# Elasticity

The second thing that you need to know about the F – x graph is how the spring constant can be determined from the graph. Spring constant is the ratio of force exerted on a spring to the extension or compression of the spring. Therefore, it is equal to the gradient of the graph.

Let’s say we need 20N to stretch the string for 5cm. The gradient of the graph is equal to 20N over 5cm, which is equal to 4 N/cm. (gradient of a straight line is equal to the height over the width of  the triangle.). Therefore, the 8spring constant of this spring is 4N/cm. Well, when finding the spring constant, make sure that you 8do not include the curve of the graph. Spring constant can only be determined from the straight line of the graph.

In previous lesson, we have discuss potential energy. There are 2 types of potential energy: the gravitational potential energy and the elastic potential energy. In this post, we will focus on the elastic potential energy.

Elastic potential energy is the energy stored in elastic materials as the result of their stretching or compressing. For example, if we compress a spring, elastic potential energy will be stored in the spring. If we stretch the spring, elastic potential energy will be stored or even we bent the spring, potential energy will be stored. Elastic potential energy can be stored in rubber bands, bungee chords, trampolines, springs, an arrow drawn into a bow.

The equation for elastic potential energy is ½ Fx or ½ kx2, where F is the force used to stretch or compress the elastic object, x is the extension or compression caused by the force, k is the spring constant.

For example, let’s say we hang a 2kg mass on a spring, the spring extends 10cm. What is the elastic potential energy stored in the spring?

The force exerts on the spring is the weight of the mass, which is 20N.  The extension is 0.1m. The elastic potential energy, EP = ½ Fx. Plug in the numbers. The elastic potential energy is equal to 1J.

Take notes that 1 Joule is equal to 1 Nm but not 1 Ncm. Therefore we need to convert the unit of extension into meter if we want to use Joule as the unit of energy, to make the unit consistent.

In previous lesson, we have discussed power. In this lesson, we are going to discuss elasticity. At first, we will discuss Hooke’s law. Under Hooke’s law, we will discuss what is elastic limit, the equation deduced from Hooke’s law, the spring constant, the 2 graphs related to Hooke’s law: the F-x graph and the x-F graph, and the work done to stretch or compress a spring and the potential energy stored inside the spring when it is stretched or compressed.

After finishing Hooke’s law, we will proceed to system of springs, where we will discuss 2 types of arrangement of spring: parallel and series arrangement. After that, we will proceed to the factors affecting the stiffness of a spring, including the length, the diameter of the spring and the diameter of the wire of the spring, material used to make the spring and arrangement of springs, such as parallel and series arrangement. These are the factors that will affect the stiffness of a spring or a system of springs.

Elasticity of an object is the ability of the object to return to its original shape when the force exerted on it is removed.

For example, when forces are applied on a balloon, the shape of the balloon changes. However, when the forces are removed, the balloon return to it’s original shape. In such a case, we say the balloon is elastic.

Elasticity of objects is discussed in Hooke’s law. There are many things that you need to know about Hooke’s law.

1. First, of course you need to know what is stated in Hooke’s law.
2. Second, you need to know what is elastic limit of an object.
3. Third, you need to know the equation deduced from Hooke’s law and what is spring constant.
4. Forth, you need to know how to analyse the Force-Extension graph.
5. Fifth, you need to know how to find work done and elastic potential energy by using equation or from the Force-Extension graph.
6. Last but not least, you need to know the difference between the Force-Extension graph and the Extension-Force graph.

Hooke’s law states that the extension of a spring is directly proportional to the applied force provided the elastic limit is not exceeded.

For example, there is a spring of length 10cm. When a 10N force exerts on the spring, it extends 2cm. When the force is increased 3 times to 30N, the extension also increases 3 times and becomes 6cm and when the force is increased 5 times to 50N, the extension also increases 5 times and becomes 10cm. This is what we mean  “the extension of a spring is directly proportional to the applied force”.

However, this is only true if the extension does not exceed the elastic limit of the spring.

