When a weight of 1 kg was suspended from a spring with a stiffness of 500 N / m. its length became 12 cm.

When a weight of 1 kg was suspended from a spring with a stiffness of 500 N / m. its length became 12 cm. To what length the spring will stretch if another weight of 1 kg is suspended from it.

1) Let’s calculate the add. elongation when another 1 kg was suspended from the spring.

Ft = Fcont., Where Ft – add. gravity (Fт = m * g, where m – additional weight (m = 1 kg), g – free fall acceleration (g = 10 m / s2)), Fcont. – elastic force (Fel. = K * ∆l, where k – stiffness (k = 500 N / m), ∆l – additional elongation).

m * g = k * ∆l.

∆l = m * g / k = 1 * 10/500 = 0.02 m.

2) Calculate the length of the stretched spring:

l = ∆l + l1, where l1 is the beginning. spring length (l1 = 12 cm = 0.12 m).

l = ∆l + l1 = 0.02 + 0.12 = 0.14 m.