Fig. 4.1 shows a spring with no load and then with a 2.0 N weight attached. The spring extends by 4.0 cm when the 2.0 N weight is added. (a) Calculate the s...
Question 1 Report
Fig. 4.1 shows a spring with no load and then with a 2.0 N weight attached.
The spring extends by 4.0 cm when the 2.0 N weight is added.
(a) Calculate the spring constant of the spring. [2]
(b) Calculate the extension when a 5.0 N weight is hung from the spring. [2]
(c) State the law that describes the relationship between force and extension for a spring. [1]
Answer Details
(a) Spring constant
The spring constant \(k\) relates the force applied to the extension produced. The defining equation is:
\(k = \dfrac{F}{x}\)
The force is \(F = 2.0\,\text{N}\) and the extension is \(x = 4.0\,\text{cm} = 0.040\,\text{m}\) (always convert to metres for SI units).
\(k = \dfrac{2.0}{0.040} = 50\,\text{N/m}\)
[2 marks]: one for correct substitution, one for correct answer with unit.
(b) Extension with a 5.0 N weight
Rearranging \(F = kx\) gives:
\(x = \dfrac{F}{k} = \dfrac{5.0}{50} = 0.10\,\text{m} = 10\,\text{cm}\)
This assumes the spring still obeys Hooke's law at this load (it has not reached its limit of proportionality).
[2 marks]: one for correct substitution, one for correct answer.
(c) The law
Hooke's law states that the extension of a spring is directly proportional to the applied force, provided the limit of proportionality is not exceeded. The data in parts (a) and (b) follow this law because doubling the force from 2.0 N to 4.0 N would double the extension from 4.0 cm to 8.0 cm.
[1 mark]
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