A 4.5-kg brick is suspended by a light string from a 2.0-kg pulley. The brick is released from rest and falls to the floor below. The pulley may be considered a solid disk of radius 1.5 m. What is the linear acceleration of the brick and the tension in the string?
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F=ma
mg-T=ma
(4.5)(9.8)-T=4.5a
Torque=I(alpha fish)
I=1/2MR^2
TR=1/2MR^2(a/R)
1.5T=(1/2)(2.0)(1.5^2)(a/1.5)
T=1a
substitute for T
44.1-a=4.5a
44.1=5.5a
a=8.02m/s^2
substitute for a
T=8.02N
To find the linear acceleration of the brick and the tension in the string, we can apply Newton's laws of motion.
1. Calculate the gravitational force acting on the brick:
The gravitational force can be calculated using the equation F = m * g, where m is the mass and g is the acceleration due to gravity, which is approximately 9.8 m/s^2.
F_gravity = (mass of brick) * g = 4.5 kg * 9.8 m/s^2 = 44.1 N
2. Calculate the moment of inertia of the pulley:
The moment of inertia of a solid disk can be calculated using the equation I = (1/2) * m * r^2, where m is the mass of the pulley and r is the radius.
I_pulley = (1/2) * (mass of pulley) * (radius of pulley)^2 = (1/2) * 2.0 kg * (1.5 m)^2 = 2.25 kg * m^2
3. Calculate the torque acting on the pulley due to the falling brick:
The torque can be calculated using the equation τ = m * a * r, where τ is the torque, m is the mass of the brick, a is the linear acceleration, and r is the radius of the pulley.
τ = (mass of brick) * a * (radius of pulley) = 4.5 kg * a * 1.5 m
4. Apply Newton's second law for rotation:
The sum of the torques acting on the pulley must equal the moment of inertia times the angular acceleration (τ = I * α). Since the pulley is assumed to roll without slipping, the linear acceleration of the brick is equal to the angular acceleration (a = α).
Therefore, (mass of brick) * a * (radius of pulley) = 2.25 kg * m^2 * a
Solving for a, we get:
a = (2.25 kg * m^2) / (4.5 kg * 1.5 m) = 0.5 m/s^2
So, the linear acceleration of the brick is 0.5 m/s^2.
5. Calculate the tension in the string:
To calculate the tension, we can use Newton's second law for linear motion:
ΣF = m * a, where ΣF is the net force acting on the brick.
Considering the forces acting on the brick, we have the gravitational force acting downwards and the tension force acting upwards.
Tension - F_gravity = (mass of brick) * a
Tension = F_gravity + (mass of brick) * a
Tension = 44.1 N + (4.5 kg * 0.5 m/s^2)
Tension = 46.1 N
Therefore, the tension in the string is 46.1 N.
To find the linear acceleration of the brick and the tension in the string, we can use Newton's second law of motion and the concept of rotational motion.
1. Start by finding the net force acting on the system. Since the brick is falling, the net force will be the gravitational force acting on it.
- The gravitational force on the brick can be calculated using the formula: F_gravity = m * g, where m is the mass of the brick and g is the acceleration due to gravity (9.8 m/s²).
- The mass of the brick is given as 4.5 kg, so the gravitational force on the brick is F_gravity = 4.5 kg * 9.8 m/s² = 44.1 N.
2. Next, consider the pulley. As the brick falls, the pulley will rotate due to the tension in the string.
- The torque on the pulley can be calculated using the formula: τ = I * α, where τ is the torque, I is the moment of inertia of the pulley, and α is the angular acceleration.
- The moment of inertia of a solid disk can be calculated using the formula: I = (1/2) * m * r², where m is the mass of the pulley and r is its radius.
- The mass of the pulley is given as 2.0 kg, and the radius is given as 1.5 m, so the moment of inertia of the pulley is I = (1/2) * 2.0 kg * (1.5 m)² = 2.25 kg·m².
- Since the brick is falling, the angular acceleration of the pulley can be related to the linear acceleration of the brick using the formula: α = a / r, where a is the linear acceleration and r is the radius of the pulley.
- The radius is given as 1.5 m, so α = a / 1.5 m.
3. Now, let's analyze the forces acting on the system. There are two tensions in the system: one on the side with the brick, and one on the side with the pulley.
- The tension in the string on the side with the brick is the same as the gravitational force acting on the brick, since they are in opposite directions. So, the tension in the string is T1 = F_gravity = 44.1 N.
- The tension in the string on the side with the pulley can be related to the torque on the pulley using the formula: T2 * r = τ, where T2 is the tension in the string and r is the radius of the pulley.
4. Now, we can write the equations of motion for the system.
- For the brick: F_net = m * a
- For the pulley: T2 * r - T1 * r = I * α
5. Substitute the values into the equations and solve for a and T2.
- For the brick: 44.1 N = 4.5 kg * a
- For the pulley: T2 * 1.5 m - 44.1 N * 1.5 m = 2.25 kg·m² * (a / 1.5 m)
Simplify the equations:
- For the brick: a = 44.1 N / 4.5 kg ≈ 9.80 m/s²
- For the pulley: T2 * 1.5 m - 44.1 N * 1.5 m = 1.50 kg·m² * a
Substitute the value of a into the pulley equation and solve for T2:
- T2 * 1.5 m - 44.1 N * 1.5 m = 1.50 kg·m² * 9.80 m/s²
- T2 * 1.5 m = 14.7 N - 14.7 N
- T2 = 0 N
6. The linear acceleration of the brick is approximately 9.80 m/s² and the tension in the string is 0 N. This means that the string is slack, and the brick falls freely without any resistance from the string.
Therefore, the linear acceleration of the brick is approximately 9.80 m/s² and the tension in the string is 0 N.