Under the action of a force of 100N, the spring lengthened by 20mm. How much the spring will lengthen if the elastic force is 325N.

In accordance with Hooke’s law, the force F lengthens the spring with the stiffness coefficient k by the value x:

F = k * x.

For the two considered forces acting on one spring, we have:

F1 = k * x1;

F2 = k * x2.

We divide one equation by another and express the required elongation x2:

F1 / F2 = (k * x1) / (k * x2);

F1 / F2 = x1 / x2;

x2 = x1 * F2 / F1;

x2 = 20 * 325/100 = 65 mm.

Thus, when a force of 325 N is applied to it, this spring will lengthen by 65 mm.