Stretching & Compressing using Force
Elastic deformation: the object returns to the original shape & size when the force is removed
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Inelastic Deformation: The object is changed permanently when the force is removed
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Hooke’s Law:
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.
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Some springs are easy to stretch. They have a low spring constant (unit: N/m)
Some springs are difficult to stretch. They have a high spring constant .
To find the force needed to stretch (or compress) a spring:
force = spring constant × extension
This relationship also applies to the compression of an elastic object, where ‘extensions’ would be the compression of the object.
Question: The extension of a spring was 0.12 m when the force was 3.0 N
Calculate the spring constant of the spring.
force = spring constant × extension
3.0 = spring constant x 0.12
spring constant = 3.0 / 0.12
The spring constant is 25 N/m
Question: A mass is suspended from a spring. Explain how a student could test if the spring was behaving elastically. (2 marks)
remove the mass
if the spring returns to the original length it is behaving elastically
A force that stretches (or compresses) a spring does work and 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.
On a graph of force on the y-axis and extension of the x-axis:
- The gradient = spring constant
- The area under the graph = work done to stretch/compress the spring
Calculate work done in stretching (or compressing) a spring (up to the limit of proportionality) using the equation:
elastic potential energy = 0.5 × spring constant × extension²
Question: Describe the relationship between work done and elastic potential energy in stretching a spring. (2 marks)
- work done is equal to elastic potential energy
- as long as the spring does not go past the limit of proportionality
Question: A different spring has a spring constant of 13.5 N / m.
Calculate the elastic potential energy stored in the spring when its extension is 12 cm.
elastic potential energy = 0.5 × spring constant × extension²
elastic potential energy = 0.5 x 13.5 x 0.122
the elastic potential energy is 0.097 Joules
Question: A student investigated how the extension of a spring depends on the force applied to the spring.
The results from the investigation are plotted on the following graph.
(i) The graph shows that the student has made an error throughout the investigation.
What error has the student made? (2 marks)
the student measured / recorded the length of the spring (and not extension)
so extension does not equal zero when force = 0