Determine the stiffness of the spring, which, under the action of a force of 2N, has lengthened by 5 cm.

To determine the stiffness of the spring, it is necessary to use Hooke’s law: = -k * Δl, where Fcont. Is the elastic force of the taken spring (according to Newton’s third law, the elastic modulus is equal to the modulus of the force stretching the spring (F = 2N)), Δl is the value of the spring elongation (5 cm or 0.05 m).

Let us express and calculate the stiffness of the spring taken: Δl = F / Δl = 2 / 0.05 = 40 N / m.

Answer: The stiffness of the spring taken is 40 N / m.