A 58.2 g object is attached to a horizontal spring with a spring constant of 11.8 N/m and released from rest with an amplitude of 24.4 cm.
What is the velocity of the object when it is halfway to the equilibrium position if the surface is frictionless?
U = potential energy of spring = (1/2) k x^2
calculate U for x = .244 meter
calculate U for x = .122 meter
decrease in U = (1/2) m v^2
4+3
To find the velocity of the object when it is halfway to the equilibrium position, we can use the principle of conservation of energy.
1. First, let's find the potential energy (PE) of the system when the object is halfway to the equilibrium position. At this point, the spring is stretched or compressed by half of the amplitude, which is 12.2 cm.
PE = 0.5 * k * Δx^2
where k is the spring constant and Δx is the displacement from the equilibrium position.
PE = 0.5 * 11.8 N/m * (0.122 m)^2 (Note: 1 cm = 0.01 m)
2. Next, let's find the total mechanical energy (E) of the system. Since the object is released from rest, the initial kinetic energy (KE) is zero.
E = PE + KE
Therefore, at the halfway point, E = PE.
3. Now, let's find the velocity (v) of the object when it's halfway to the equilibrium position. We can use the equation for potential energy to find this:
PE = 0.5 * m * v^2
where m is the mass of the object.
Rearranging the equation:
v^2 = 2 * PE / m
v = sqrt(2 * PE / m)
Substituting the values into the equation:
v = sqrt((2 * 0.5 * 11.8 N/m * (0.122 m)^2) / 0.0582 kg)
Calculating the value:
v ≈ 0.741 m/s
Therefore, the velocity of the object when it is halfway to the equilibrium position is approximately 0.741 m/s.