Answer:
1.44%
Explanation:
The potential energy of a spring is given by
[tex]U=\dfrac{1}{2}kx^2[/tex]
The kinetic energy is
[tex]E=\dfrac{1}{2}kA^2[/tex]
Now x = 12 A
Applying in the first equation
[tex]U=\dfrac{1}{2}k(12A)^2\\\Rightarrow U=\dfrac{1}{2}k144A^2\\\Rightarrow U=144\dfrac{1}{2}kA^2\\\Rightarrow U=144E[/tex]
The percentage would be 14400%
at [tex]x=\dfrac{1}{2}A[/tex] (makes more sense)
[tex]U=\dfrac{1}{2}k(\dfrac{1}{2}A)^2\\\Rightarrow U=\dfrac{1}{2}k\dfrac{1}{4}A^2\\\Rightarrow U=\dfrac{1}{4}\times \dfrac{1}{2}kA^2\\\Rightarrow U=0.25E[/tex]
The percentage is 25 % of the kinetic energy
