Forces - 5.3 Forces and Elasticity (GCSE Physics AQA)

Forces and Elasticity

Forces Acting On Objects

Forces can cause stretching, bending or compression of an object. You need to understand each of these three effects, and give examples of forces causing each.

Stretching, bending and compression require multiple forces to be acting on an object. For example, if you just pushed a spring, then it would move in a certain direction. But if you pushed both ends of the spring, the spring would compress.

Stretching an Object

• Stretching is caused by pulling forces. To stretch a spring, we have to pull on it. This can be done by adding some small weights to the end of the spring, making it change in length.
• Pulling causes a spring to stretch. As the weights are added onto the spring, the length of the spring will increase. The spring has stretched from its original length to a new, longer length. This is called extension.

Compressing an Object

• Compression is caused by pushing forces. To compress a spring, we have to push it. This can be done by pushing the two ends of the spring together, making it change in length.
• Pushing causes a spring to squash. As the two ends are pushed together, the length of the spring will decrease. The spring has been compressed from its original length to a new, shorter length. This is called compression.

Bending an Object

• Bending forces can be push or pull. A force that causes a bend can either be a push force (Fig 2) or a pull force (Fig 3).
• The centre of an object will bend the most. When a force causes an object to bend, the centre of the object will be the ‘bendiest’ part. This causes the ends of the object to move closer towards each other. This is shown in Fig 16; the centre of the object is bending, whilst two ends move towards each other (in this case, they are moving upwards).

Changing the Shape of Stationary Objects

Applying One vs., Multiple Forces

For AQA exams, we need to learn the effects of different forces on stationary objects. We discuss above that stretching, bending and compressing require multiple forces.

• Applying a single force causes movement. When a single force is applied to a stationary object, the object will move in the direction of the force. There is no change of shape, since only one force has been applied.
• Applying multiple forces causes change of shape. When multiple (two or more) forces are applied to a stationary object, the object will change shape. The second force will prevent the object from moving. Also, the second force will act with the first force to change the shape of the object (see Fig 4).

• Changes of shape include stretching, compressing and bending. We previously discussed examples of stretching, compressing and bending in the section ‘Forces Acting On Objects’. Now, we know that all of these changes require two or more forces to be applied.

Elastic and Inelastic Deformation

• Deformation is a change in shape. An object is said to be deformed when it has changed shape and length, due to the effect of forces. There are two types of deformation: elastic and inelastic.
• Elastic deformation is reversible. If an object completely returns to ‘normal’ once all forces are removed, it is said to be elastically deformed. By normal, we mean that the object has completely returned to its original shape, size and length.
• Inelastic deformation is irreversible. if an object doesn’t completely return to ‘normal’ once all forces are removed, it is said to be inelastically deformed. Another term used to describe inelastic deformation is plastic deformation.

The Relationship between Force and Extension

Hooke’s Law

Hooke’s law is used to describe the relationship between force applied to an elastic object and the extension of the elastic object. It is a linear relationship, where force is directly proportional to the extension. If an object doesn’t obey Hooke’s law, there is a non-linear relationship between force and extension.

Where:

• Force, F, in newtons (N)
• spring constant, k, in newtons per metre (N/m)
• extension, e, in metres (m)

Extension is the difference between the original length and the new, longer length.

We will learn about the spring constant below.

The Spring Constant

Linear Cases

The spring constant is used to describe the ‘stiffness’ of a spring. If the spring constant is very high, it means that the spring is very stiff.

For AQA exams, we need to calculate the spring constant in linear cases.

The spring constant can be found by rearranging Hooke’s Law.

Hooke’s Law:

We want to find k, so we can rearrange the equation to give:

Calculating The Spring Constant

Question: A force of 10N is applied to a spring that is 50cm long. The spring extends to a value of 70cm. Calculate the spring constant, with units.

1. Write down the equation for Hooke’s Law.

F = k e

2. Find the extension.
The units of extension are in metres, so we will have to convert from cm to m in this example.

20cm = 0.2m and 70cm = 0.7m

The extension is 0.7 – 0.2 = 0.5m

3. Rearrange the formula.

F / e = k

4. Substitute for numbers.
Using the information given in the question, put in the value for force. The value for extension is the number we have just calculated.

5. Give the correct units.
In this question, we are asked to find the spring constant with its units.

Spring Constant = 20 N/m

Relationship Between Force and Extension

Direct Proportionality

• Extension occurs due to force. When we apply force to an elastic object (such as a spring) it will extend. The extension of an elastic object can be described by Hooke’s law.
• Force and extension are directly proportional. When we increase the force, the the extension will also increase by the same amount. This is called direct proportionality, which is often obeyed by elastic objects such as springs.
• Direct proportionality forms a straight line graph. As a graph is plotted of force (on the y axis) against extension (on the x axis), a straight line will start to form. This straight line will go through the origin, since when there is no force applied, there will be no extension.

