A ball weighing 0.64 kg is fixed on a spring with a stiffness of 0.4 kN / m. How much should the spring be stretched

A ball weighing 0.64 kg is fixed on a spring with a stiffness of 0.4 kN / m. How much should the spring be stretched from the equilibrium position for the ball to pass the equilibrium position at a speed of 1 m / s?

Given:

m = 0.64 kilograms is the mass of the ball attached to the spring;

k = 0.4 kN / m = 400 N / m – spring rate;

v = 1 m / s – the required speed of the ball when passing the equilibrium position.

It is required to determine dx (meter) – how much you need to stretch the spring from the equilibrium position.

Let the potential energy of the spring and the kinetic energy of the ball in the equilibrium position be equal to zero.

Then, by stretching the ball by a certain amount dx, we will increase the potential energy of the spring, which, when the ball passes the equilibrium position, completely transforms into the kinetic energy of the ball’s motion, that is:

m * v ^ 2/2 = k * dx ^ 2/2;

m * v ^ 2 = k * dx ^ 2;

dx ^ 2 = m * v ^ 2 / k;

dx = (m * v ^ 2 / k) ^ 0.5 = (0.64 * 12/400) 0 ^, 5 = (0.64 / 400) ^ 0.5 =

= (0.64 / 400) ^ 0.5 = 0.0016 ^ 0.5 = 0.04 meters = 4 centimeters.

Answer: The spring needs to be stretched 4 centimeters from the equilibrium position.