Question

A spring 40 mm long spring is stretched by applying a force. If 10 N force is required to stretch the spring through one mm, then work done in stretching the spring through 40 mm is.

Answer 1

Hint: To find the value of the work done, we find the work done in terms of spring constant and spring extension where the spring constant is denoted as k and spring extension is denoted as x.
\[W=\dfrac{1}{2}k{{x}^{2}}\]
And to find the value of spring constant k, we use the formula of k = F/x where F is the force required to stretch the spring and x is the spring displacement.

Complete step by step answer:
Assuming the spring hanging with the applied force of 10N and is stretched with the spring extension of \[1\text{ }mm\], converting the value of the spring extension length from mm to m we get the spring constant as value \[0.001\text{ }mm\].
Now using the spring extension length and force applied, we get the spring constant as:
\[k\text{ }=\text{ }\dfrac{F}{x}\]
With placing the value of F as 10N and x as 0.001 mm, we get the spring constant as:
\[k\text{ }=\text{ }\dfrac{10}{0.001}\]
\[\Rightarrow k\text{ }=\text{ }100000\text{ }N/m\]
Now after getting the value of the spring constant we will find the work done in terms of spring constant and spring displacement as:
\[W=\dfrac{1}{2}k{{x}^{2}}\]
\[\Rightarrow \dfrac{1}{2}\times 100000\times \dfrac{40}{1000}\times \dfrac{40}{1000}\]
\[W=8J\]

Therefore, the work done by the spring under the influence of spring constant is \[8J\].


Note:
The formula \[F\text{ }=-kx\] is also known as Hooke's Law where the length changes by x dimension meaning when the spring is compressed the force is decreased and when the spring is expanded the force is increased.