Determine the stiffness of the spring if, under the action of a force of 150 N, it is stretched by 20 cm.

Initial data: F (value of the acting force) = 150 N; Δl (amount of tension (deformation) of the spring) = 20 cm = 0.2 m.

The spring rate can be determined according to Hooke’s law: F = Fcont. = k * Δl, whence k = F / Δl.

Let’s make a calculation: k = 150 / 0.2 = 750 N / m.

Answer: The stiffness of the spring under the given conditions is 750 N / m.

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