Find the stiffness of the dynamometer spring if, under the action of a force of 10n, the spring is lengthened by 4cm.

These tasks: F (the force that acted on the dynamometer spring) = 10 N; Δx (spring elongation) = 4 cm (in SI system Δx = 0.04 m).

We express the stiffness of the spring of the dynamometer under consideration from the formula: F = -Fcont. = k * Δx, whence k = F / Δx.

Let’s perform the calculation: k = 10 / 0.04 = 250 N / m.

Answer: The spring of the dynamometer in question has a stiffness of 250 N / m.