A spring of negligible mass stretches 2.20 cm from its relaxed length when a force of 10.5 N is applied. A 1.1 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. cm and released from rest at t=0.
(g2) Determine the acceleration of the particle when t=0.50 s
To determine the acceleration of the particle at t = 0.50 s, we need to find the force acting on the particle at that time.
1. First, let's calculate 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 resting position.
F = -kx
Here, F is the force applied, k is the spring constant, and x is the displacement from the relaxed length of the spring.
Rearranging the equation, we get:
k = -F / x
Plugging in the values, we have:
k = -10.5 N / 0.0220 m = -477.3 N/m
2. Next, we need to determine the equilibrium position (x = 0) when the particle is at rest. From the given information, we know that when the spring is stretched by 2.20 cm, a force of 10.5 N is applied. Therefore, the relaxed length of the spring is the equilibrium position.
3. Now, we can write the equation of motion for the particle using Newton's second law:
F_net = ma
The net force acting on the particle is the force of the spring, given by Hooke's Law (F = -kx), and the force of gravity (F = mg). Since the particle is on a horizontal surface, there is no friction acting on it.
F_net = -k(x - x_eq) + mg
Rearranging the equation, we get:
ma = -kx + kx_eq + mg
The mass of the particle (m) is given as 1.1 kg, the spring constant (k) is -477.3 N/m (negative sign indicates the force opposes displacement), the displacement of the particle (x) is 0.05 m, and the equilibrium position (x_eq) is 0.
4. Substitute the values into the equation and solve for acceleration (a):
(1.1 kg)a = -(-477.3 N/m)(0.05 m) + (1.1 kg)(9.8 m/s^2)
Simplifying the equation, we get:
1.1a = 23.87 N + 10.78 N
1.1a = 34.65 N
Finally, divide both sides by 1.1 to solve for acceleration (a):
a = 34.65 N / 1.1 kg ≈ 31.50 m/s^2
Therefore, the acceleration of the particle at t = 0.50 s is approximately 31.50 m/s^2.