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.
(c) What is the total energy of the system?
To find the total energy of the system, we need to consider both the potential energy and the kinetic energy.
1. Potential Energy:
The potential energy of the spring is given by the equation:
PE_spring = (1/2)kx^2
Where k is the spring constant and x is the displacement from the relaxed length.
Given that the spring stretches 2.20 cm and the force applied is 10.50 N, we can find the spring constant:
F = kx
10.50 N = k * 0.022 m
k = 10.50 N / 0.022 m
k ≈ 477.27 N/m
Now we can calculate the potential energy of the spring when the particle is at x=5.0 cm:
PE_spring = (1/2) * 477.27 N/m * (0.05 m)^2
PE_spring ≈ 0.5965 J
2. Kinetic Energy:
The kinetic energy of the particle is given by the equation:
KE_particle = (1/2)mv^2
Where m is the mass of the particle and v is its velocity.
At the maximum displacement, the particle is momentarily at rest (v=0), so the initial kinetic energy is zero.
3. Total Energy:
The total energy of the system is the sum of the potential energy and the kinetic energy:
Total energy = PE_spring + KE_particle
Total energy = 0.5965 J + 0 J
Total energy ≈ 0.5965 J
Therefore, the total energy of the system is approximately 0.5965 Joules.
To find the total energy of the system, we need to consider both the potential energy and the kinetic energy.
1. Potential Energy:
The potential energy of the spring can be calculated using Hooke's Law, which states that the potential energy stored in a spring is proportional to the displacement from its equilibrium position.
The potential energy stored in the spring (Us) is given by the formula:
Us = (1/2) * k * x^2
Where:
- Us is the potential energy stored in the spring
- k is the spring constant
- x is the displacement from the equilibrium position
In this case, the spring constant (k) can be determined using the force applied and the displacement from the relaxed length:
k = F / x
Substituting the given values:
k = 10.50 N / (2.20 cm) = 4.77 N/cm
Now, we can calculate the potential energy stored in the spring when the particle is displaced at x = 5.0 cm:
Us = (1/2) * (4.77 N/cm) * (5.0 cm)^2 = 59.63 N·cm = 0.5963 J
2. Kinetic Energy:
The kinetic energy (KE) of the particle can be calculated using the formula:
KE = (1/2) * m * v^2
Where:
- KE is the kinetic energy of the particle
- m is the mass of the particle
- v is the velocity of the particle
Since the particle is released from rest, its initial velocity (v) is zero. At any given time, the velocity of the particle can be calculated using the equation of motion: v = ω * A * sin(ωt + φ)
Where:
- ω is the angular frequency of the oscillation
- A is the amplitude of the oscillation
- t is the time
- φ is the phase constant (assumed to be zero here)
Since the particle is released from rest, the angular frequency (ω) can be determined using the spring constant and the mass of the particle:
ω = √(k / m)
Substituting the given values:
ω = √(4.77 N/cm / 1.100 kg) = 2.016 rad/s
Now, we can find the velocity of the particle when it is at x = 5.0 cm:
v = ω * A * sin(ωt + φ) = 2.016 rad/s * 0.05 m * sin(0 rad) = 0 m/s
Since the particle has zero velocity at x = 5.0 cm, its kinetic energy is also zero:
KE = (1/2) * (1.100 kg) * (0 m/s)^2 = 0 J
3. Total Energy:
The total energy (E) of the system is the sum of the potential energy (Us) and the kinetic energy (KE):
E = Us + KE = 0.5963 J + 0 J = 0.5963 J
Therefore, the total energy of the system is 0.5963 Joules.