How much will the spring lengthen if, upon its deformation, an elastic force of 80N occurs? The spring rate is 10kN / m
| July 26, 2021 | education
Task data: Fcont. (the elastic force arising from the deformation of the taken spring) = 80 N; k (coefficient of spring stiffness) = 10 kN / m (10 * 10 ^ 3 N / m).
The elongation of the taken spring is expressed from the formula: Fcont. = | k * Δx | and Δx = Fcont. / k.
Let’s perform the calculation: Δx = 80 / (10 * 10 ^ 3) = 8 * 10 ^ -3 m = 8 mm.
Answer: The taken spring should have lengthened by 8 mm.
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