The spring under a load of 25n is lengthened by 2mm find the elongation of the spring under a load of 100n.

Data: F1 (first spring load) = 25 N; Δx1 (first elongation) = 2 mm; F2 (second spring load) = 100 N.

Since the load acted on the same spring, its second elongation can be determined from the equality: F1 / Δx1 = k = F2 / Δx2 and Δx2 = F2 * Δx1 / F1.

Let’s calculate: Δx2 = 100 * 2/25 = 8 mm.

Answer: With a load of 100 N, the taken spring will lengthen by 8 mm