Determine the work of the elastic force of a weightless spring if, under the action of a load weighing 1.3 kg

Determine the work of the elastic force of a weightless spring if, under the action of a load weighing 1.3 kg, a vertically hanging spring stretches from an undeformed state by 1 cm.

m = 1.3 kg.

g = 10 N / kg.

x = 1 cm = 0.01 m.

A -?

The work of the elastic force A is equal to the change in potential energy ΔEp, taken with the opposite sign: A = – ΔEp.

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

We express the stiffness of the spring k from the condition of the balance of the load: Fcont = m * g.

Fпр = k * x – Hooke’s law.

k * x = m * g.

k = m * g / x.

En = m * g * x ^ 2/2 * x = m * g * x / 2.

A = – m * g * x / 2.

A = – 1.3 kg * 10 N / kg * 0.01 m / 2 = – 0.065 J.

Answer: the work of the elastic force is A = – 0.065 J.