# Work, Energy and Power

A LevelAS LevelAQA

## Work, Energy and Power

One way in which energy can be transferred is when work is done. Work can be done when pushing an object, for example. Power is the rate at which this energy is transferred when work is done.

## Work Done

Work done is defined as “the amount of energy transferred when a force causes an object to move”. The equation for work done is:

$W = F \times s$

• $W$ is the work done in joules $\left(\text{J}\right)$.
• $F$ is the force in Newtons $\left(\text{N}\right)$.
• $s$ is the distance in metres $\left(\text{m}\right)$.

We can see from this equation that an alternative unit for energy is $\text{Nm}$.

When an object is moved over a certain distance, work is done to overcome the forces of friction, air resistance and any other drag forces. For this calculation to be applied, the measurement of distance must be parallel to the direction the force is applied.

Example: If an object is moved $20 \: \text{m}$ with a constant force of $1500 \: \text{N}$, how much work is done?

\begin{aligned} W &= F \times s \\ W &= \textcolor{bd0000}{1500} \times \textcolor{7cb447}{20} = \boldsymbol{30000} \: \textbf{J}\end{aligned}

However, often the force being applied is not parallel to the direction of motion. To calculate the work done, we need to resolve the force into its components to find the component in the direction of motion. This gives us 2 new equations:

$W = F \cos \theta \times s$

$W = F \sin \theta \times s$

Example: Calculate the work done moving the object below.

[2 marks]

$W = F \cos \theta \times s$

$\boldsymbol{W = 1500 \times \cos 25 \times 20 }$

$\boldsymbol{W = 27190} \: \textbf{J}$

A LevelAS LevelAQA

## Power

Power and work done are related.

Power is the rate of energy transfer from one form of energy to another or from one object to another. It is also the rate of work done. Therefore it can defined using the equation:

$P = \dfrac{\Delta W}{\Delta t}$

• $P$ is the power in watts $\left(\text{W}\right)$.
• $\Delta W$ is the work done in joules $\left(\text{J}\right)$.
• $\Delta t$ is the time taken for the work to be done, in seconds $\left(\text{s}\right)$.

Example: A cyclist does $12000 \: \text{J}$ of work on a bicycle to move for $30$ seconds. What is the cyclist’s power?

\begin{aligned} P &= \dfrac{\Delta W}{\Delta t} \\ \\ P &= \dfrac{12000}{30} = \boldsymbol{400} \: \textbf{W} \end{aligned}

Alternatively, we can use another equation to calculate power if we know the force and velocity of a moving object:

$P = F \times v$

• $v$ is the velocity of the object in metres per second $\left(ms^{-1}\right)$.

This is because:

$P = \dfrac{\Delta W}{\Delta t} = \dfrac{Fs}{\Delta t} = Fv$

Example: A man is moving house and carries a box, using a force of $200 \: \text{N}.$ He moves the box $20$ metres in $10$ seconds. Calculate the power.

[2 marks]

$V = \dfrac{s}{t} = \dfrac{20}{10} = \boldsymbol{2} \: \textbf{ms}^{-1}$

$P = F \times v = 200 \times 2 = \boldsymbol{400} \: \textbf{W}$

A LevelAS LevelAQA

## Force-Displacement Graphs

A force-displacement graph can be used to show how the force applied to an object changes over the displacement it has been moved. This is useful as forces are not always constant.

For example, a person pushing an object along the floor will need to vary their force to overcome changes in friction with the floor surface.

Work done can be calculated using a force-displacement graph by calculating the area under the graph. This is the equivalent of $\text{Work} = \text{Force} \times \text{Displacement}$.

Example: The graph below shows how the force of carrying an object changed with displacement. Calculate the work done carrying the object $3$ metres.

[3 marks]

Firstly, we need to split the graph into sections so that we can calculate the area of each section. This is because work done is equal to the area under a force-displacement graph.

$\text{Area of rectangle} = 16 \times 0.5 = \boldsymbol{8} \: \textbf{J}$

$\text{Area of triangle} = \dfrac{1}{2} \times 16 \times 2.5 = \boldsymbol{20} \textbf{J}$

$\text{Total work done} = 8 + 20 = \boldsymbol{28} \: \textbf{J}$

A LevelAS LevelAQA

## Efficiency

The efficiency of a system or a device is a measure of its ability to transfer the input energy into useful output energy. This can be expressed as a decimal between 0 and 1 or as a percentage from 0 to 100%. The equation for calculating efficiency as a percentage is:

$\text{Efficiency} = \dfrac{\text{Useful energy output}}{\text{Total energy input}} \times 100$

Or we can also use:

$\text{Efficiency} = \dfrac{\text{Useful power output}}{\text{Total power input}} \times 100$

No device or system can be $100 \%$ efficient but the more efficient a device, the less energy is wasted. If you wanted to calculate efficiency as a decimal instead of a percentage, simply don’t multiply by $100$.

Example: A television requires a power input of $200 \: \text{W}$. It transfers this power into $75 \: \text{W}$ of sound, $75 \: \text{W}$ of light and $50 \: \text{W}$ of heat. Calculate the efficiency of the TV.

[2 marks]

$\text{Total input energy} = 200 \: \text{w}$

$\text{Total useful power output} = 75 + 75 = 150 \: \text{W}$

$\text{Efficiency} = \dfrac{\text{Useful power output}}{\text{Total power input}} \times 100$

$\text{Efficiency} = \boldsymbol{\dfrac{150}{200} \times 100 = 75 \% }$

A LevelAS LevelAQA

## Work, Energy and Power Example Questions

Firstly, convert $1$ day to seconds:

$1 \times 24 \times 60 \times 60 = \boldsymbol{86400} \: \textbf{s}$

\begin{aligned} P &= \dfrac{\Delta W}{\Delta t} \\ \\ P &= \dfrac{3.1 \times 10^7 \: \text{J}}{86400} = \boldsymbol{360} \: \textbf{W} \end{aligned}

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A diagram can be drawn to represent the information in the question:

\begin{aligned} W &= F \cos \theta \times s \\ W &= 2700 \cos 60 \times 150 \\ W &= \boldsymbol{202500} \: \textbf{J} \: \text{or} \: \boldsymbol{0.2} \: \textbf{MW} \end{aligned}

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\begin{aligned} \textbf{Efficiency} &\boldsymbol{= \dfrac{\textbf{Useful power output}}{\textbf{Total power input}} \times 100} \\ \\ \text{Useful Power output} &= \dfrac{\text{Efficiency}}{100} \times \text{Total power output} \\ \\ \text{Useful Power output} &= \boldsymbol{\dfrac{85 }{100} \times 5000} \\ \\ \text{Useful Power output} &= \boldsymbol{4250} \: \textbf{W} \end{aligned}

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