The spring has lengthened under the action of the suspended load.

The spring has lengthened under the action of the suspended load. Find the extension of the spring if its stiffness is 100n / m and the weight of the load is 1kg

The given tasks: k (coefficient of cruelty of the taken spring) = 100 N / m; m (suspended load) = 1 kg.

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

We express the elongation of the taken spring under the action of the load suspended from it from the formula: Ft = -Fcont. = k * Δx, whence Δx = Fт / k = m * g / k.

Let’s calculate: Δx = 1 * 10/100 = 0.1 m.

Answer: The deformation of the taken spring under the action of the load will be 0.1 m.