What work does the elastic force do when the length of a deformed spring with a stiffness of 80

What work does the elastic force do when the length of a deformed spring with a stiffness of 80 N / m changes from 10 to 8 cm?

k = 80 N / m.

x1 = 10 cm = 0.1 m.

x2 = 8 cm = 0.08 m.

A -?

The work of the elastic force A is equal to the change in the potential energy of the spring ΔEp: A = ΔEp.

ΔEp = En2 – En1.

The potential energy of the spring, En, is determined by the formula: En = k * x ^ 2/2, where k is the stiffness of the spring, x is the elongation of the spring.

ΔEp = k * Δx ^ 2/2.

Δx = x2 – x1.

Δx = 0.08 m – 0.1 m = 0.02 m.

A = 80 N / m * (0.02 m) 2/2 = 0.016 J.

If the spring was compressed by the elastic force, then A = 0.016 J.

If the spring was compressed by an external force, then A = – 0.016 J.

Answer: during deformation, the elastic force performed the work A = 0.016 J.