The spring, stretched 2 cm when released, shrank 1 cm. Determine the work of the elastic force if the spring rate is 40 N / m.

Potential energy of a spring with stiffness k, stretched from the equilibrium state by an amount x:
Ep = k x ^ 2/2.
Let us first determine the potential energy of the spring, corresponding to the tension x1 = 2 cm = 0.02 m:
Ep1 = k x1 ^ 2/2 = 40 × 0.02 ^ 2/2 = 0.0080 J.
On the other hand, for x2 = 1 cm = 0.01 m:
Ep2 = k x2 ^ 2/2 = 40 × 0.01 ^ 2/2 = 0.0020 J.
The sought-after work of the elastic force of the spring when it contracts from x1 = 2 cm to x2 = 1 cm is equal to the difference:
A = Ep1 – Ep2 = 0.0080 – 0.0020 = 0.006 J.
Answer: 0.006 J.