Force 1H stretches the spring by 1 cm. What force is needed to stretch two of the same springs

Force 1H stretches the spring by 1 cm. What force is needed to stretch two of the same springs by 1 cm .. A) Connected sequentially B) Connected in parallel

Data: F1 (initial tensile force) = 1 N; Δx (required deformation of the taken spring) = 1 cm = 0.01 m.

1) The stiffness of each of the springs (Hooke’s law): k = F1 / Δx = 1 / 0.01 = 100 N / m.

2) Serial connection:

a) rigidity: 1 / kposl = 1 / k + 1 / k = 2 / k; kposl = k / 2 = 100/2 = 50 N / m;

b) the required force: F = kposl * Δx = 50 * 0.01 = 0.5 N.

2) Parallel connection:

a) hardness: kpair = k1 + k2 = 2 * k = 200 N / m;

b) the required force: F = kpair * Δx = 200 * 0.01 = 2 N.