Work, energy and power

What you'll learn

What work done means and how it links forces to energy transfer.
How to use $Ek=12mv2E_k = \frac{1}{2}mv^2$ and $Ep=mgΔhE_p = mg\Delta h$ in mechanics problems.
How to calculate work from a force-displacement graph.
How power and efficiency describe energy transfer in real systems.

The starting point: forces, displacement and energy

A force is a push or pull on an object, measured in newtons (N). A displacement is a change in position in a stated direction, measured in metres (m).

In mechanics, forces can change an object’s motion, but they can also transfer energy. This is where the idea of work done comes in.

Definition

Energy

Energy is a conserved quantity measured in joules (J). It can be stored in different ways, such as in a moving object, a raised object, or a stretched spring, and transferred between stores.

Work done by a force

When a force moves an object through a distance, energy is transferred. That energy transfer is called work done.

Definition

Work done

Work done is the energy transferred by a force when its point of application moves through a displacement. For a constant force acting in the direction of motion:

W = F s

where $W$ is work done in joules (J), $F$ is force in newtons (N), and $s$ is displacement in metres (m).

One joule is the work done when a force of 1 N moves an object 1 m in the direction of the force. So:

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

If the force is at an angle to the displacement, only the component of the force parallel to the displacement does work:

Fs\cos\theta

Diagram showing work done by an angled force using the parallel component of force

Key Idea

Only the parallel component matters

A force does positive work if it has a component in the direction of motion, negative work if it opposes the motion, and no work if it is perpendicular to the motion.

Example

Work done by an angled pull

A student pulls a sledge with a force of 45 N at $30∘30^\circ$ above the horizontal for 12 m. Calculate the work done by the pulling force.

Resolve the force in the direction of motion: $NF_{\parallel}=F\cos\theta=45\cos30^\circ=39 \text{ N}$ .
Multiply the parallel component by the displacement: $JW=F_{\parallel}s=39 \times 12 = 4.7 \times 10^2 \text{ J}$ .
Compare with the maximum possible value: it is less than $45 \times 12$ because not all of the force acts horizontally.

Common Mistake

Using the whole force

If the force is not parallel to the displacement, do not use $W = F s$ with the full force. Use the parallel component, or use $W=Fscos⁡θW=Fs\cos\theta$ .

Work done from a force-displacement graph

The equation $W = F s$ works when the force is constant. If the force changes during the motion, the work done is found from the area under a force-displacement graph.

Force-displacement graphs showing work done as area under the graph

The area has units of N m, which are the same as joules. This is especially useful for springs, where the force often increases as the extension increases.

Definition

Elastic potential energy

Elastic potential energy is energy stored when an object is stretched or compressed and can return to its original shape.

Example

Using the area under a force-displacement graph

A spring is stretched from 0 to 6.0 cm. The force rises uniformly from 0 to 18 N. Calculate the elastic energy stored.

Convert the extension into SI units: 6.0 cm = 0.060 m.
Use the area of the triangular force-extension graph: $JW=\frac{1}{2} \times 0.060 \times 18 = 0.54 \text{ J}$ .
Interpret the energy transfer: assuming no energy is dissipated, 0.54 J is stored as elastic potential energy.

Common Mistake

When the force changes

Use the area under a force-displacement graph when the force is not constant. The simple equation $W = F s$ only applies to a constant force component.

Kinetic energy and gravitational potential energy

A moving object has kinetic energy.

Definition

Kinetic energy

Kinetic energy is the energy stored by an object because it is moving:

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

where $m$ is mass in kg and $v$ is speed in m s $^{-1}$ .

An object in a gravitational field can have gravitational potential energy because of its height.

Definition

Gravitational potential energy

The change in gravitational potential energy near Earth’s surface is:

ΔEp=mgΔh\Delta E_p = mg\Delta h

where $m$ is mass in kg, $g$ is gravitational field strength, usually 9.81 N kg $^{-1}$ , and $Δh\Delta h$ is the change in height in m.

The zero level for gravitational potential energy is chosen by you or the question. Usually, only the change in height matters.

Example

Comparing kinetic and gravitational potential energy

A 1.2 kg ball is moving at 4.0 m s^-1. It is then lifted by 0.80 m. Compare its kinetic energy with the increase in gravitational potential energy.

Calculate the kinetic energy: $JE_k=\frac{1}{2}mv^2=\frac{1}{2}\times1.2\times4.0^2=9.6 \text{ J}$ .
Calculate the increase in gravitational potential energy: $J\Delta E_p=mg\Delta h=1.2\times9.81\times0.80=9.4 \text{ J}$ .
Compare the values: the two energy changes are very similar, so lifting the ball by 0.80 m stores about the same energy as its motion at 4.0 m s^-1.

