Materials Tutorial 2 - Hooke’s Law


If we load a spring, we find that the extension (code e) or stretch is proportional to the force (code F).  If we double the force, we double the stretch.  This is called Hooke's Law.

   F µ e


Ţ F = ke

The constant of proportionality is called the spring constant (or force constant) and is measured in newtons per metre (N m-1). 



Remember to convert the extension into metres and the load into Newtons before working out the spring constant.


In the AQA syllabus, the extension is given the code Dl, i.e. the change in length.  In some text books, you may see it given as x.  In these notes, you will see all three used.


Question 1

When a 500 g mass is placed on a spring, it stretches by 12 cm. 


What is its spring constant in N m-1?            



We can plot this as a graph:

We can see that the graph is a straight line and that the gradient gives us the spring constant.  That is why we have the extension on the horizontal axis.  We say that force is directly proportional to the extension:


The same is true if we apply a squashing force, a compression force.


Springs in series and parallel

It may seem strange to bring in terms more familiar in electric circuits, but we can have springs arranged in series and parallel.  The picture below shows a single spring, then two springs in series.


If we load the first spring with a weight W, we see that it extends by e.  Now we attach a second identical spring in series, and put on the weight W.   The same force acts through each spring, so each spring stretches by e.  Therefore the total stretch is 2e.


Since k = F/e, we can easily see that the spring constant halves.


Question 2

When a 500 g mass is placed on a spring, it stretches by 12 cm.  A second identical spring is now placed below the first.


How far do the two springs in series stretch?


What is the new spring constant in N m-1?            




The picture below shows the same two springs in parallel:

The springs are identical to the ones before.  This time the weight W is shared out equally between the two.  Since each spring has 0.5 W acting on each one, the stretch is 0.5 e.  The spring constant of the parallel springs is therefore k = W ÷ 0.5e = 2k.


Question 3

When a 500 g mass is placed on a spring, it stretches by 12 cm.  A second identical spring is now placed alongside the first.


How far do the two springs in parallel stretch?


What is the new spring constant in N m-1



Cars can be modelled as a block of mass m, placed on 4 springs each of spring constant k.



Question 4

A car of mass 1600 kg is placed on four identical springs.  Each spring is seen to be compressed by a distance of 5 cm.  What is the spring constant of each spring?

Use g = 9.8 N kg-1




Elastic Strain Energy

The energy is the area under the force-extension graph.  How do we achieve this result?


If we stretch the spring by a tiny amount dl, we do a tiny job of work:



This is shown by the little rectangle.


Do it again, we get another little rectangle:

Now fill in all the little rectangles:

All the little rectangles give area under the graph to give:



A neater graph is shown here:


So we can use this result to say:

This result can also be obtained using the  process of integration, which is part of the branch of mathematics called calculus.  The expression will use e as the code for extension:



You are not expected to know this for AS, although you will be if you study Physics at University.  All we need to say here is that the energy is the area of the triangle.


We can derive two further expressions from this result:


Question 5

Show that E = ˝ ke2.                                                                                   


Question 6

Use a similar method to show that E = 1/2 F2/k


Question 7

A car of mass 1600 kg is placed on four identical springs.  Each spring is seen to be compressed by a distance of 5 cm.  What is the energy in each spring?

Use g = 9.8 N kg-1



When we stretch the spring, we have to do a job of work.   If we release the spring, we can recover that energy, which is called the elastic strain energy.  Ideally we recover all of it but in reality a certain amount is lost as heat.  This lost energy is called hysteresis.  This is shown on the graph in the yellow triangle.

From all hysteresis graphs, we should note:


Hysteresis, please, not hysteria.



A rubber band is an example of a material that does not obey Hooke's Law.  The force extension graph looks like this:

Rubber is an example of an elastomer.  It is made up of long polymer chains that are tangled up with cross-links (disulphide bridges) between the chains.  The chains with no load are rather like spaghetti.  When the load is small, a large extension is seen because the chains are simply untangling. 

Once the chains are untangled, then the bonds between atoms are stretched.  This requires a greater force with a small extension.

We can see from the graph that there is a large hysteresis.  This is because work has to be done to break cross-links as the elastomer is stretched.  When the elastomer is relaxed, the cross-links reform, and give out heat as they do so. 


Natural rubber is quite soft and flexible as there are relative few disulphide bridges.  Hard rubber is vulcanised, which means that more disulphide bridges are added.


Energy Changes in Springs

When a spring is stretched (in tension) or squashed (in compression), it stores energy as elastic potential energy.  The energy can then be released to do a job of work, for example, to lift a mass against gravity.



The extension of a spring is directly proportional to the force (Hooke’s Law).  Consider a mass, m, put onto a spring of spring constant k so that so that it stretches by an extension l.


The energy changes are as follows:

The force on the spring = mg, and the stretching tension = kl.



 mg = kl


Suppose the spring is pulled down by a distance x below the rest position.  Now the stretching force becomes:


 k(l + x)


This is also the tension in the spring acting upwards. 


So the restoring force, Fup, is given by:

Fup = k(l + x) – mg. 


This is because mg is the weight, which always acts downwards.


Since kl = mg, we can write:


Fup = kl + kx – kl = kx


The height to which the mass will bounce is also x, the displacement by which the mass was pulled down.


The total energy is:


The total energy in the system remains the same.  But it is converted from potential to kinetic and back again. 


When released, kinetic energy is formed.


If we consider the speed, we can write:



Therefore the speed is directly proportional to the extension.


We can show the energy interchange in this graph:



The shape of this graph is sinusoidal, because the mass-spring system oscillates (bounces) in simple harmonic motion.  You will study this in the A-level year.


Like everything in Physics, it isn’t perfect.  Energy is lost as heat, so the amplitude of the bounce decreases.  The energy lost can be shown to be a constant fraction of the total.


Question 8

A spring has a 300 g mass added to it.  It has a spring constant of 35 N m-1

a. Calculate the amount by which it is stretched if the mass hangs freely.

b. The spring is then displaced a further 10 mm.  Calculated the force applied. 

c. Calculate the energy in the stretched spring.

Use g = 9.8 N kg-1

Give your answers to an appropriate number of significant figures.