Two springs of stiffness of 1kN / m and 2kN / m were connected in parallel and stretched

Two springs of stiffness of 1kN / m and 2kN / m were connected in parallel and stretched at the ends with a force of 300N. what kind of work did you do?

Initial data: k1 (stiffness of the first spring) = 1 kN / m (1000 N / m); k2 (stiffness of the second spring) = 2 kN / m (2000 N / m); F (force with which springs connected in parallel were stretched) = 300 N.

1) The stiffness of the resulting connection of the springs: k = k1 + k2 = 1000 + 2000 = 3000 N / m.

2) The amount of deformation of the springs: Δx = F / k = 300/3000 = 0.1 m.

3) Work with tension of the obtained connection of springs: A = 0.5 * k * Δx2 = 0.5 * 3000 * 0.12 = 15 J.

Answer: When stretching the springs, 15 J.