Final answer:
The velocity of the cylinder when it moves downward 0.2 m from its equilibrium position is approximately 1.10 m/s.
Step-by-step explanation:
To determine the velocity of the cylinder when it moves downward 0.2 m from its equilibrium position, we can use the principle of conservation of mechanical energy. When the cylinder is at its equilibrium position, it has potential energy stored in the spring. The potential energy is given by the equation PE = 1/2 * k * x^2, where k is the stiffness of the spring and x is the displacement from the equilibrium position.
Given that the stiffness of the spring is 120 N/m and the displacement is 0.2 m, we can calculate the potential energy as follows: PE = 1/2 * 120 * (0.2)^2 = 2.4 J.
Since mechanical energy is conserved, the potential energy at the equilibrium position is converted into kinetic energy when the cylinder moves downward. The kinetic energy is given by the equation KE = 1/2 * m * v^2, where m is the mass of the cylinder and v is its velocity.
Using the given mass of 4 kg, we can rearrange the equation to solve for v: v = sqrt(2 * KE / m) = sqrt(2 * 2.4 / 4) = sqrt(1.2) ≈ 1.10 m/s.