Hooke's law establishes that the elongation of a spring is directly proportional to the modulus of the force [tex]F[/tex] applied to it, as long as the spring is not permanently deformed:
[tex]F=k (x-x_{o})[/tex] (1)
Where:
[tex]k[/tex] is the elastic constant of the spring. The higher its value, the more work it will cost to stretch the spring.
[tex]x_{o}[/tex] is the length of the spring without applying force.
[tex]x[/tex] is the length of the spring with the force applied.
According to this, we have a spring where only the force due gravity is applied.
In other words, the force applied is the weigth [tex]W[/tex] of the block:
[tex]W=m.g[/tex] (2)
Where [tex]m=250g=0.25kg[/tex] is the mass of the block and [tex]g=9.8\frac{m}{s^{2}}[/tex] is the gravity acceleration.
[tex]W=(0.25kg)(9.8\frac{m}{s^{2}})[/tex] (3)
[tex]W=2.45N[/tex] (4)
Knowing the force applied [tex]W[/tex] and [tex]x=18cm=0.18m[/tex] and [tex]x_{o}=0[/tex], we can substitute the values in equation (1) and find [tex]k[/tex]:
[tex]W=k (x-x_{o})[/tex] (5)
[tex]2.45N=k (0.18m-0m)[/tex] (6)
Finally:
[tex]k=13.61\frac{N}{m}[/tex]
