When suspending a 5 kg load, the dynamo meter spring stretched to the maximum scale division.

When suspending a 5 kg load, the dynamo meter spring stretched to the maximum scale division. Spring rate 1kN / m. What work was done while stretching the spring?

m = 5 kg.
g = 10 m / s2.
k = 1 kN / m = 1000 N / m.
A -?
Mechanical work A, which is performed by stretching the spring, is determined by the formula: A = ΔU, where ΔU is the change in the potential energy of the spring.
The change in the potential energy of the spring ΔU is determined by the formula: ΔU = k * x ^ 2/2, where k is the stiffness of the spring, x is the elongation of the spring.
Let us write down the equilibrium condition: Ft = Fcont.
Fт = m * g.
Fcont = k * x.
m * g = k * x.
x = m * g / k.
A = k * (m * g) ^ 2 / k ^ 2 * 2 = (m * g) ^ 2 / k * 2.
A = (5 kg * 10 m / s2) ^ 2/1000 N / m * 2 = 1.25 J.
Answer: when the spring was stretched, work A = 1.25 J.