A body weighing 10 kg was suspended from a spring with a stiffness of 500 N / m. How many centimeters

A body weighing 10 kg was suspended from a spring with a stiffness of 500 N / m. How many centimeters will the spring stretch?

The spring force will be balanced by the force of gravity, the value of which will correspond to the weight of the body.
Fcont. = Fт., Where Fcont. = k * ∆l (k is the stiffness of the spring, k = 500 N / m; ∆l = the amount of deformation (elongation)), Ft = P = m * g (m is the body mass, m = 10 kg, g is the acceleration of the free fall (we take g = 10 m / s ^ 2)).
Let us express and calculate the deformation of the spring:
∆l = m * g / k = 10 * 10/500 = 0.2 m = 20 cm.
Answer: The spring will stretch 20 cm.