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Question: A spring 40 mm long spring is stretched by applying a force. If 10 N force is required to stretch th...

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.

Explanation

Solution

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=12kx2W=\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 mm1\text{ }mm, converting the value of the spring extension length from mm to m we get the spring constant as value 0.001 mm0.001\text{ }mm.
Now using the spring extension length and force applied, we get the spring constant as:
k = Fxk\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 = 100.001k\text{ }=\text{ }\dfrac{10}{0.001}
k = 100000 N/m\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=12kx2W=\dfrac{1}{2}k{{x}^{2}}
12×100000×401000×401000\Rightarrow \dfrac{1}{2}\times 100000\times \dfrac{40}{1000}\times \dfrac{40}{1000}
W=8JW=8J

Therefore, the work done by the spring under the influence of spring constant is 8J8J.

Note:
The formula F =kxF\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.