If a spring with a coefficient of elasticity of 10H \ m lengthened by 20 cm after the load was suspended

If a spring with a coefficient of elasticity of 10H \ m lengthened by 20 cm after the load was suspended, then the mass of the load is …

Initial data: k (coefficient of elasticity of the spring) = 10 N / m; Δx (elongation of the spring under the action of the load) = 20 cm (0.2 m).

Reference values: g (acceleration due to gravity) ≈ 10 m / s2.

The mass of the load suspended from the spring is determined from the equality: k * Δх = F = m * g, whence m = k * Δх / g.

Calculation: m = 10 * 0.2 / 10 = 0.2 kg (200 g).

Answer: A weight of 200 grams was suspended from the spring.