What work will the force perform when the spring is elongated with a cruelty of 350 N / m from 4 to 6 cm?

Given: k (coefficient of rigidity of the taken spring) = 350 N / m; x1 (original length of the spring) = 4 cm (in SI x1 = 0.04 m); x2 (length after extension) = 6 cm (in SI x2 = 0.06 m).

We calculate the work of the elastic force when lengthening the taken spring by the formula: A = ΔEp = k * x2 ^ 2/2 – k * x1 ^ 2/2 = k * (x2 ^ 2 – x12) / 2.

Calculation: A = 350 * (0.06 ^ 2 – 0.04 ^ 2) / 2 = 0.35 J.

Answer: When the taken spring was lengthened by the elastic force, a work of 0.35 J.