FORCES
4.
Elasticity and energy stored in a spring,
experimental investigations and
calculations
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Doc Brown's school physics revision notes: GCSE
physics, IGCSE physics, O level physics, ~US grades 8, 9 and 10
school science courses or equivalent for ~14-16 year old students of
physics
What is an elastic material? Why does stretching a spring involve
doing work? How is energy stored in an elastic
material? Does a compressed, bent or stretched
elastic material always return to its original shape on releasing the tension? How do you calculate the potential
energy stored in a spring? What is the relationship between the
extension of a spring when increasing weights are connected to it? How can we investigate the relationship
between a spring's extension and weights added to it?
Sub-index for this page
(a)
Introduction to subjecting materials to
physical forces
(b)
Deformation of a
material by bending, stretching or compressing
(c)
Work is done in
stretching or compressing a material
(d)
An experiment to investigate the force applied to a spring
and resulting
extension
(e)
What happens if you keep on increasing the force applied to an elastic material?
(f)
Particle model of an elastic energy store
(g)
Elastic potential energy
- energy stored and work done - calculations
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(a) Introduction
to subjecting materials to physical forces
When a material is subjected to two equal and
opposite forces they may change the relative positions of the particles i.e. may
change its shape.
This is what happens if you stretch a spring or rubber band,
squash moulding clay or bend a strip of plastic or metal.
When springs or elastic material are
stretched, elastic potential energy is stored in the system.
When the forces causing the stretching is
removed, the spring or elastic returns to its original length (shape)
The more an elastic material is
stretched, the greater the elastic potential energy store.
If the forces of compression/stretching are removed, different
materials behave in different ways.
The spring or rubber band are likely to
return to their original shape and length, displaying their elastic properties.
The clay will stay in its new shape.
The plastic strip like a ruler may return
to its original shape.
A strip of the soft metal like lead will be permanently
deformed but a strip of strong steel, if not bent too far, is likely to return
to its original shape - this happens with a clock spring in a clockwork clock.
Materials which have a tendency to return to
their original shape are called
elastic.
Those that do not are referred to as inelastic
(non-elastic). The more elastic a material, the greater its ability to regain
its original shape.
Engineers designing structures like bridges
need to understand the different ways forces operate. When the applied forces
stretch materials the material is under
tension. If the applied forces squeeze or compress the material,
it is described as being under
compression. In a bridge, the supports on either side are under
compression but the main body of the bridge carrying traffic is under tension -
this could be the roadway and/or the cables in a suspension bridge.
TOP OF PAGE and
sub-index
(b) Deformation of a
material by bending, stretching or compressing
If you want to bend, stretch or compress an
elastic object you must apply a force.
As explained above an elastic material
is one that can be deformed in shape by applying a force and returns to its
original shape if the forces are removed - springs and rubber bands clearly
exhibit elasticity.
If an object doesn't return to its original length
and shape it is inelastic.
The
extension of a simple spring by stretching it with an applied force can be used
to illustrate these points.
Initially the extension of a stretched
spring is proportional to stretching force applied (e.g. adding increasing
weight). This simple law is obeyed until you reach the limit of
proportionality (L on
the graph) - in other words the graph is linear up to point L.
After that, between point L and point D, the
stretching is greater than expected - non-linear graph, but the spring will
still return to its original length - the spring is still behaving elastically, but
only for a relatively small further increase in the applied force.
However, if you overstretch an elastic
material it may only partially contract return to its original shape on
removing the applied force, so the spring is permanently deformed. The point
at which this first happens is beyond what is called the
elastic limit (point D on
the graph).
Beyond the elastic limit D, the
greater the stretching force applied the greater the permanent deformation
(from point D onwards) - the less the object returns to its original shape -
this is seen on the graph as it curves over in the extension direction!
When the object doesn't return to its
original shape after removing the forces it is called
plastic deformation.
With some materials, the elastic limit L, is
so low that you see little elastic behaviour and permanent deformation sets in
quickly with a relatively small applied force. In these cases, the force -
extension graph is a curve (non-linear) with virtually no linear portion at the
start.
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sub-index
(c) Work is done in
stretching or compressing a material
In the process of bending, stretching
or compressing, energy is transferred in the process, so work is done.
You are having to do work against an
opposing force e.g. a stretched spring or rubber band, a squeezed rubber
ball want to return to their original shape
In order to deform a material to be
bent, stretched or compressed two forces must be operating, often in opposite
directions.
If only one force was involved the
material would stay the same shape and just change position.
Bending occurs when you wind up a
mechanical clock. Strictly speaking on the 'outer' surface to the right the
force is one of tension (stretching) and the 'inner' surface to the left,
experiences compression (think of bending quite a thick plate, ok?).
