W} \quad (\text{or } 20\text{ kW})

4. Efficiency

No machine is perfect. Whenever energy is transferred, some of it is always dissipated to the surroundings, usually as heat or sound due to friction. Efficiency is a measure of how much of the total energy (or power) put into a system actually ends up doing the job you want it to do.

\text{Efficiency} = \frac{\text{useful output power}}{\text{input power}}

(You can also calculate efficiency using energy instead of power: $energy\text{Efficiency} = \frac{\text{useful output energy}}{\text{total input energy}}$ )

Efficiency can be written as a decimal (between 0 and 1) or multiplied by 100 to be expressed as a percentage.

Tip

Sanity check

Efficiency can never be greater than 1 (or 100%). If your calculation gives you an efficiency of 1.2, you have probably put the larger input number on the top of the fraction by mistake!

Practical context: Lifting a mass with a motor

A common AQA practical setup involves using a small electric motor to lift a mass $m$ through a vertical height $h$ . You measure the time $t$ it takes.

Useful output power: The work done is the gain in gravitational potential energy ( $m g h$ ). So the useful output power is $mght\frac{mgh}{t}$ .
Total input power: You measure the voltage $V$ and current $I$ supplied to the motor. The electrical input power is $P = I V$ .

When you do this experiment, you must be aware of errors:

Random errors: Variation in your reaction time when starting and stopping the stopwatch. You can reduce the impact of this by repeating the experiment and taking an average.
Systematic errors: The mass balance used to weigh the mass might not be zeroed properly (a zero error), or the ruler measuring the height might be consistently read from the wrong angle (parallax error).

Example

Efficiency of an electric hoist

An electric motor is used to lift a 2.0 kg mass through a vertical height of 1.5 m in a time of 4.0 s. The motor is connected to a 12 V supply and draws a current of 1.5 A. Calculate the efficiency of the motor as a percentage. Take $9.81\text{ N kg}^{-1}$ .

Calculate the useful work done (gain in potential energy):

\Delta W = mgh = 2.0 \times 9.81 \times 1.5 = 29.43\text{ J}

Calculate the useful output power:

P_{\text{out}} = \frac{\Delta W}{t} = \frac{29.43}{4.0} = 7.3575\text{ W}

Calculate the total electrical input power:

P_{\text{in}} = IV = 1.5 \times 12 = 18.0\text{ W}

Use the efficiency formula:

\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{7.3575}{18.0} = 0.40875

Convert to a percentage and state to an appropriate number of significant figures (2 s.f. based on the data provided):

\text{Efficiency} = 41\%

Exam technique

In the exam

Check the angle: When using $\cos \theta$ , explicitly draw the force vector and the direction of motion on the diagram if they aren't already there. Ensure $θ\theta$ is the angle directly between them.
Watch for constant speed: If a question mentions a vehicle moving at "steady speed" or "constant velocity", immediately think of two things: resultant force is zero (Driving Force = Resistive Forces), and you can use $P = F v$ .
Graph units matter: When estimating the area under a force-displacement graph, always check the axis prefixes. If force is in kN and displacement is in cm, your area calculation needs to account for $10^3$ and $10^{-2}$ to get an answer in standard Joules.

Self review

Check yourself

Can you explain why carrying a heavy box horizontally at a constant height requires zero work done on the box according to physics definitions?
When deriving $P = F v$ , what two basic equations are combined?
If an electric motor's efficiency drops when lifting heavier loads, what might be increasing to cause more energy to be dissipated?
How would you structure an experiment to find the useful power output of a student running up a flight of stairs?

Work, energy and power

Welcome to the foundation of mechanics! Almost every physical process you will study in A-Level Physics—from circuits to quantum mechanics—comes back to energy.

What you'll learn in this topic:

How to calculate the work done by a force, even when it acts at an angle to the direction of motion.
How to find the work done when a force is constantly changing (variable forces).
How to link power, force, and velocity for moving objects.
How to calculate the efficiency of real-world systems, such as an electric motor lifting a mass.

1. Work Done (Energy Transferred)

Definition

Work Done

Work done is the energy transferred when a force moves an object through a distance. One Joule (1 J) of work is done when a force of 1 Newton moves an object by 1 metre in the direction of the force.

To account for this, we use the general equation:

\cos \theta

Where:

$F$ is the magnitude of the applied force.
$s$ is the displacement of the object.
$θ\theta$ is the angle between the force vector and the direction of displacement.

Block pulled at an angle

Common Mistake

Using the wrong angle

Example

Pulling a sledge

Identify the given values: $120\text{ N}$ , $45\text{ m}$ , and $θ=30∘\theta = 30^\circ$ .
State the formula for work done at an angle:

\cos \theta

Substitute the values into the equation:

\times 45 \times \cos(30^\circ)

Calculate the result:

\approx 120 \times 45 \times 0.866 \approx 4676\text{ J}

State the final answer with appropriate significant figures and units: The work done is 4700 J (to 2 s.f.).

2. Variable Forces

Instead, we look at a force–displacement graph.

Force-displacement graph

Key Idea

Area under the graph

For any force-displacement graph, the area under the line or curve represents the total work done (energy transferred).

3. Power

If work done is how much energy is transferred, power is how fast that energy is transferred.

