What is the stiffness of the spring if, under the action of a force of 4 N, the spring is lengthened by 0.02 m?
| June 19, 2021 | education
Initial data: F (force acting on the spring) = 4 N; Δl (elongation, deformation of the spring) = 0.02 m.
We express the required spring stiffness from Hooke’s law: F = k * Δl, whence k = F / Δl.
Calculation: k = 4 / 0.02 = 200 N / m.
Answer: 4) The spring rate was 200 N / m.
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