Elastic limit is the maximum force that can be applied to the spring before it become permanently deformed.

For example, there is a spring of length 10cm. If a force applied on the spring, it will extend. If the force is lower than 50N, when the force is removed, the length of the spring will come back to 10cm.

Let’s see another example. If the force applied exceed 50N, let’s say 80N. The spring will extend even longer. When the force is removed, the length of the spring does not return to its original length. Instead, it is longer than its original length. If this is the case, we say the spring is permanently deformed.

So, if the force applied on the spring is less than 50N, the spring will return to its original length if the force is removed. If the force exceed 50N, the spring will permanently deformed and will never return to its original length even though the force is removed.

The 50N of force is called the elastic limit of the spring. It is the maximum force that we can applied on an elastic object and the object is still not permanently change its length or shape.

In conclusion, if the force applied on an elastic object is lower than the elastic limit, the object maintain its elasticity. If the force exceeds the elasticity, the object loses it’s elasticity and the shape change permanently.

Well, Hooke’s law states that the extension of a spring is directly proportional to the applied force. In mathematics, we write F directly proportional to x, where F is the force applied on the spring and x is the extension. When writing in equation, we need to multiply a constant to the variable. Let’s say the constant is k, the equation will be F = kx. This is the equation of Hooke’s Law.

We call the “k” the spring constant of the elastic object. Spring constant is the ratio of force exerted on a spring to the extension or compression of the spring.  In equation, we write k = F/x. For example, if a force of 10N exerts on a spring and the spring extends for 5cm, then the spring constant of this spring is equal to 10N/5cm, equal to 2N/cm. When we plot a graph of the force exerted on a spring against the extension of the spring, we will probably get something like this. In SPM, there are a few things that you need to know about this graph.

First of all, we can divide this graph into 2 parts. In the first part, the graph is a straight line. This indicates that the force is directly proportional to the extension of the spring. This is the range where the Hooke’s Law applied. The second part of the graph is a curve. This shows that in this part, the force is no longer directly proportional to the extension. The point where the straight line turn into the curve is the elastic limit. As what we have discussed in previous slide, elastic limit is the maximum force that can be applied to the spring before it become permanently deformed. From the graph, we can see that when the spring is extended between its original length and the elastic limit, the Hooke’s law is still applied. Beyond the elastic limit, Hooke’s law is no longer applied.

The second thing that you need to know about the F – x graph is how the spring constant can be determined from the graph. Well, spring constant is the ratio of force exerted on a spring to the extension or compression of the spring. Therefore, it is equal to the gradient of the graph.

Let’s say we need 20N to stretch the string for 5cm. The gradient of the graph is equal to 20N over 5cm, which is equal to 4 N/cm. (gradient of a straight line is equal to the height over the width of  the triangle.). Therefore, the spring constant of this spring is 4N/cm. Well, when finding the spring constant, make sure that you do not include the curve of the graph. Spring constant can only be determined from the straight line of the graph.

The third thing that you need to know regarding the F – x graph is how the work done to stretch the spring can be determined from the F – x graph. Well, in an F – x graph, the work done is equal to the area between the graph and x-axis.

Well, if you still remember the work – energy relationship, then you probably know that the amount of energy stored in a spring is equal to the work done to stretch the spring. In other words, the elastic potential energy stored in the spring is also equal to the area between the graph and x-axis.

In previous lesson, we have discuss potential energy. There are 2 types of potential energy: the gravitational potential energy and the elastic potential energy. In this slide, we will focus on the elastic potential energy.

Elastic potential energy is the energy stored in elastic materials as the result of their stretching or compressing. For example, if we compress a spring, elastic potential energy will be stored in the spring. If we stretch the spring, elastic potential energy will be stored or even we bent the spring, potential energy will be stored. Elastic potential energy can be stored in rubber bands, bungee chords, trampolines, springs, an arrow drawn into a bow.

The equation for elastic potential energy is ½ Fx or ½ kx2, where F is the force used to stretch or compress the elastic object, x is the extension or compression caused by the force, k is the spring constant.

