What force acts on a spring with a stiffness of 10 N / m if its absolute elongation is 2 cm?
| March 17, 2021 | education
Initial data: k (coefficient of stiffness of the given spring) = 10 N / m; Δl (absolute elongation (deformation value) of the spring) = 2 cm.
SI system: Δl = 2 cm = 2/100 m = 0.02 m.
The force acting on the spring is calculated using the following formula (Hooke’s law): F = k * Δl.
Let’s perform the calculation:
F = 10 * 0.02 = 0.2 N.
Answer: A force of 0.2 N. acts on this spring.
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