What is the stiffness of a spring if, under the action of a force of 250N, it lengthens by 5mm?
| September 16, 2021 | education
Initial data: F (force under the action of which the spring is elongated) = 250 N; Δl (spring elongation, deformation value) = 5 mm = 5 * 10 ^ -3 m.
The spring rate can be determined using Hooke’s law: F = k * Δl, whence k = F / Δl.
Let’s make a calculation: k = 250 / (5 * 10 ^ -3) = 50 * 10 ^ 3 N / m = 50 kN / m.
Answer: The spring rate is 50 kN / m.
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