What is the stiffness of the spring if it stretches 7.5 cm under the action of a force of 12 Newtons
| August 3, 2021 | education
The given tasks: F (the force under the action of which the spring was stretched) = 12 N; Δl (spring tension value) = 7.5 cm.
SI system: Δl = 7.5 cm = 7.5 / 100 = 0.075 m.
The stiffness of the spring under consideration is determined using Hooke’s law: Fcont = -F = -Δl * k, whence k = F / Δl.
Let’s make calculations: k = F / Δl = 12 / 0.075 = 160 N / m.
Answer: The stiffness of the spring in question is 160 N / m.
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