Stretching happens when you use a spring
balance to weigh something or put a rubber band round an object.
Older (still do?) forms of car suspension
use steel springs which compress on meeting a bump in the road to absorb the
energy of the impact. Rubber shock absorbers in the under carriage of a car
have the same effect.
The large sturdy spring S of the wheel suspension of a Land Rover,
which is expected to absorb the impact energy of some pretty hefty bumps! On
compression, the stiff spring will store a great deal of elastic energy, if only
for a brief moment in time!
Key for the 5 photographic diagrams: S =
suspension spring; H = the pipe conveying the hydraulic brake fluid
(see hydraulics)
D = the brake drum and disc on which the brake pads in
casing P are forced into contact with the smooth disc by hydraulic
pressure when you press the brake pedal.
A double spring suspension on a red van, conveniently jacked
up!
'Spring' photographs by courtesy of Mark Raw of M T R
Autotech Ltd garage, Castleton, North Yorkshire, England
TOP OF PAGE and
sub-index
(d) An experiment to investigate the force applied to a spring
and resulting
extension
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.
TOP OF PAGE and
sub-index
(e)
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.
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sub-index
(f) Particle model of an elastic energy store
Solids have a fixed shape due to the
strong attractive forces between the particles.
When the material is stretched so it
cannot immediately return to its original shape, the forces of attraction
still try to restore the particles to their original positions.
In this process work is done on the
particles in this separation and so potential energy is stored.
If the restriction is removed (applied
force released), particles can 'spring' back into position releasing
the elastic potential energy - a good example is wound up a clock
spring.
Note: If the applied force is too strong
and the interparticle forces are weakened too much, the particles cannot
return to their original positions - so the shape is permanently changed -
plastic deformation.
A
particle picture comparing elastic and inelastic materials
In terms of spacing in the lattice, the
particles (atoms, ions or molecules) in a solid at equilibrium in terms of their
spacing. These fixed 'balanced' positions are determined by the balancing of
various forces of particle attraction (opposite charges) and repulsion (like
charges).
When you apply a force to a solid object you
are pressing/squeezing the particles closer together giving rise to an opposing
force of repulsion. To continue any compression requires a greater and greater
force. When the compression force is removed the particles repel each other and
move back to their original position if it is a truly elastic material.
When an object is stretched you are pulling
the particles apart from their normal stable positions and the attractive forces
between the particles try to resist the stretching - force of tension produced.
Therefore you are doing work on the system and storing energy in it as you
stretch the material. If the stretching force is removed, and the material is
truly elastic, the object will return to its original shape and length. Rubber
materials have molecules that can actually be stretched at the particle level,
straining the chemical bonds. These bonds can relax back to normal length and
the object e.g. a rubber band or a steel spring return to their original shape
and length.
However, if the force is great enough, some
of the bonds are broken e.g. with an overstretched rubber band, or, the
particles shift position changing the structure in some way e.g. an over
stretched steel spring where layers of atoms can slide over each other. This
produces a permanent deformation when you go beyond the elastic limit of
proportionality and Hooke's law no longer applies.
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sub-index
(g) Elastic potential energy
- energy stored and work done - calculations
Elastic potential energy is the energy stored when
some material is stretched or compressed and the energy released when the
constriction is released eg the wound up spring of a clockwork clock, a pulled elastic
rubber band, stretched coiled metal
spring, the compressed spring in a an animal trap, stretched bow
before the arrow is released.
Since elastic potential energy
is a form of stored energy, it does nothing until it is released and
converted into another form of energy.
You can calculate the work done in
stretching or compressing a spring = energy transferred = the energy added to
the elastic potential energy store.
The amount of elastic potential
energy stored in a stretched spring can be calculated using the
equation:
elastic potential energy = 0.5 × spring constant ×
(extension)2,
Ee = 1/2
k e2
(assuming the limit of proportionality has not been
exceeded)
elastic potential energy store,
Ee, in joules,
J
spring constant,
k, in newtons per metre,
N/m
extension,
e, in metres,
m
This equation is only valid within the
limit of proportionality, that is when the spring obeys Hooke's Law.

Using the experiment results from above we
can calculate the stored elastic potential energy for the 0.06 m
extension of the spring with a load of 3.0 N.
You can actually do this calculation in two
ways.
(a) Using the equation Ee =
1/2
k e2 described above
k = 50 N/m, e = 0.06 m
Ee
= 0.5 x 50 x 0.062 =
0.09 J
BUT, you can also do the stored energy
calculation from the graph of results using the graph above and the principle
explained by the graph below.

