A spring with a stiffness of 200 N / m is stretched by an applied force of 3 cm. What is the elongation

A spring with a stiffness of 200 N / m is stretched by an applied force of 3 cm. What is the elongation of a spring with a stiffness of 300 N / m under the same force?

According to Hooke’s law, discovered in 1660, the deformation of the spring x is proportional to the applied force F:
x = F / k,
where k is the coefficient of stiffness (or simply stiffness) of the spring, N / m.
Let’s define the effort F:
F = k x = 200 × 0.03 = 6.0 H.
Under the action of the same force F = 6.0 N, the second spring with a stiffness coefficient k = 300 N / m will lengthen by the amount:
x = 6/300 = 0.02 m = 2 cm.
Answer: it will lengthen by 2 cm.