When the buffer spring is compressed by 3 cm, an elastic force of 6 kN occurs.

When the buffer spring is compressed by 3 cm, an elastic force of 6 kN occurs. How much will this force increase if you compress the spring another 2 cm?

Data: Δx1 (initial compression of the buffer spring) = 3 cm (0.03 m); F1 (initial elastic force) = 6 kN (6 * 10 ^ 3 N); Δx (additional spring compression) = 2 cm (0.02 m).

1) Rigidity coefficient of the buffer spring: k = F1 / Δx1 = 6 * 10 ^ 3 / 0.03 = 200 * 10 ^ 3 N / m.

2) Second spring compression: Δx2 = Δx1 + Δx = 0.03 + 0.02 = 0.05 m.

3) Elastic force at the second compression: F2 = k * Δx2 = 200 * 10 ^ 3 * 0.05 = 10 * 10 ^ 3 N.

4) Increase in elastic force: ΔF = F2 – F1 = 10 * 10 ^ 3 – 6 * 10 ^ 3 = 4 * 10 ^ 3 (4 kN).

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