(b) Calculating the area under the graph of
force versus extension.
This is the area under the shaded triangle of
the graph of force versus extension.
Using the same values as above: x = 0.06 m, y
= 3.0 N
area under graph = xy / 2
Ee = 0.06 x 3.0 /
2 = 0.09 J
Comments on the above two calculations:
(i) Work done = energy = force x
distance = N x m = J, but don't forget the halving in calculation (b).
(ii) Method (b) does not directly involve the
spring constant k.
(iii) Again, it should be pointed out that these
calculations are only valid if the spring is not stretched beyond the elastic
limit (limit of proportionality).
Remember, force must be in newtons and
extension in metres (cm/100 = m, mm/1000 = m).
At a higher mathematical level what you are doing via the graph is called an integration, here of the work
equation W = F x d.
BUT, you cannot use this simple equation to calculate the
stored energy because for this equation to be valid the force must be constant,
but for a spring the force must be steadily increased to increase the extension
so please use Ee =
1/2
k e2 to calculate elastic potential energy unless given
sufficient graphical data!
Questions on spring
extension and elastic potential energy
(overlap with energy 0b,
need some graph based questions)
Q1A student carried out an experiment by putting weights on the end of a spring.
After each weight was added, the length of the spring was carefully measured.
The 'idealised' results are summarised below.
Weight added to spring (N) |
2 |
4 |
6 |
8 |
10 |
Length of spring (cm) |
23 |
27 |
31 |
35 |
39 |
You can solve these problems with or without a graph.
(i) What is the spring extension per newton weight added?
For every 2 N weight added, the spring extends by 4 cm.
Therefore the spring extends
2 cm per N.
(ii) What would be the spring length with an added weight of
7N?
Interpolating the graph/data
33 cm,
mid-way between 31 and 35 cm.
(iii) What is the real length of the spring with no added
weight on?
Extrapolating back from 23 cm for 2 N gives 23 - 4 =
19 cm for
no added weight.
(iv) What would be the spring length be with an weight of
15N?
This weight is 5 N beyond the data limit, but you can use
extrapolation with data or graph.
The spring extends 2 cm per N weight added.
Therefore the spring will extend another 5 x 2 = 10 cm,
so the predicted length of the spring is 30 + 10 =
49 cm
(v) What assumption are you making in doing the (iii)
and (iv) calculations?
You are assuming the string is
perfectly elastic
with no deformation from a linear relationship between
spring extension and added weight.
Q2 A
spring is fixed firmly in a vertical position. When a mass of 120.0
g is attached to the spring it extends in length by 3.2 cm.
(a) Assuming the gravitational field strength is 9.8 N/kg,
calculate the spring constant k.
100 g is equivalent to a weight (force) of 9.8 x 120/1000
= 1.176 N
The extension e = 3.2/100 = 0.032 m. Force F
= 1.176 N
(The equation F = ke has already been dealt with in
detail further up this page)
F = ke, so the spring constant k =
F ÷ e = 1.176 ÷ 0.032 =
36.8 N/m (3 sf)
(b) Calculate the extra elastic potential energy
stored in the spring as a result of the added weight.
Using the equation for elastic stored
energy: Ee =
1/2
k e2 (dealt with in detail already on this
page)
Ee =
1/2
k e2 =
1/2
x 36.75 x 0.032 x 0.032 =
0.019 J
(c) If an extra 200 g mass is placed on
the spring, how much longer will it get?
200 g equates to a weight of 9.8 x
200/1000 = 1.96 N, k = 36.75 N/m
F = ke, so the spring extension e =
F ÷ k = 1.96/36.75 =
0.053 m (5/3 cm)
(d) What force is needed to extend the
spring by 30 cm?
force required = F = ke = 36.75 x
30/100 =
11.0 N (1 dp, 3 sf)
(d) What important assumption have you
made concerning the calculations in (a) to (c)?
You have assumed the spring behaves
truly elastic i.e. within the elastic limit (limit of
proportionality).
Q3 A spring with spring
constant of 5.00 N/m is stretched for an extra 10.0 cm.
How much extra energy is stored in the
elastic potential energy store of the spring by this extension.
Eepe = 1/2
k e2, 10.0 cm ≡
10.0 / 100 = 0.10 m
Eepe = 0.5 x 5.0 x 0.102
Eepe =
0.025 J
Q4 A
spring has a spring constant of 2000 N/m.
(a) If the elastic potential energy store
of the spring is 50.0 J, how far is the spring compressed?
