Determine the elongation of a spring with a stiffness of 500 N / m if it stretched under a force of 0.2kN?

Data: k (rigidity of the taken spring) = 500 N / m; F (value of the tensile force of the spring) = 0.2 kN (0.2 * 10 ^ 3 N).
The desired elongation of the taken spring can be calculated using Hooke’s law: F = -Ftr = k * Δl and Δl = F / k.
Let’s perform the calculation: Δl = 0.2 * 10 ^ 3/500 = 0.4 m (40 cm).
Answer: Under the action of a force of 0.2 kN, the taken spring was stretched 40 centimeters.