# Work, Energy and Power

## Work, Energy and Power Revision

**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?

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}

**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?

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}

**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}

**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 \% }

## Work, Energy and Power Example Questions

**Question 1:** A wind turbine does 3.1 \times 10^7 \: \text{J} of work converting kinetic energy into electrical energy in one day. What is the power output of the wind turbine?

**[2 marks]**

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}**Question 2:** A winch pulls an object with a force of 2700 \: \text{N} at 60 \degree to the horizontal. It moves 150 \: \text{m} horizontally. Calculate the work done.

**[3 marks]**

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}

**Question 3:** A bicycle motor produces 5 \: \text{kW} of power. The motor is 85 \% efficient at converting its power into kinetic energy. What is the maximum useful power output of the bicycle?

**[3 marks]**

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