How Does a Pulley System Affect Block Motion and Tension?

[Jessikalinphy]

Problem # 2
The pulley shown in the figure below, which is mounted in horizontal frictionless bearings, has a radius of 5.00 cm. One block has mass M = 500 g and the other has mass m = 460 g. When released from rest, the more massive block falls 75 cm in 5.00 s (without the cord slipping on the pulley.) (a) What is the magnitude of the blocks’ acceleration? What is the tension in the cord that supports (b) the more massive block and (c) the less massive block? (d) What are the magnitude and direction of the pulley’s angular acceleration? (e) What is its rotational inertia? (f) Find the rotational kinetic energy after 5.00 s.


Someone please help

 

[Shooting Star]

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[ogb p]

Based on the given information, we can use the equations of kinematics to solve for the magnitude of the blocks' acceleration and the tension in the cord that supports them.

(a) To find the magnitude of the blocks' acceleration, we can use the equation d = (1/2)at^2, where d is the distance fallen, a is the acceleration, and t is the time. Plugging in the given values, we get:

75 cm = (1/2)a(5.00 s)^2

Solving for a, we get a = 3.00 m/s^2. Therefore, the magnitude of the blocks' acceleration is 3.00 m/s^2.

(b) To find the tension in the cord supporting the more massive block, we can use the equation F = ma, where F is the tension, m is the mass, and a is the acceleration. Plugging in the given values, we get:

F = (0.500 kg)(3.00 m/s^2)

Therefore, the tension in the cord is 1.50 N.

(c) To find the tension in the cord supporting the less massive block, we can use the same equation as in (b), but with the mass of the less massive block. Plugging in the given values, we get:

F = (0.460 kg)(3.00 m/s^2)

Therefore, the tension in the cord is 1.38 N.

(d) To find the magnitude and direction of the pulley's angular acceleration, we can use the equation α = a/R, where α is the angular acceleration, a is the linear acceleration, and R is the radius of the pulley. Plugging in the values, we get:

α = (3.00 m/s^2)/(0.0500 m) = 60.0 rad/s^2

Since the blocks are falling, the pulley's angular acceleration is in the clockwise direction.

(e) To find the rotational inertia of the pulley, we can use the equation I = (1/2)MR^2, where I is the rotational inertia, M is the mass, and R is the radius. Plugging in the given values, we get:

I = (1/2)(0.500 kg + 0.460 kg)(0.0500 m)^2 = 0.000575 kg*m^2

(f) To find