Two blocks are connected by a light, inextensible string over a frictionless pulley. Block A has a mass of 3 kg and hangs vertically. Block B has a mass of 2 kg and rests on a frictionless horizontal table.
What is the acceleration of the system in m/s²?
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We have a pulley system with two blocks connected by a string. Block A has a mass of 3 kilograms and hangs vertically. Block B has a mass of 2 kilograms and rests on a frictionless horizontal table. The string is light and inextensible, and the pulley is frictionless. We need to find the acceleration of this system.
Now let's analyze the forces acting on each block. For Block A hanging vertically, we have its weight of 29.4 Newtons acting downward and tension T acting upward. For Block B on the horizontal table, tension T acts horizontally, while the normal force and weight cancel each other vertically since the table is frictionless.
Now we apply Newton's second law to each block. For Block A moving vertically, the net force equals mass times acceleration: 29.4 minus T equals 3a. For Block B moving horizontally, tension T equals mass times acceleration: T equals 2a. Since the string is inextensible, both blocks have the same magnitude of acceleration.
Now let's solve for the acceleration. We have two equations: 29.4 minus T equals 3a, and T equals 2a. Substituting the second equation into the first gives us 29.4 minus 2a equals 3a. Rearranging, we get 29.4 equals 5a. Therefore, a equals 29.4 divided by 5, which equals 5.88 meters per second squared.
To summarize what we've learned: We applied Newton's second law to analyze forces on both blocks, used the constraint that the string gives both blocks the same acceleration, and solved the resulting system of equations to find that the acceleration is 5.88 meters per second squared. This systematic approach can be applied to solve any similar pulley system problem.