To store an energy of 260J, one must stretch the spring to a length of 0.57 meter.
Given the data in the question;
- Spring constant; [tex]K = 1600N/m[/tex]
- Energy stored in spring; [tex]E = 210J[/tex]
From Hooke's law:
The force needed to stretch a spring is directly proportional to the amount of stretch.
Elastic potential energy is equal to the work done to stretch a spring.
Work done = ΔP.E = [tex]\frac{1}{2}kx^2[/tex]
Where k is the spring constant and x is the displacement
We substitute our given values into the equation
[tex]260J = \frac{1}{2}\ *\ 1600N/m\ *\ x^2\\\\260N.m = 800N/m\ *\ x^2\\\\x^2 = \frac{260N.m}{800N/m}\\\\x^2 = 0.325 m^2\\\\x = \sqrt{0.325 m^2}\\\\x = 0.57m[/tex]
Therefore, to store an energy of 260J, one must stretch the spring to a length of 0.57 meter.
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