# Hooke's law

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## Key Points

When a force is applied to an object it can change its size and shape.

The force will either stretch or compress the object. Some objects, like springs, obey

**Hooke’s law**.This law describes the relationship between the force applied and the spring’s extension or compression.

## Deformation

occurs when we apply a force to an object to change its shape. Applying a force to an object can cause it to either:

stretch - the object increases in length

compress - the object decreases in length

In general, the greater the force applied, the more deformed an object will become.

However, not all materials behave in the same way. **Elastic materials** will return to their original shape once the force is removed, whereas **inelastic materials** will change shape permanently and may even break.

## Hooke's law

When you apply a force to a material it can extend. The is the amount the length has increased by.

## More about Hooke's law

Watch the video of Brian Cox explaining Hooke's law.

## Investigating Hooke's law

To investigate , you can add masses to a spring and measure the length of the spring when the of the masses is increased.

This experiment investigates Hooke's law.

**Aim of the experiment**

To investigate how adding to a spring affects its extension.

**Method**

- Set up the apparatus as in the diagram below.

Add a 10 g mass to the holder and record the spring length.

Add another 10 g mass and record the new spring length.

Take away the previous spring length from the new length to calculate the extension (the difference).

Repeat by adding 10 g masses until 100 g is reached.

**Variables**

The is the mass (or the weight of the mass).

The is the extension.

include using the same spring for all the measurements.

**Expected results**

Remember to subtract the original length of the spring from each of your length measurements, so that you are recording the of the spring, and not its total length.

Remember that to calculate the force applied to the spring due to the weight of the mass, you need to convert the mass in grams into kilograms (divide by 1000) and then multiply this mass by the gravitational field strength (10 N/kg), using this equation:

\(Weight = mass \times gravitational~field~strength \)

For example, the first mass added has a mass of 10 g. To convert this to kg, divide by 1000:

10 ÷ 1000 = 0.01

Now work out the weight:

\(Weight = mass \times gravitational~field~strength \)

\(weight = 0.01 \times 10\)

\(weight = 0.1~N\)

**Record your results in a table**

Mass used (g) | Force (N) | Spring length (mm) | Extension (mm) |
---|---|---|---|

0 | 0 | 20 | 0 |

10 | 0.1 | 25 | 5 (extension = spring length - original spring length) |

20 | 0.2 | 30 | 10 |

30 | 0.3 | 35 | 15 |

40 | 0.4 | 40 | 20 |

50 | 0.5 | 46 | 26 |

The spring extended 5 mm each time the force was increased by 0.1 N. This follows Hooke’s law which states that the extension of an elastic object (like a spring) is to the force added.

When the dependent variable is directly proportional to the independent variable, the graph will have a straight line that goes through the origin (0,0).

Elastic objects will usually obey Hooke’s law. However, some will also reach their elastic limit. This is where so much force has been applied that the object will no longer return to its original length when the force is removed and has been permanently deformed.

## Try the experiment online

Try out this experiment in Atomic Labs. Go to the Physics lab and try the **Extension of a spring** experiment.