Question

A spring $40\;{\text{mm}}$ long spring is stretched by applying a force. If $10\;{\text{N}}$ force is required to stretch the spring through one mm, then work done in stretching the spring through $40\;{\text{mm}}$ is:
A) $24\;{\text{J}}$
B) $8\;{\text{J}}$
C) $56\;{\text{J}}$
D) $64\;{\text{J}}$

Answer 1

Hooke’s law is relevant in stretching a spring. The force applied to stretch the spring will be proportional to the displacement. Hence the work done can be found from the spring constant which is force per unit displacement.

Complete step by step answer:
Given the force applied to stretch the spring through one mm is $10\;{\text{N}}$.
From the Hooke’s law, the force is proportional to the distance moved from the rest.
$F = - kx$
Where, $k$ is the spring constant and $x$ is the distance moved from rest. The negative sign is only because the restoring force formed in the spring is opposite to the applied force. Thus the spring constant is given as,
$k = \dfrac{F}{x}$
Substitute the values in the above expression,

$$k = \dfrac{{10\;{\text{N}}}}{{1\;{\text{mm}}}} \\\ = \dfrac{{10\;{\text{N}}}}{{{\text{0}}{\text{.001}}\;{\text{m}}}} \\\ = {10^4}\;{\text{N/m}} \\\$$

Thus the spring constant is ${10^4}\;{\text{N/m}}$.
The expression for the work done to stretch the spring is given as,
$W = \dfrac{1}{2}k{x^2}$
We can substitute the values for finding the stretch of $40\;{\text{mm}}$.
$W = \dfrac{1}{2} \times {10^4}\;{\text{N/m}} \times {\left( {40\;{\text{mm}}} \right)^2} \\\ = \dfrac{1}{2} \times {10^4}\;{\text{N/m}} \times {\left( {{\text{0}}{\text{.04m}}} \right)^2} \\\ = 8\;{\text{J}} \\\$
Thus the work done in stretching the spring through $40\;{\text{mm}}$ is $8\;{\text{J}}$.
The answer is option B.

Note: We have to note that the work done to change one object from one place to another place will be the change in potential energies of the object in two places. Thus the work done in stretching a spring will be the potential energy stored in the spring. This is also called elastic potential energy.