What work does the elastic force do when the spring deformation changes with a stiffness of 400 N / m from 7 to 3 cm?

These tasks: k (coefficient of rigidity of the taken spring) = 400 N / m; x1 (original length of the spring) = 7 cm (in SI x1 = 0.07 m); x2 (final length) = 3 cm (in SI x2 = 0.03 m).

The work done by the elastic force during deformation of the spring taken is determined by the formula: A = ΔEp = k * x1 ^ 2/2 – k * x2 ^ 2/2 = k * (x1 ^ 2 – x2 ^ 2) / 2.

Let’s make a calculation: A = 400 * (0.07 ^ 2 – 0.03 ^ 2) / 2 = 0.8 J.

Answer: With the deformation of the taken spring, the elastic force performed a work of 0.8 J.