A system comprised blocks,a light frictionless pulley,and connecting ropes is shown in the figure.The 9.0-kg block is on a perfectly smooth horizontal table.The surfaces of the 12-kg block are rough,with between the block and the table.If the 5.0-kg block accelerates downward when it is released,find its acceleration. A)1.0 m/s² B)1.2 m/s² C)1.4 m/s² D)1.6 m/s² E)1.8 m/s²

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Quizplus · Accepted Answer

Since the 9.0-kg block is on a horizontal table, it does not affect the system, and we only need to consider the motion of the 5.0-kg block and the 12.0-kg block. The force acting on the 5.0-kg block is its weight, which is 5g N, where g is the acceleration due to gravity. The net force on the 5.0-kg block is its weight minus the tension in the rope, which is connected to the 12.0-kg block. Therefore, we have:
5g - T = 5a
where T is the tension in the rope and a is the acceleration of the 5.0-kg block.
The force acting on the 12.0-kg block is the tension in the rope, which is pulling it upward, and the force of friction, which is pulling it backward. Therefore, we have:
T - 11g = 12a
where 11g is the force of friction and a is the acceleration of the 12.0-kg block.
Since the pulley is light and frictionless, the tension in the rope is the same on both sides of the pulley. Therefore, we can eliminate T from the equations above:
5g - 5a = 12a + 11g
Simplifying and solving for a, we get:
a = (5g - 11g) / (5 + 12) = -6g/17
Since the acceleration is negative, it means that the 5.0-kg block is moving upward. We need to take the absolute value to obtain the magnitude of the acceleration:
|a| = 6g/17 = 1.0 m/s^2 (approximately)

A System Comprised Blocks,a Light Frictionless Pulley,and Connecting Ropes Is