Tip

Watch the symbol W

The symbol $W$ can mean work done, but weight is also often written as $W = m g$ . Use the units to separate them: work done is in joules (J), weight is in newtons (N).

Conservation of energy

The principle of conservation of energy says energy cannot be created or destroyed. It can only be transferred from one store to another.

In ideal mechanics problems, energy transfers may be treated as perfectly useful. For example, if air resistance is ignored, gravitational potential energy lost can become kinetic energy gained.

In real systems, some energy is usually dissipated. Dissipated energy is transferred to less useful stores, often heating the surroundings.

Key Idea

Energy accounting

A useful way to write many mechanics problems is:

dissipated\text{initial energy} = \text{final useful energy} + \text{energy dissipated}

The work-energy principle connects forces directly to energy:

force=ΔEk\text{work done by the resultant force} = \Delta E_k

For a constant resultant force, this also links back to dynamics. Using $F = ma$ and $v^2=u^2+2as$ :

\frac{1}{2}m(v^2-u^2)

So the work done by the resultant force changes the kinetic energy.

Example

Finding speed with energy losses

A 0.60 kg trolley starts from rest and rolls down a ramp, dropping vertically by 0.75 m. Work done against resistive forces is 1.3 J. Calculate its final speed.

Set up the energy account: $mgΔh=Ek+Edissipatedmg\Delta h = E_k + E_{\text{dissipated}}$ , so $Ek=mgΔh−EdissipatedE_k = mg\Delta h - E_{\text{dissipated}}$ .
Substitute the values: $JE_k = 0.60\times9.81\times0.75 - 1.3 = 3.1 \text{ J}$ .
Use kinetic energy to find the speed: $s−1v=\sqrt{\frac{2E_k}{m}}=\sqrt{\frac{2\times3.1}{0.60}}=3.2 \text{ m s}^{-1}$ .

Common Mistake

Forgetting dissipated energy

If the question mentions friction, air resistance, braking, or work done against resistive forces, do not assume all gravitational potential energy becomes kinetic energy.

Power

Power tells you how quickly energy is transferred or work is done.

Definition

Power

Power is the rate of energy transfer:

\frac{E}{t}

or, for work done:

\frac{W}{t}

Power is measured in watts (W), where 1 W = 1 J s $^{-1}$ .

If a constant force acts in the direction of motion, then:

\frac{Fs}{t}

Since speed is displacement per unit time, $v=stv=\frac{s}{t}$ , this becomes:

P = F v

This is very useful for motors, lifts, vehicles and athletes.

Definition

Efficiency

Efficiency compares useful output energy or power with total input energy or power:

input\eta = \frac{\text{useful output}}{\text{total input}}

To give efficiency as a percentage, multiply by $100%100\%$ .

Example

Calculating power and efficiency

A motor takes in 600 W of electrical power and lifts a 50 kg mass vertically at a steady speed of 0.80 m s^-1. Calculate the useful output power and the efficiency.

Use steady speed to identify the lifting force: the upward force balances the weight, so $NF=mg=50\times9.81=4.9\times10^2 \text{ N}$ .
Calculate the useful output power: $WP_{\text{out}}=Fv=490.5\times0.80=3.9\times10^2 \text{ W}$ .
Compare output power with input power: $η=PoutPin×100%=392600×100%=65%\eta=\frac{P_{\text{out}}}{P_{\text{in}}}\times100\%=\frac{392}{600}\times100\%=65\%$ .

Tip

Average or instantaneous power

Use $P=EtP=\frac{E}{t}$ for average power over a time interval. Use $P = F v$ when you know the force and speed at that moment, or for steady motion.

Exam technique

In the exam

Start by deciding the energy transfer: work done, kinetic energy, gravitational potential energy, elastic energy, or dissipated energy.
Convert all distances, extensions and heights into metres before calculating energy.
For force-displacement graphs, find the area under the graph and remember that N m = J.
If resistive forces are mentioned, include energy dissipated instead of assuming perfect energy transfer.
For efficiency, compare useful output with total input using either energies or powers, but do not mix mismatched time intervals.

Self review

Check yourself

Why does a force perpendicular to the motion do no work?
When would you use the area under a force-displacement graph instead of $W = F s$ ?
A motor lifts a mass at constant speed. Why can you use $P = F v$ with $F = m g$ ?

Work, energy and power

What you'll learn

The starting point: forces, displacement and energy

Work done by a force

Work done from a force-displacement graph

Kinetic energy and gravitational potential energy

Conservation of energy

Power

Recap questions

Work, energy and power Revision Guide