Definition

Power

Power is the rate of doing work, or the rate of energy transfer. It is measured in Watts (W). One Watt is equal to one Joule per second ( $s−11\text{ W} = 1\text{ J s}^{-1}$ ).

The fundamental equation for power is:

\frac{\Delta W}{\Delta t}

Where $ΔW\Delta W$ is the work done (or energy transferred) and $Δt\Delta t$ is the time taken.

Power for moving objects

There is an incredibly useful shortcut for calculating the power of a vehicle (like a car or a train) that is moving at a constant speed against resistive forces.

We know that $W = F s$ . If we substitute this into the power equation, we get:

\begin{aligned} P &= \frac{Fs}{t} \\ P &= F \left( \frac{s}{t} \right) \end{aligned}

Since velocity $\frac{s}{t}$ , we can replace that part of the equation to arrive at:

P = F v

Where $F$ is the driving force and $v$ is the velocity. This equation is essential for AQA exams. Note that for this formula to work, the force and the velocity must be in the same direction.

Example

Car engine power

Recognise that because the car is moving at a constant velocity, the driving force from the engine must perfectly balance the resistive forces. Therefore, the driving force $800\text{ N}$ .
Identify the velocity: $25\text{ m s}^{-1}$ .
State the formula linking power, force, and velocity:

P = F v

Substitute the values:

\times 25

Calculate the final answer and include units:

20000\text{ W} \quad (\text{or } 20\text{ kW})

4. Efficiency

\text{Efficiency} = \frac{\text{useful output power}}{\text{input power}}

(You can also calculate efficiency using energy instead of power: $energy\text{Efficiency} = \frac{\text{useful output energy}}{\text{total input energy}}$ )

Efficiency can be written as a decimal (between 0 and 1) or multiplied by 100 to be expressed as a percentage.

Tip

Sanity check

Efficiency can never be greater than 1 (or 100%). If your calculation gives you an efficiency of 1.2, you have probably put the larger input number on the top of the fraction by mistake!

Practical context: Lifting a mass with a motor

A common AQA practical setup involves using a small electric motor to lift a mass $m$ through a vertical height $h$ . You measure the time $t$ it takes.

Useful output power: The work done is the gain in gravitational potential energy ( $m g h$ ). So the useful output power is $mght\frac{mgh}{t}$ .
Total input power: You measure the voltage $V$ and current $I$ supplied to the motor. The electrical input power is $P = I V$ .

When you do this experiment, you must be aware of errors:

Random errors: Variation in your reaction time when starting and stopping the stopwatch. You can reduce the impact of this by repeating the experiment and taking an average.
Systematic errors: The mass balance used to weigh the mass might not be zeroed properly (a zero error), or the ruler measuring the height might be consistently read from the wrong angle (parallax error).

Example

Efficiency of an electric hoist

Calculate the useful work done (gain in potential energy):

\Delta W = mgh = 2.0 \times 9.81 \times 1.5 = 29.43\text{ J}

Calculate the useful output power:

P_{\text{out}} = \frac{\Delta W}{t} = \frac{29.43}{4.0} = 7.3575\text{ W}

Calculate the total electrical input power:

P_{\text{in}} = IV = 1.5 \times 12 = 18.0\text{ W}

Use the efficiency formula:

\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{7.3575}{18.0} = 0.40875

Convert to a percentage and state to an appropriate number of significant figures (2 s.f. based on the data provided):

\text{Efficiency} = 41\%

Exam technique

In the exam

Check the angle: When using $\cos \theta$ , explicitly draw the force vector and the direction of motion on the diagram if they aren't already there. Ensure $θ\theta$ is the angle directly between them.
Watch for constant speed: If a question mentions a vehicle moving at "steady speed" or "constant velocity", immediately think of two things: resultant force is zero (Driving Force = Resistive Forces), and you can use $P = F v$ .
Graph units matter: When estimating the area under a force-displacement graph, always check the axis prefixes. If force is in kN and displacement is in cm, your area calculation needs to account for $10^3$ and $10^{-2}$ to get an answer in standard Joules.

Self review

Check yourself

Can you explain why carrying a heavy box horizontally at a constant height requires zero work done on the box according to physics definitions?
When deriving $P = F v$ , what two basic equations are combined?
If an electric motor's efficiency drops when lifting heavier loads, what might be increasing to cause more energy to be dissipated?
How would you structure an experiment to find the useful power output of a student running up a flight of stairs?

Work, energy and power

1. Work Done (Energy Transferred)

2. Variable Forces

3. Power

Power for moving objects

4. Efficiency

Practical context: Lifting a mass with a motor

Recap questions

Work, energy and power

1. Work Done (Energy Transferred)

2. Variable Forces

3. Power

Power for moving objects

4. Efficiency

Practical context: Lifting a mass with a motor

Recap questions

Work, energy and power

1. Work Done (Energy Transferred)

2. Variable Forces

3. Power

Power for moving objects

4. Efficiency

Practical context: Lifting a mass with a motor

Recap questions

Work, energy and power Revision Guide

Work, energy and power

1. Work Done (Energy Transferred)

2. Variable Forces

3. Power

Power for moving objects

4. Efficiency

Practical context: Lifting a mass with a motor

Recap questions

Work, energy and power Revision Guide