When a force is applied to an object it will try to make the object move, however, if the object is unable to move the force can stretch, bend or compress the object instead. These changes in the shape of an object are known as deformations, or to deform.
Try it: Take a sheet of paper, now push it along a clean desk with one finger, there is nothing stopping the paper from moving so it does so in the direction of the force. Now place an obstacle in the way of the paper, something too heavy for the paper to move. Now push the paper, when it hits the obstacle the paper will bend and twist, it has been deformed.
The reason for the deformation is the presence of a second force that resists or prevents the object from moving. This force is in the opposite direction to and opposes the movement.
If this spring does not have a force at the top to prevent
it from moving, when the load is added it will simple fall.
But, if the spring is held by a clamp at the top the second restraining force allows the load to stretch the spring.
Elastic and Inelastic
When an object is deformed, (stretched, bent or compressed), if it returns to its original shape once the forces are removed, it is an elastic deformation. If it does not return to its original shape it is an inelastic deformation.
The spring in the diagram below will stretch when a load is applied to it, but once the load is removed the spring will return to its original length and shape. Elastic bands also show elastic deformation.
The crumple zone at the front of the car, in the image above, will not return to the original shape once the forces of the crash are removed. This is an example of an inelastic deformation.
Springs can show elastic and inelastic deformation. If the load in within the limits of the springs design the stretch will be elastic. If it is overstretched, the load is too high, the spring will remain stretched when the load is removed. The point at which a spring will only just return to its original shape is called its elastic limit.
When a force is applied to a spring it will start to stretch or compress, depending on the type of spring. The change in length of a spring is proportional to the force applied. For any given spring if you double the force on the spring you will double the change in length, (the extension).
This is true within the elastic limit of the spring, once this limit is exceeded the spring will not return to its original shape.
This relationship between the linear extension of a spring and the load or force applied is known as Hooke’s Law.
This law can be expressed as an equation:
Force on the spring = Spring constant x extension.
Or _ F = k x _
_F _: Force in Newtons
_k _: Spring constant in N/m
_x _: Extension of the spring in m.
The spring constant is a property of the spring and depends on the material used, the stiffness of the coils, the length and number of coils. It is a measure of how easy or hard it is to stretch the spring. The higher the value of k, the more force is needed to stretch or compress the spring.
As the force is applied the spring extends in a linear fashion, above the elastic limit the extension in non-linear, as shown by the curve of the line.
Calculating spring constant, k
To calculate the spring constant for an individual spring, the force applied and the change in length, (extension), need to be measured experimentally. Measuring the length of the spring with no load and with a known load allows the extension to be calculated.
Example: A spring has a length of 10 cm with no load, with a load of 200g it stretches to 35 cm, what is the spring constant for this spring?
Load : 200 g = 0.2 kg
The load will be equal to the weight of this mass
W = mg = 0.2 x 10N/kg = 20 N
|10 cm = 0.1 m 35 cm = 0.35 m
|= 0.35 - 0.1 = 0.25 m
|F = kx
|so k = F ÷ x
|k = 20 ÷ 0.25
|= 80 N/m
Energy in Springs
When an elastic object, like spring, is stretched it gains Elastic Potential Energy, energy that is converted to kinetic energy when the extension force is removed and the spring returns to its original shape.
The force to extend the spring has to move the spring and this requires __Work __to be done. As energy is always conserved, the work done must equal the Elastic Potential Energy stored in the stretched spring.
Energy transferred to spring = ½ x spring constant x extension ²
- A spring with a spring constant of 80 N/m, is extended by 25 cm. How much energy has been transferred to the spring?
Note that the extension must be converted to meters.
- A Spring with a spring constant of 150 N/m has 200 J of work done to it to stretch it. How far will this spring extend?
- What is the spring constant of a spring that is stretched from 5cm to 10cm by a 20N force?
- Your answer should include: 1N/m / 1
Explanation: Extension : 10cm - 5cm = 5cm = 0.05m F = kx so k = F x k = 20 x 0.05 = 1 N/m