A 9.3 kg mass is attached to a hanging vertical spring with k=49 N/m. If the mass is released when the spring is unstretched, how fast will it be moving as it falls past the equilibrium point?
To determine the speed at which the 9.3 kg mass will be moving as it falls past the equilibrium point, we need to consider the potential energy at the starting position and convert it to kinetic energy at the equilibrium point.
Here are the steps to solve this problem:
Step 1: Calculate the potential energy at the starting position.
The potential energy stored in a spring is given by the equation:
Potential Energy (PE) = 0.5 * k * x^2
where k is the spring constant and x is the displacement from the equilibrium point. In this case, the spring is unstretched, so x = 0.
Potential Energy (PE) = 0.5 * 49 N/m * (0 m)^2
PE = 0 J
Step 2: Convert potential energy to kinetic energy.
At the equilibrium point, all the potential energy is converted to kinetic energy. Therefore, the initial potential energy is equal to the final kinetic energy.
Potential Energy (PE) = Kinetic Energy (KE)
0 J = 0.5 * m * v^2
where m is the mass and v is the velocity.
Step 3: Solve for velocity.
Rearrange the equation to solve for v:
v^2 = (2 * PE) / m
v^2 = (2 * 0 J) / 9.3 kg
v^2 = 0 m^2/s^2
Taking the square root of both sides, we get:
v = 0 m/s
Therefore, the mass will be moving with a velocity of 0 m/s as it falls past the equilibrium point.