A 5 kg mass attached to a spring passes its equilibrium point moving at 390 m/s. If the spring constant is 1450 N/m, what is the maximum displacement of the spring from its equilibrium position?
To find the maximum displacement of the spring from its equilibrium position, we can use the principle of conservation of mechanical energy.
Initially, the mass is at its equilibrium position, so its potential energy is zero. The total mechanical energy is equal to the sum of its potential and kinetic energy.
At the equilibrium point, all the mechanical energy is in the form of kinetic energy. The kinetic energy of the mass can be calculated using the equation:
KE = (1/2) * m * v^2
where KE is the kinetic energy, m is the mass, and v is the velocity.
Substituting the given values, we have:
KE = (1/2) * 5 kg * (390 m/s)^2
= 1/2 * 5 * 152100 m^2/s^2
= 760500 J
Since the potential energy is zero when the mass is at its equilibrium position, the total mechanical energy of the system can be considered as the kinetic energy at the maximum displacement position. At the maximum displacement, all the mechanical energy is in the form of potential energy.
The potential energy of a spring can be calculated using the equation:
PE = (1/2) * k * x^2
where PE is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position.
We can rearrange the equation to solve for x:
x = sqrt(2 * PE / k)
Substituting the known values, we have:
x = sqrt(2 * 760500 J / 1450 N/m)
= sqrt(525.86 m)
≈ 22.94 m
Therefore, the maximum displacement of the spring from its equilibrium position is approximately 22.94 meters.