A body weighing P = 20H is suspended on a spring with a stiffness of k = 500 N / m. What is the elongation of the spring.

The force of elasticity of the spring will be balanced by the action of the force of gravity, the value of which will correspond to the weight of the body suspended on the spring.
Fcont. = Fт.
Fcont. = k * ∆l, where k is the stiffness of the spring k = 500 N / m; ∆l = amount of deformation of the spring (elongation).
Ft = P, where P is the body weight (P = 20 N).
Let us express and calculate the deformation of the spring:
∆l = P / k = 20/500 = 0.04 m = 4 cm.
Answer: The elongation of the spring is 4 cm.