For example, let’s say we hang a 2kg mass on a spring, the spring extends 10cm. What is the elastic potential energy stored in the spring?

The force exerts on the spring is the weight of the mass, which is 20N.  The extension is 0.1m. The elastic potential energy, EP = ½ Fx. Plug in the numbers. The elastic potential energy is equal to 1J.

Take notes that 1 Joule is equal to 1 Nm but not 1 Ncm. Therefore we need to convert the unit of extension into meter if we want to use Joule as the unit of energy, to make the unit consistent. In SPM, for questions about the graph of elasticity, most probably we will see the F – x graph. However, sometime, you may encounter the x – F graph, especially in physics paper 3. Therefore, it is important for us to know what’s the differences between these 2 graphs.

Well, in previous slide, we have learned that the gradient of the F – x graph is equal to the spring constant. In x – F graph, the gradient is not the spring constant anymore. Instead, the gradient of the graph is equal to the reciprocal of the spring constant, k.

How about the work done to stretch the spring? In both graphs, the work done to stretch the spring is equal to the area between the graph and x-axis.

If 2 or more springs is joined and worked together, we call it a system of springs. Spring can be joined in series or parallel, or combination of these 2. For example, these 2 springs are in series, these are parallel, these are combination of series and parallel.

Well, a system of springs can act just like a single huge spring. It obeys Hooke’s Law. It also has its own spring constant. Let’s call it the spring constant of the system. How springs joined together will affect the spring constant of the system.

We will discuss this in the very next post.

Let’s say we have a spring. If the spring is pulled by a 12N force, it extend 2cm. The spring constant of the spring is k = 12N/2cm, which is equal to 6N/cm.

If we join 2 similar springs in series, they become a system of springs. What will happen if we exert a 12N force on the spring system, as what we do before? Well, this 12N force will pull the lower spring and make it extend 2cm. At the same time, the force is also pulling the upper spring, and makes it extend for 2cm as well. As a result, the spring system extends 4cm in total, and the spring constant of the system, k is equal to 12N/4cm, which is equal to 3N/cm. If we add another similar spring to the system, make it 3 springs in series, and again pull it by a force of 10N. The system will extend 6cm, and the spring constant will become 2N/cm.

In conclusion, when we join springs in series, the extension of the system of spring will be x’ = nx, where n is the number of springs and x is the extension of a single spring. The spring constant of the system of spring will be  k’ = k/n, where k is the spring constant of a single spring and n is the number of springs. The more springs connected in series, the lower the spring constant and hence the lower the stiffness of the spring system.

What will happen if the springs are connected in parallel? Well, if the springs are connected parallel, the 12N force will be shared equally among the 2 springs, each of the springs withstand a pull of 6N. As a result, both springs extend 1cm instead of 2cm, and the extension of the spring system is also 1cm. The spring constant, k is equal to 12N/1cm, which is equal to 12N/cm. The spring constant is doubled compare with a single spring.

If 3 springs are connected parallel and pulled by a 12N force, the extension will become 2/3 cm and the spring constant of the system will become 18N/cm.

In conclusion, when we join springs in parallel, the extension of the system of springs will be x’ = x/n, where n is the number of springs and x is the extension of a single spring. The spring constant of the system of springs will be  k’ = n x k, where k is the spring constant of a single spring and n is the number of springs. The more springs connected in parallel, the greater the spring constant and hence the stiffer the system of springs.

Well, the weight of the 3kg load is 30N. When pulled by a force of 10N, a single spring will extend 2cm. Therefore, when pulled by a force of 30N, the spring will extend 6cm.

The 2 springs in series will extend 2 x 6 = 12cm in total. The 2 springs in parallel will extend 6cm ÷ 2 = 3cm, in total. The total extension of the system is 12cm + 3cm = 15cm. The length x = initial length + extension = 3 x 10cm + 15 cm = 45cm. The 10cm is the initial length of the springs.