**
4.3a An experiment to investigate the force applied to a spring
and resulting
extension**

How can we investigate the relationship
between a spring's extension and weights added to it?

What is the relationship between the
extension of a spring when increasing weights are connected to it?

If weights are attached to a firmly fixed
suspended spring, the spring will elongate depending on the value of the weight
attached. The greater the weight, the greater the spring is extended. The extra
length the spring attains is called the spring extension and this phenomena can
be systematically investigated using the simple apparatus described below.

When a weight is added and the spring is
static, the weight of the mass (and the spring itself) is counterbalanced by the
force of tension in the spring.

A
metre rule is fixed in a vertical position using a stand and several clamps.
Preferably with the linear scale pointing downwards!

The metre rule scale can be read in mm or cm.

From the top of the 1 m ruler a spring is
suspended with a hook-base is added to which extra mass can be added e.g. in 50g
increments (0.5 N force increase increment).

Fix a pointer onto the hook to which the
weights will be attached. If you can't fix a pointer on, just use the
base of the weight hook and sight it horizontally onto the scale.

You take the initial reading with no extra
weight on (other than the weight of the spring itself plus hook) and take the initial
reading on the scale. This first reading with no extra load applied is the
crucial starting point for all the successive measurements.

You then add an extra mass and take the scale
reading in mm/cm. From each successive reading you must subtract the initial
reading to obtain the true extension of the spring (I haven't actually shown
this in the table of results below).

Record a minimum of five observations carefully in a
prepared table and convert the mm/cm scale readings to the extension in m.

Below is a typical
**table of results**, already
corrected by subtracting the 'initial' reading.

50 g load
= 0.05 kg ~ force/weight of 0.5 N

For simplicity I've taken gravity as **
10 N/kg**, therefore every 50 g mass added equals an incremental weight
increase of 0.5 N.

From the data table you plot a graph of
total force (= tension) versus the total extension in the spring length (graph
sketched on the left).

(the tension in the spring equals the
force created downwards by the weight of mass).

Draw the line of best fit from the 0,0
graph origin.

If the spring is truly elastic a linear
graph is obtained.

This means a simple linear equation
describes the behaviour of the spring under these conditions.

The experiment is a simple proof and
demonstration that **the extension of a spring or any elastic material is directly
proportional to the force applied** (the load or weight in newtons).

This relationship is expressed with the
simple equation:

**
force = a spring constant x extension**

**
F = ke**

where **F** = the
applied force in newtons (N), **e** = the spring extension in metres (m)

and **k** is the spring (elastic) constant in N/m.

**The stiffer the spring** (or any
material being stretched)** the greater the spring constant**.

and you would see a steeper gradient
of the graph line, or a smaller gradient for a weaker spring,

the point illustrated by the
'theoretical' purple lines on the above graph.

This linear equation relationship between
force applied and the extension (or compression) of an elastic material is also
known as **Hooke's Law** of proportionality.

This can be stated as:

**
The extension of a spring or wire
(or any elastic object) is
proportional to the load (force applied)**

or

If the deformation of a material is proportional to the force applied, the
material is truly elastic and is said to obey Hooke's Law (a law of
proportionality).

From the graph you can calculate the spring
constant e.g. rearranging the equation (Hooke's Law equation)

**k** = F/e = gradient of the graph =
3.0/0/0.06 = **50 N/m**

**Five important points to note:**

This equation works for compression where e is the difference between the
full length and compressed length

The spring constant varies with the
material of the spring, the size and number of coils of the spring.

The stronger/stiffer the spring the greater the
value of the spring constant k.

This spring system is the basis for simple instruments used to measure the
weight of an object like a fish you have caught!

A
force meter is used for experiments in a laboratory - school, college,
university or in the engineering industry where it is used e.g. to test the
strength of materials.

Above is an illustration of a simple instrument for
weighing objects.

It is essentially a 'force meter' calibrated to read
in g and kg.

(so it takes into account gravity at the Earth's
surface, but it would be any good on the Moon or Mars with their
different strength of gravitational fields)

Prior to taking a reading the pointer should be
adjusted to read zero.

You place the object on the hook which stretches the
spring and read off its weight on the calibrated scale.

**Extending the investigation and alternative graphs**

Different groups in a class can look at different
springs, or if time permits each group of students can look at several
springs. The class results can be pooled and
graphs drawn.

Instead of plotting force
versus extension (which I prefer) you can plot extension versus
force.

Since F = ke, e = F / k, so the
gradient will be 1 / k, the reciprocal of the spring constant.

In the 'idealised' right-hand graph of spring
extension versus force, springs A and C results did not go beyond the
limit of proportionality.

However, the
results for spring B showed a deviation from linearity and the graph
curves upwards from point L, the limit of proportionality.

**
4.3b
What happens if you keep on increasing the force applied to an elastic material?**

In the above experiment, if you add even more
weights to the spring then the resulting graph of results may not be linear for
the higher weight readings.

This is because the spring is overstretched beyond its
elastic limit (the limit of proportionality).

**
Beyond point L Hooke's Law is no longer obeyed.**

In other words the non-linear section of the graph
is beyond **L, the limit of proportionality**
- the spring stretches more than you expect and the graph begins to curve over.

From zero force to L Hooke's Law is obeyed -**
the linear section of elasticity**.

After that, between point L and point D, the
stretching is greater than expected - non-linear graph, but the spring will
return to its original length - **the spring is still behaving elastically**, but
only for a relatively small further increase in the applied force.

Just because an object behaves
elastically, it doesn't mean that Hooke's Law is obeyed.

An elastic band is 'elastic' but it
doesn't obey Hooke's Law!

Eventually at point
**D, called the elastic limit**,
the force is too great for spring will not return to its original length -
permanent deformation beyond the limit of elasticity.

This happens with a repeatedly stretched
elastic band - eventually it breaks!

From point D onwards the spring behaves with
**plastic deformation**.

On the right-hand graph, an alternative
representation of the graphical data, I've indicated
the permanent extension showing the spring will NOT return to its original
length.

Sub-index of physics notes: FORCES
4. Elastic potential energy

**
Keywords, phrases and learning objectives for elastic potential energy**

Be able to describe an experiment to investigate the force applied to a
spring and resulting extension.

Explain the experimental procedure to validate
whether a stretched material obeys Hooke's Law - processing data
with graphs and calculations.

Know what happens if the material (e.g. a spring) is
stretched beyond the elastic limit - interpret a graph to aid your
explanation.

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Sub-index of physics notes: FORCES
4. Elastic potential energy