How does this pulley work ---**1. Question Stem:** 8. Calculate accelerations a₁ a₂ and value of tension T. **2. Options:** The options are partially cropped and unreadable. The visible parts are: * a. 20g * b. 40c **3. Chart/Diagram Description:** * **Type:** A mechanical system diagram illustrating a system of masses, pulleys, and a string. * **Main Elements:** * **Horizontal Surface:** A flat, horizontal surface is present. It is explicitly labeled "Smooth", indicating that friction should be neglected. * **Mass m₁:** A rectangular block labeled "m₁" is placed on the smooth horizontal surface. * **Fixed Pulley:** A circular fixed pulley is attached to the right end of the smooth horizontal surface. * **Movable Pulley:** A circular movable pulley is located below the fixed pulley. * **Mass m₂:** A rectangular block labeled "m₂" is attached to the axle of the movable pulley, hanging vertically downwards. * **String System:** A continuous string connects m₁, the pulleys, and a fixed support: * One end of the string is attached to mass m₁ and extends horizontally to the right. * This horizontal segment passes over the fixed pulley. * From the fixed pulley, the string extends vertically downwards and passes underneath the movable pulley. * From the movable pulley, the string extends vertically upwards and is attached to a fixed horizontal bar (support) positioned above and to the right of the fixed pulley. * **Tension Labels:** * The horizontal segment of the string (between m₁ and the fixed pulley) is labeled "T" with arrows indicating the tension pulling m₁ to the right. * Both vertical segments of the string passing around the movable pulley (one coming down from the fixed pulley, the other going up to the fixed support) are labeled "T" with arrows indicating tension forces along the string. * **Directional Arrows:** * An arrow on the horizontal string indicates a rightward direction of pull on m₁. * Arrows on the vertical string segments show the direction of the string (downwards from the fixed pulley, upwards to the support). * An arrow originating from m₂ points vertically downwards, indicating its weight or the direction of its potential motion.

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This compound pulley system consists of mass m1 on a smooth frictionless horizontal surface, connected by a string that passes over a fixed pulley. The string then supports a movable pulley with mass m2 attached to it. The key feature is that the movable pulley has two string segments supporting it, each carrying tension T, which creates a mechanical advantage in this system. Now we establish free body diagrams for both masses. For mass m1 on the horizontal surface, tension T pulls it rightward, while normal and weight forces act vertically and cancel each other. For mass m2 attached to the movable pulley, its weight m2g acts downward. The key insight is that the movable pulley is supported by two string segments, each carrying tension T, so the total upward force is 2T. We establish coordinates with rightward positive for m1 and downward positive for m2. The string length constraint creates a specific relationship between the accelerations. When mass m1 moves a distance x to the right, the string segment over the fixed pulley increases by x. This extra string length must come from somewhere in the system. Due to the movable pulley configuration, the pulley and mass m2 move downward by only half that distance, x over 2. This geometric constraint gives us the acceleration relationship: a2 equals a1 over 2, where a1 is the acceleration of m1 and a2 is the acceleration of m2. Now we apply Newton's second law to both masses. For mass m1 in the horizontal direction, the sum of forces equals mass times acceleration, giving us T equals m1 times a1. For mass m2 and the movable pulley system in the vertical direction, we have the weight m2g acting downward and the tension force 2T acting upward, since two string segments support the pulley. This gives us m2g minus 2T equals m2 times a2. We then substitute our constraint relationship a2 equals a1 over 2 into the second equation, yielding m2g minus 2T equals m2 times a1 over 2. Now we solve the system of equations simultaneously. Starting with T equals m1 times a1 and m2g minus 2T equals m2 times a1 over 2, we substitute the first equation into the second. This gives us m2g minus 2 times m1 times a1 equals m2 times a1 over 2. Rearranging, we get m2g equals a1 times the quantity 2m1 plus m2 over 2, which simplifies to a1 times 4m1 plus m2, all over 2. Solving for a1, we get a1 equals 2m2g over 4m1 plus m2. Then a2 equals a1 over 2, which gives us m2g over 4m1 plus m2. Finally, the tension T equals m1 times a1, which is 2m1 times m2 times g, all over 4m1 plus m2. We can verify our solution by checking that all original equations are satisfied.