A horizontal spring has a spring constant k = 23.2 N/m. A 0.165 kg mass is attached to the spring and is free to slide on a frictionless surface. The mass is pulled 26.2 cm from the equilibrium point. What is the speed (m/s) of the mass as it passes the equilibrium point?
To find the speed of the mass as it passes the equilibrium point, we can start by determining the potential energy stored in the spring when it's pulled 26.2 cm from the equilibrium point.
The formula for potential energy stored in a spring is given by:
U = (1/2)kx²
where U is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position.
Given that the spring constant k = 23.2 N/m and the displacement x = 26.2 cm = 0.262 m, we can substitute these values into the formula:
U = (1/2) * 23.2 * (0.262²)
U = (1/2) * 23.2 * 0.068644
U = 0.7909472 J
Next, we can use the principle of conservation of mechanical energy to determine the speed of the mass as it passes the equilibrium point. At its highest position (maximum displacement), all of the potential energy is converted into kinetic energy.
The formula for kinetic energy is given by:
K = (1/2)mv²
where K is the kinetic energy, m is the mass, and v is the velocity.
Since the mass m = 0.165 kg, and the potential energy U is equal to the kinetic energy K, we can equate the equations:
U = K
(1/2)kx² = (1/2)mv²
Substituting the known values:
0.7909472 = (1/2) * 0.165 * v²
Now, we can solve for v² by rearranging the equation:
v² = (0.7909472 * 2) / 0.165
v² = 9.571222
v ≈ √9.571222
v ≈ 3.097 m/s
Therefore, the speed of the mass as it passes the equilibrium point is approximately 3.097 m/s.