The work required to stretch it from 11 cm to 25 cm is equal to 109.76 joule.
We are given a spring that has a natural length of 11 cm. Then it is stretched to a length of 36 cm with a force of 28 N. In order to find the work required to stretch the spring from 11 cm to 25 cm, we first need to find the spring constant.
Recall that k, the spring constant, is equal to force F over displacement x.
[tex]k = \frac{F}{x}\\\\k=\frac{28}{(36-11)} \\\\k=1.12[/tex]
We have found the spring constant! It's equal to 1.12. Next, we're going to calculate the elastic potential energy.
[tex]U=\frac{1}{2}kx^2\\\\U=\frac{1}{2}(1.12)(25-11)^2\\\\U=\frac{1}{2}(1.12)(14)^2\\\\U=109.76[/tex]
We have confirmed that the work required to stretch the spring from 11 cm to 25 cm is equal to 109.76 J.
Learn more about Hooke's law and spring constant here: https://brainly.com/question/13946502
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