How much will the spring lengthen under a load of 15 N, if under a load of 10 N, the spring lengthens by 5 cm?

Task data: F2 (second spring load) = 15 N; F1 (first spring load) = 10 N; Δx1 (spring deformation under the first load) = 5 cm (0.05 m).

The extension of the spring under a load of 15 N is expressed from the equality: F1 / Δx1 = k (spring stiffness) = F2 / Δx2, whence Δx2 = F2 * Δx1 / F1.

Let’s perform the calculation: Δx2 = 15 * 0.05 / 10 = 0.075 m (75 mm).

Answer: Under a load of 15 N, the spring will lengthen by 75 mm.