The spring to which the weight of 1 kg is suspended has a length of 14 cm.

The spring to which the weight of 1 kg is suspended has a length of 14 cm. If an additional weight of 2 kg is suspended from it, the length of the spring will increase to 18 cm. What is the length of the spring if both weights are removed from it?

m1 = 1 kg.

l1 = 14 cm = 0.14 m.

g = 10 m / s2.

m2 = 2 kg.

l2 = 18 cm = 0.18 m.

l0-?

When the load is in balance, the force of gravity is balanced by the elastic force of the spring.

Let us write down the condition for the balance of weights on the spring for both cases:

m1 * g = k * (l1 -l0), (m1 + m2) * g = k * (l2 -l0), where k is the stiffness of the spring.

Subtract the first from the second equation: (m1 + m2) * g – m1 * g = k * (l2 -l0) – k * (l1 -l0).

m1 * g + m2 * g – m1 * g = k * l2 – k * l0 – k * l1 + k * l0.

m2 * g = k * (l2 – l1).

k = m2 * g / (l2 – l1).

k = 2 kg * 10 m / s2 = / (0.18 m – 0.14 m) = 500 N / m.

m1 * g = k * (l1 – l0).

m1 * g = k * l1 – k * l0.

l0 = (k * l1 – m1 * g) / k.

l0 = (500 N / m * 0.14 m – 1 kg * 10 m / s2) / 500 N / m = 0.12 m.

Answer: in an unstretched state, the spring has a length l0 = 0.12 m.