A spring of negligible mass stretches 2.20 cm from its relaxed length when a force of 10.50 N is applied. A 1.100 kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is displaced from the origin to x=5.0 cm and released from rest at t=0.
(d) What is the maximum velocity of the system?
To find the maximum velocity of the system, we need to use the principles of conservation of energy. The initial potential energy stored in the stretched spring will be converted into kinetic energy when the particle is released.
First, let's find the spring constant (k) using Hooke's Law. Hooke's Law states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.
F = -kx
where F is the force applied, k is the spring constant, and x is the displacement from the relaxed length.
Given that the force applied is 10.50 N and the displacement is 2.20 cm (or 0.022 m), we can rewrite Hooke's Law equation as:
10.50 N = -k * 0.022 m
Rearranging the equation, we get:
k = -10.50 N / 0.022 m
k = -477.27 N/m
Now that we have the spring constant, we can calculate the potential energy stored in the spring when the particle is at its maximum displacement.
Potential energy (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.
Given that the displacement (x) is 0.05 m (or 5.0 cm), and the spring constant (k) is -477.27 N/m, we can calculate the potential energy:
PE = (1/2) * (-477.27 N/m) * (0.05 m)^2
PE = 0.5966 J
Since the total mechanical energy of the system is conserved, the potential energy at the maximum displacement will be equal to the kinetic energy at the maximum velocity.
Kinetic energy (KE) = Potential energy (PE)
0.5 * m * v^2 = 0.5966 J
where KE is the kinetic energy, m is the mass of the particle, and v is the velocity.
Given that the mass (m) is 1.1 kg, we can solve for the maximum velocity (v):
0.5 * 1.1 kg * v^2 = 0.5966 J
v^2 = (0.5966 J) / (0.5 * 1.1 kg)
v^2 = 1.0824 J/kg
v = √(1.0824 J/kg)
v ≈ 1.04 m/s
Therefore, the maximum velocity of the system is approximately 1.04 m/s.