The spring is stretched by 1 = 10 cm. It is held in a stretched state by applying a force F = 100 N.

The spring is stretched by 1 = 10 cm. It is held in a stretched state by applying a force F = 100 N. What work A was done when the spring was stretched?

x = 10 cm = 0.1 m.

F = 100 N.

A -?

When the spring is stretched, external forces perform work A equal to the change in the potential energy of the spring ΔEp: A = ΔEp.

We express the potential energy of the spring by the formula: En = k * x2 / 2, where k is the stiffness of the spring, x is the elongation of the spring.

At the initial moment of time, the elongation of the spring is x0 = 0, therefore, ΔEp = En – En0 = k * x2 / 2 – k * x02 / 2 = k * x2 / 2.

We express the spring stiffness k according to Hooke’s law: F = k * x.

k = F / x.

A = F * x2 / 2 * x = F * x / 2.

A = 100 N * 0.1 m / 2 = 5 J.

Answer: when the spring was stretched, work A = 5 J.