# 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.