Eepe = 1/2
k e2, rearranging gives e2 = 2Eepe / k,
e = √(2Eepe / k) and e = the compression
e = √(2Eepe / k) = √(2 x
50 / 2000) = √0.063 =
0.224 m (22.4 cm, 3 s.f.)
(b) What force is needed to compress
the spring by 15.0 cm in length?
F = ke = 2000 x 15/100 =
300 N
Q5 It takes 5.0 J
of work to stretch a spring 20 cm.
How much extra work must be done to
stretch it another 20 cm?
(i) You need to work out the spring
constant. 20 cm ≡
20 / 100 = 0.20 m
Eepe = 1/2
k e2, rearranging gives k = 2Eepe / e2
k = (2 x 5.0) / (0.20 x 0.20) =
250 N/m
(ii) Then work out the total work to
stretch the spring a total of 40 cm.
The total work done on the spring
equals its elastic potential energy store when fully stretched 40 cm
(which is 0.40 m). Since you now know the spring constant, you use
the same equation again, but solving for the total elastic potential
energy.
Eepe = 1/2
k e2 = 0.5 x 250 x 0.402 =
20 J
(iii) You then subtract (i) from (ii)
to get the extra work done.
Therefore the extra work done =
20 - 5 =
15 J
Q6 A spring stores an
extra 20 J of elastic potential energy when stretched an extra 40 cm.
Calculate the spring constant.
40 cm = 0.40 m.
Eepe = 1/2
k e2, rearranging:
k = Eepe x 2 / e2
= 20 x 2 / 0.402 =
250
N/m
Q7 A stretched string has
a total length of 60 cm and a spring constant of 240 N/m.
If the stretched spring is storing 20
J of energy, what is the length of the unstretched spring to the nearest
cm?
Eepe = 1/2
k e2, rearranging and 60 cm = 0.60 m
e = √(2Eepe / k) =
√(2 x 20 / 240) = √(1/6) = 0.408 m
0.408 m = 40.8 cm, ~41 cm = extra
length added to the stretched string
Therefore original length of spring =
60 - 41 =
~19
cm
Q8 A spring
has a spring constant of 20.0 N/m.
What would the extension be
in cm, if 0.50 J of work was done on stretching the spring?
Elastic potential energy formula: Eepe = 1/2
k e2
rearranging: e = √(2Eepe
/ k) = √(2 x 0.50 / 20) = 0.224 m =
22.4 cm
Q9 For this
question you need to know the formula for gravitational potential
energy (GPE) and gravitational field constant g = 9.8 m/s2
or 9.8 N/kg). Watch out for units, remember in the end to work in J,
m and kg.
A fun toy consists of a
spring and a funny head on the end.
When the toy spring is
compressed 5.0 cm, and released, it leaps vertically up into the
air.
The toy spring has a mass of
20 g.
When released it leaps up to
a maximum height of 75 cm.
If we assume all the elastic
potential energy (EPE) is transferred to the GPE store of the
toy spring, deduce the spring constant of the spring.
GPE = mgh = (20 /
1000) x 9.8 x (75 / 100) = 0.147 J
Neglecting air resistance we
can say GPE = EPE for the energy store transfer
Elastic potential energy
formula: Eepe = 1/2
k e2
rearranging gives: k = Eepe x 2 / e2
= 0.147 x 2 / (5.0 /100)2 =
118 N/m
(3 sf)
|
TOP OF PAGE and
sub-index
What next?
Forces revision notes index
FORCES 1. What are contact forces &
non-contact forces?, scalar & vector quantities, free body force diagrams
FORCES
2. Mass and the effect of gravity force on it - weight, (mention of work done and
GPE)
FORCES 3. Calculating resultant forces using vector
diagrams and work done
FORCES
4.
Elasticity and energy stored in a spring
FORCES 5. Turning forces and moments - from spanners
to wheelbarrows and equilibrium situations
FORCES 6. Pressure in liquid fluids and hydraulic
systems
FORCES 7. Pressure & upthrust in liquids, why do
objects float or sink in a fluid?, variation of atmospheric pressure with
height
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Know and
understand that a force acting on an object may cause a change
in shape of the object. Know and understand that a force applied
to an elastic object such as a spring will result in the object
stretching and storing elastic potential energy. Calculation of
the energy stored when stretching an elastic material is not
required. Know and understand that for an object that is able to
recover its original shape, elastic potential energy is stored
in the object when work is done on the object to change its
shape.
Know and understand that the extension of an elastic object is
directly proportional to the force applied, provided that the
limit of proportionality is not exceeded: Be able to use the
equation: F = k x e (F = ke) F is the force in newtons, N k is
the spring constant in newtons per metre, N/m e is the extension
in metres,
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