The spring, to which a weight of 500 g is attached, was stretched by 10 cm and released.

The spring, to which a weight of 500 g is attached, was stretched by 10 cm and released. Spring stiffness 100n / m. The spring began to oscillate. Determine the speed of the load when it comes to equilibrium.

Data: m (mass of the weight attached to the spring of the pendulum) = 500 g (in SI m = 0.5 kg); Δx (stretching) = 10 cm (in SI Δx = 0.1 m); k (spring rate) = 100 N / m.

To determine the speed of the load at the moment of equilibrium passing, we apply the law of conservation of energy: k * Δx ^ 2/2 = m * V ^ 2/2, whence V = √ (k * Δx ^ 2 / m).

Let’s perform the calculation: V = √ (k * Δx ^ 2 / m) = √ (100 * 0.1 ^ 2 / 0.5) = 1.41 m / s.

Answer: When passing the equilibrium point, the load will have a speed of 1.41 m / s.