A force of 20 N acts on the spring, under which it lengthened by 4 cm. Determine its stiffness.
| August 9, 2021 | education
Given: F (value of the force that acts on the spring) = 20 N; Δl (elongation of the spring under the action of a given force) = 4 cm (0.04 m).
The stiffness of the spring taken can be calculated using Hooke’s law: F (acting force) = -Fcont. (elastic force) = k * Δl whence k = F / Δl.
Let’s make the calculation: k = 20 / 0.04 = 500 N / m.
Answer: The calculated spring rate is 500 N / m.
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