Limit of Proportionality

The limit of proportionality is when the spring stops obeying Hooke’s Law. It is the point where the force applied to the spring is too large for the spring to handle. As a result, the extension of the spring increases drastically, and the two variables are no longer in a proportional relationship.

We learnt before about elastic and non-elastic deformation. After the limit of proportionality, the spring has non elastic deformation, and will not return to its original length once the load is removed.

For AQA exams, there are several important things to remember about the limit of proportionality:

You need to be able to describe the difference between a linear and non-linear relationship between force and extension:

• Linear relationships obey Hooke’s Law. Previously, we mentioned that there is a linear relationship between the force and extension of elastic objects. This linear relationship can be described by Hooke’s Law.
• Non linear relationships do not obey Hooke’s Law. Past the limit of proportionality, elastic objects will not obey Hooke’s law. This means that there is no longer a non-linear relationship between force and extension.

The Relationship Between Force and Compression

Compression vs. Extension

• Compression and extension are opposites. We know that extending a spring means increasing its length, so compressing a spring will decrease its length. The amount of force applied will be the same in both scenarios, but the direction of the force will be different (Fig 9).
• Extension is the difference between the original length and the new, longer length.
• Compression is the difference between the original length and the new, shorter length.

Compressing Elastic Objects

Similar to extension, we can use Hooke’s law to describe the compression of an elastic object. This means that there is a directly proportional relationship between the force on an object and the compression it experiences.

The equation linking force and compression is:

Where:

• Force, F, in newtons (N)
• spring constant, k, in newtons per metre (N/m)
• compression, c, in metres (m)

Work Done in a Spring

Energy Transfers

• Forces do work on springs. When a force acts on a spring to compress or stretch it, the force is doing work. We previously mentioned that doing work is simply another way to talk about the transfer of energy.
• Energy is transferred into elastic potential energy in the spring. As the force acts on an the spring, all of it’s energy will be transferred into elastic potential energy (EPE). This elastic potential energy will become stored in the spring and transferred into another form of energy later on.
• Usually the work done is equal to the EPE stored. When an object is elastically deformed, all of its energy will be transferred into EPE. Therefore the work done will be equal to the EPE stored. However, if the object is inelastically deformed (i.e. we have passed the limit of proportionality), this isn’t the case.

Calculating Work Done in a Spring

When a spring changes shape, for example when stretched or compressed, we can calculate the amount of work done and energy transferred to EPE.

Remember, we can only calculate the work done up to the limit of proportionality (the point at which force and extension are no longer in a proportional relationship).

This is the equation for calculating the work done when compressing or stretching a spring:

Where:

• Ee is elastic potential energy, measured in joules, J
• k is the spring constant, measured in Newtons per metre, N/m
• e is the extension, measured in metres, m.

The area under the force-extension graph is equal to the potential energy stored. This is the graph with the straight line not the curved line, because if we go beyond the limit of proportionality we cannot calculate elastic energy (or we can, but you don’t need to know about it!).

Question: A spring is stretched by 50cm. The spring constant for the spring is 30 N/m. What is the elastic energy (in J) stored in the spring?

1. Convert the extension into the correct units.
The spring constant is in newton-metres, so our extension needs to be in metres.

50cm = 0.5m.

2. Calculate the elastic energy

Ee = 0.5 ke²

E = 0.5 x 30 x 0.5 = 7.5J

Question: What is the relationship between work done and elastic` potential energy in a spring which is being stretched?

Mark 1 – Work done is proportional to elastic potential energy stored in the spring.

Mark 2 – Until the limit of proportionality.

Investigating Force and Extension of a Spring

Method

1. Set up a clamp stand with a spring. Attach a spring to the end of a clamp stand. Make sure that there are no weights hanging off the end of the spring.
2. Measure the unextended spring. Place a ruler just behind the spring, so that you can read off a value for the length of this unextended spring. Make a note of this value, ensuring that the unit you record is metres.
3. Add a weight to the spring. Next, attach a weight to the other end of the spring. The spring will extend.
4. Measure the extension. Wait until the spring has stopped moving and carefully read off a new value for spring length. Make sure that you are reading off your values at eye level to prevent any errors from occurring.
5. Record your results. Record your results in a table with the following headings: