The spring lengthened by 2 cm under the action of a force of 5 N. Determine the stiffness.
| March 23, 2021 | education
To calculate the stiffness of a given spring, we use the formula (we use Hooke’s law): Fcont = F = k * Δx, whence we express: k = F / Δx.
Variables: Δx – deformation during elongation (Δx = 2 cm = 0.02 m); F – acting force (F = 5 N).
Let’s make a calculation: k = F / Δx = 5 / 0.02 = 250 N / m.
Answer: The spring has a stiffness of 250 N / m.
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