Answer :
The speed of the block after it has descended is 3 meters is 3.54 m/s.
To determine the speed of the block after it has descended 3 meters, we need to consider the energy conservation principle. The initial potential energy of the block is equal to the final kinetic energy of the block plus the work done on the block by the pulleys.
The initial potential energy of the block is given by:
U = mgh
where m is the mass of the block, g is the acceleration due to gravity, and h is the height of the block.
The final kinetic energy of the block is given by:
K = (1/2) × m × v²
where m is the mass of the block and v is the velocity of the block.
The work done on the block by the pulleys is given by:
W = F × d
where F is the force exerted on the block by the pulleys and d is the distance traveled by the block.
The force exerted on the block by the pulleys is equal to the tension in the rope. The tension in the rope can be calculated using the following formula:
T = (m × g) / (1 + (Ia/Ib))
where T is the tension in the rope, m is the mass of the block, g is the acceleration due to gravity, Ia is the moment of inertia of pulley A, and Ib.
where I is the moment of inertia, m is the mass of the pulley, and r is the radius of the pulley.
Substituting the given values into the formulas above, we get:
U = 5 × 9.81 × 3 = 146.1 J.
Ia = (1/2) × 15 × (77)² = 70,245 [tex]mm^4[/tex]
Ib = (1/2) × 3 × (31)² = 2,871 [tex]mm^4[/tex]
T = (5 × 9.81) / (1 + (70,245/2,871)) = 33.4 N
W = T × 3 = 100.2 J
K = (1/2) × 5 × v²
Substituting the values for U, W, and K into the energy conservation equation, we get:
146.1 J = 100.2 J + (1/2) × 5 × v²
v² = (146.1 J - 100.2 J) / (1/2) × 5
v² = 45.9 J / 2.5
v = √(45.9 J / 2.5) = 3.54 m/s
Therefore, the speed of the block after it has descended 3 meters is approximately 3.54 m/s, rounded to two decimal places.
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