The student assembled the installation under the action of a load weighing 0.4 kg, the spring stretched by 0.1 m

The student assembled the installation under the action of a load weighing 0.4 kg, the spring stretched by 0.1 m, the potential energy of the spring when elongated is

Task data: m (mass of the load acting on the spring) = 0.4 kg; Δx (spring tension) = 0.1 m.

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

The potential energy that the spring will acquire after elongation is determined by the formula: En = k * Δx ^ 2/2 = (F / Δx) * Δx ^ 2/2 = F * Δx / 2 = m * g * Δx / 2.

Let’s make a calculation: Ep = 0.4 * 10 * 0.1 / 2 = 0.2 J.

Answer: After elongation, the spring will acquire a potential energy of 0.2 J.