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 can use the principle of conservation of mechanical energy. The system begins with only potential energy, which is converted into kinetic energy as the particle begins to move.
From the given information, we know that the spring stretches 2.20 cm and a force of 10.50 N is applied. We can use Hooke's Law to find the spring constant, k.
Hooke's Law:
F = k * x
10.50 N = k * 0.022 m
k = 10.50 N / 0.022 m
k ≈ 477.27 N/m
Now we can find the potential energy stored in the spring when the particle is at x = 5.0 cm.
Potential Energy:
U = (1/2) * k * x^2
U = (1/2) * 477.27 N/m * (0.05 m)^2
U ≈ 0.596 N
Since energy is conserved, at the maximum velocity, all potential energy is converted into kinetic energy.
Kinetic Energy:
K = (1/2) * m * v^2
0.596 N = (1/2) * 1.100 kg * v^2
Solving for v:
v^2 = (2 * 0.596 N) / 1.100 kg
v^2 = 1.083 N / 1.100 kg
v^2 ≈ 0.984 m^2/s^2
Taking the square root of both sides, we get:
v ≈ ± 0.992 m/s
The maximum velocity of the system is approximately 0.992 m/s in either direction.
To find the maximum velocity of the system, we need to analyze the energy in the system.
Given information:
- Force applied to stretch the spring, F = 10.50 N.
- Displacement of the spring from its relaxed length, x = 2.20 cm.
- Mass of the particle, m = 1.100 kg.
- Displacement of the particle from the origin, x = 5.0 cm.
We'll start by finding the spring constant, k.
The formula relating the force, displacement, and spring constant is:
F = k * x
Rearranging the formula to solve for k:
k = F / x
Plugging in the values:
k = 10.50 N / 0.022 m (converting 2.20 cm to meters)
k = 477.27 N/m
Now, let's find the potential energy stored in the spring, which will be converted to kinetic energy of the particle:
The potential energy in a spring is given by the formula:
PE = (1/2) * k * x^2
Plugging in the values:
PE = (1/2) * 477.27 N/m * (0.022 m)^2
PE ≈ 0.117 J (rounded to three decimal places)
The maximum velocity of the system occurs when all the potential energy is converted to kinetic energy. So, we can equate the potential energy to the kinetic energy:
PE = KE
0.117 J = (1/2) * m * v^2
Solving for v (velocity) gives:
v^2 = (2 * PE) / m
v^2 = (2 * 0.117 J) / 1.100 kg
v ≈ 0.494 m/s (rounded to three decimal places)
Therefore, the maximum velocity of the system is approximately 0.494 m/s.