Spring rate 200 N / m. How long will the spring lengthen if a body weighing 0.5 kg is suspended from it?

k = 200 N / m.
m = 0.5 kg.
g = 9.8 N / m.
Δl -?
The force of elasticity of the spring Fcont is balanced by the force of gravity Ft of the load: Fcont = Ft.
The elastic force is determined by Hooke’s law: Fel = k * Δl, where k is the stiffness of the spring, Δl is the elongation of the spring.
The force of gravity is determined by the formula: Ft = m * g, where m is the mass of the body, g is the acceleration of gravity.
k * Δl = m * g.
Δl = m * g / k.
Δl = 0.5 kg * 9.8 N / m / 200 N / m = 0.0245 m.
Answer: the spring will lengthen by Δl = 0.0245 m.