The spring of force constant k = 400 N/m is set up horizontally with a 0.300 kg mass attached on it resting on a frictionless table. The mass is pulled so that the spring is stretched 0.100 m from the equilibrium point, and is then released from rest. Determine the speed of the system when the spring is at the equilibrium position.
max Pe = max Ke
(1/2)k Xmax^2 = (1/2) m Vmax^2
(1/2)(400)(.1)^2 = (1/2)(.3)(Vmax)^2
40/3 = Vmax^2
To determine the speed of the system when the spring is at the equilibrium position, we can use the principle of conservation of mechanical energy.
The total mechanical energy at any point is equal to the sum of the kinetic energy and the potential energy. When the mass is at the equilibrium position (0.100 m from the rest position), the potential energy of the spring is maximum and the kinetic energy is zero.
First, let's calculate the potential energy at the maximum displacement.
The potential energy stored in a spring can be calculated using the formula:
PE = (1/2)kx^2
Where:
PE is the potential energy of the spring,
k is the force constant (400 N/m),
x is the displacement from the equilibrium position (0.100 m).
Substituting the given values into the formula:
PE = (1/2)(400 N/m)(0.100 m)^2
= (1/2)(400 N/m)(0.010 m^2)
= 2 J
So, at the maximum displacement, the potential energy of the spring is 2 Joules.
Since the system is at rest at this position, all of the potential energy is converted into kinetic energy as it reaches the equilibrium position.
So, the kinetic energy (KE) at the equilibrium position is equal to the potential energy (PE).
KE = PE = 2 J
The kinetic energy can be calculated using the formula:
KE = (1/2)mv^2
Where:
KE is the kinetic energy of the system,
m is the mass (0.300 kg),
v is the speed of the system.
Substituting the given values into the formula:
2 J = (1/2)(0.300 kg)v^2
Multiplying through by 2:
4 J = (0.300 kg)v^2
Dividing through by 0.300 kg:
(4 J) / (0.300 kg) = v^2
Simplifying:
13.333 m^2/s^2 = v^2
Taking the square root of both sides:
v = √(13.333 m^2/s^2)
Using a calculator:
v ≈ 3.65 m/s
Therefore, the speed of the system when the spring is at the equilibrium position is approximately 3.65 m/s.