To solve this exercise it is necessary to apply the concepts related to kinetic energy in a spring.
By definition the kinetic energy keeping what is stipulated in the stable Hooke law that
[tex]KE_s = \frac{1}{2} kx^2[/tex]
Where,
K = Spring constant
x = Displacement
Our values are given as,
x = 0.15m
k = 2000N/m
Therefore replacing
[tex]KE_s = \frac{1}{2} kx^2[/tex]
[tex]KE_s = \frac{1}{2} (2000)(0.15)^2[/tex]
[tex]KE_s = 22.5J[/tex]
Therefore the initial kinetic energy is 22.5J