Answer:
Final elastic potential energy: [tex]\frac{1}{2}kL_0^2[/tex]
Explanation:
The elastic potential energy of a spring is given by
[tex]U=\frac{1}{2}k(L-L_0)^2[/tex]
where
k is the spring constant
[tex]L[/tex] is the length of the spring
[tex]L_0[/tex] is the natural length of the spring
Therefore, for a spring at its natural length,
[tex]L=L_0[/tex]
So its elastic potential energy is zero:
[tex]U=0[/tex]
If the spring is stretched to 2 times its normal length,
[tex]L=2L_0[/tex]
Therefore its elastic potential energy would be
[tex]U=\frac{1}{2}k(2L_0-L_0)^2=\frac{1}{2}kL_0^2[/tex]
