A weight weighing 4 kg was suspended on a spring 30 cm long.

A weight weighing 4 kg was suspended on a spring 30 cm long. When suspending the load, the length of the spring became 36.5 cm. Determine the work of stretching the spring.

l0 = 30 cm = 0.3 m.

l = 36.5 cm = 0.365 m.

g = 10 m / s2.

m = 4 kg.

A -?

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

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.

ΔEn = k * x ^ 2/2.

A = k * x ^ 2/2.

x = l – l0.

x = 0.365 m – 0.3 m = 0.065 m.

The spring stiffness k is found according to Hooke’s law: m * g = Fcont.

Fupr = k * x.

m * g = k * x.

k = m * g / x.

k = 4 kg * 10 m / s2 / 0.065 m = 615.4 N / m.

A = 615.4 N / m * (0.065 m) ^ 2/2 = 1.3 J.

Answer: work A = 1.3 J. was performed to stretch the spring.