What is the stiffness of the spring if, under the action of a force of 200 mN, it stretches by 2 mm?
| February 6, 2021 | education
Given: F (the force under which the spring is stretched) = 200 mN (200 * 10-3 N = 0.2 N); Δl (spring tension value) = 2 mm (2 * 10-3 m).
To calculate the stiffness of the spring under consideration, you need to use Hooke’s law: F = -Fcont = k * Δl, whence k = | Fcont / Δl |.
Let’s make the calculation: k = 0.2 / (2 * 10-3) = 100 N / m.
Answer: The stiffness of the given spring is 100 N / m.
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