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KS4 Forces and elasticity

You should already have worked through the Forces and their interactions and the Work done and Energy transfer sections.

Now we will look at how forces can bring about changes to the shapes of objects.

So, let's continue with the KS4 section, Forces and elasticity.

In the AQA KS4 section "Forces and Motion" you will learn about (assuming you haven't completed it yet) Isaac Newton's Laws of Motion. In these laws he describes how a Resultant Force can change the motion of an object; it can make the object speed up, slow down, start moving, stop or change direction. But there is one other thing that a force can do which Newton doesn't discuss, not because he didn't think about it, but because it doesn't concern MOTION.

A Force can also change the SHAPE or SIZE of an object.

This is true, isn't it?
If you squeeze a drink-can you crush it and change its shape, permanently.
If you get a blob of blu-tac and stretch it, you change its shape permanently.
If you bend your plastic ruler you change its shape, temporarily.
If you stretch a rubber band a small amount, you change its shape, temporarily.

So, a Force can change the shape of an object, sometimes permanently and sometimes temporarily. That is what this section is all about!

Let's start with the simple task of naming the forces involved, for example, in some of the above situations.

The 3 common shape changing forces

1. Squeezing or crushing a drink-can:
two forces are involved, acting inwards onto the object; we call these forces "compression forces".

2. Stretching a blob of blu-tac or a rubber band:
two forces are involved, acting outwards from the object; we call these forces "tension forces".

3. Bending a plastic ruler:
two forces are involved, acting inwards very much like the compression forces, but having the result of bending the object; we call these forces "bending forces".

Notice that in all cases of "compressing", "stretching" and "bending", two forces are involved; imagine trying to squash a can by only applying a force to one side!
Or trying to stretching a rubber band by pulling on only one end! It would be silly, wouldn't it.
But its an important point to note and one that examiners might ask about.

Before we move on, what would happen to a drink-can if you did only apply a force to one side?
Answer: it would be moved in the direction of the force; in other words it becomes a "forces and motion" situation, rather than a "forces and elasticity" situation.

What would happen to the rubber band if you only pulled on one end?
Answer: again, like the example above, it would be moved in the direction of the force in other words it becomes a "forces and motion" situation, rather than a "forces and elasticity" situation.

So, now that we know why TWO forces HAVE TO BE involved in shape changing, we can turn to something else touched on in the examples above; the idea of a permanent or a temporary change of shape.

Elastic or Inelastic Deformation

The title above tells you the "proper" terms for temporary and permanent changes of shape.

When we stretch a rubber band or a spring a small amount we temporarily deform it; this is an elastic deformation. The object will return to its original size once the deforming force is removed.

But, when we squash a drink-can or stretch a piece of blu-tac, we permanently deform it; this is an inelastic deformation. The object will not return to its original size once the deforming force is removed.

Exploring this idea in more detail:
Using the apparatus shown below -

If we hang 100g masses (equivalent to a weight or a force of 1 newton) from the spring and measure its extension as each 100g mass is added, then plot of the extension against the load or the force (in newtons), we get a straight line (see graph 1 below).

From the graph you can see that it is a straight line through the origin which means that the force applied to the spring and the extension of the spring are directly proportional.
Which further means that we can apply a bit of mathematics and write:

We can work out the value of k for the spring used to plot graph 1:

From the equation, you can see that k = F/e which is the same as saying that K equals the gradient of the line

So, choose a suitable value of F, say 10N, and read off its corresponding value of extension, which is 70mm or 0.07m

So, k = 10/0.07

k = 142.9 N/m

You can check that this value for k works, eg put e = .03m (30mm) into the F = k e equation and you should get a value for F close to 4N, according to the graph! Try it.

OK. So far, so good and so simple!
But what happens if we keep adding masses (and hence force) to the spring?
We find that eventually the proportional relationship between the force applied and the extension breaks down.
The point at which this happens is known as the limit of proportionality.
See graph 2 which shows what happens if we add just a little bit more force to the spring.

The graph is no longer a straight line after the limit of proportionality.

So, now that we know the whole story we can declare:

The extension of an elastic object, such as a spring, is directly proportional to the force applied, provided that the limit of proportionality is not exceeded.

And up to the limit, the elastic object obeys the equation, F = k e .

Another way of referring to the two regions of the graph is as LINEAR (the straight line section) and NON-LINEAR (the curved section).
We could say, "In the LINEAR region, only, the elastic object obeys the equation, F = k e."
Also, we could calculate the value of k using a graph in the LINEAR region (but not in the NON-LINEAR region).

Also notice that, although we have just been describing a "stretching" or a "tension" situation, the statement that we have just made and the equation apply equally to compression situations. We simply replace the word Extension for the word Compression in both the statement and the equation; although the letter "e" is still used even though it may mean "compression"!.

Before you tackle a few questions by yourself, let's have a look at one or two example calculations.

Ok, now its your turn.

Notice in Question 2 that you had to work out the Extension yourself. This is quite common, as is the question:
What is meant by "extension"?
To which your answer should be:
Extension is the "new length minus the original length";
or Extension = New Length - Original Length.

Work Done and Elastic Potential Energy

A force that stretches (or compresses) a spring Does Work and so Elastic Potential Energy is stored in the spring.
Provided the spring is not inelastically deformed, the work done on the spring and the elastic potential energy stored are equal.

So, when a man or a woman (or a boy or girl) pulls a pair of chest expander springs and holds them with arms outstreched, the Work Done in pulling them (ie the energy input) is equal to the Elastic Potential Energy, now stored in the set of springs.

The person doing the pulling Does Work and Transfers Energy to the springs where it is Stored as Elastic Potential Energy, Ee.
As long as the person holds the springs outstretched then the Elastic Potential Energy remains stored.

The same would be true if the person Compressed a spring or Bent a spring or Twisted a spring (eg as in "wound up a spring" like in a clock). Work is Done and Energy is Stored in the spring.

So, its an Energy Transfer from the person Doing the Work to the Energy Stored in the Spring or Springs.

Using this equation we can now calculate how much energy is stored in any stretched or compressed spring! All we need to know is its spring constant and the size of the extension or compression of the spring. Let' have a go at an example

Over to you.