FORCES 4. Elasticity and energy stored in a spring
Doc Brown's Physics Revision Notes
Suitable for GCSE/IGCSE Physics/Science courses or their equivalent
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?
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.
If the forces 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.
However, if you overstretch an elastic material it may only partially contract return to its original shape on removing the forces, so the object-material is permanently deformed. The point at which this first happens is called the elastic limit (see later - the elastic limit of proportionality). Beyond the elastic limit, the greater the stretching force applied the greater the permanent deformation and the less the object returns to 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.
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.
In the process of bending, stretching or compressing, energy is transferred in the process, so work is done.
In order to deform a material to be bent, stretched or compressed two forces must be operating, often in opposite directions.
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.
An experiment to investigate the force applied to a spring and the 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
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.
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:
Above is an illustration of a simple instrument for weighing objects.
It is essentially a 'force meter' calibrated to read in g and kg.
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.
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).
L is also known as the elastic limit and beyond it Hooke's Law is no longer obeyed. In other words the non-linear section of the graph is beyond L, the limit of proportionality.
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.
Elastic potential energy
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.
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
Note: (i) Work done = energy = force x distance = N x m = J
(ii) Method (b) does not involve the spring constant.
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 (A level in UK) 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!
A particle picture of 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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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
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, m
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