With a force of 80 N, the spring is stretched by 2 cm. The initial length of the spring is 15 cm.

With a force of 80 N, the spring is stretched by 2 cm. The initial length of the spring is 15 cm. What work must be done to stretch it to 20 cm?

F1 = 80 N.

x0 = 15 cm = 0.15 m.

Δx1 = 2 cm = 0.02 m.

x2 = 20 cm = 0.2 m.

A -?

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

The potential energy of the spring Ep is determined by the formula: Ep = k * x2 / 2, where k is the stiffness of the spring, x is the elongation of the spring.

ΔEn = k * Δx2 ^ 2/2 = k * (x2 – x0) ^ 2/2.

The spring stiffness k is found from Hooke’s law: k = F1 / Δx1.

The formula for determining the work will take the form: A = F1 * (x2 – x0) ^ 2/2 * Δx1.

A = 80 H * (0.2 m – 0.15 m) ^ 2/2 * 0.02 m = 5 J.

Answer: to stretch the spring, it is necessary to do work A = 5 J.