Answer:
[tex]\displaystyle \Delta x=1.74\ m[/tex]
Explanation:
Elastic Potential Energy
Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.
[tex]\displaystyle PE = \frac{1}{2}k(\Delta x)^2[/tex]
Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.
Solving for Δx:
[tex]\displaystyle \Delta x=\swrt{\frac{2PE}{k}}[/tex]
Substituting:
[tex]\displaystyle \Delta x=\sqrt{\frac{2*8600}{5700}}[/tex]
Calculating:
[tex]\displaystyle \Delta x=\sqrt{3.0175}[/tex]
[tex]\boxed{\displaystyle \Delta x=1.74\ m}[/tex]
