Answer:
The work done in stretching the spring is 0.875 J.
Explanation:
Given that,
Force = 140 N
Natural length = 60-40 = 20 cm
Stretch length of the spring = 65-60 = 5 cm
We need to calculate the spring constant
Using formula of Hooke's law
[tex]F= kx[/tex]
[tex]140=k\times20\times10^{-2}[/tex]
[tex]k=\dfrac{140}{20\times10^{-2}}[/tex]
[tex]k=700[/tex]
We need to calculate the work done
[tex]W=\int_{a}^{b}{kx}dx[/tex]
[tex]=\int_{0}^{0.05}{700x}dx[/tex]
On integration
[tex]W=700\times(\dfrac{x^2}{2})_{0}^{0.05}[/tex]
[tex]W=700\times(\dfrac{(0.05)^2}{2}-0)[/tex]
[tex]W=0.875\ J[/tex]
Hence, The work done in stretching the spring is 0.875 J.
