12th grade physics explainer video---**Question Number:** 27.
**Question Stem:** In the figure, the blocks A, B and C of mass m each have accelerations a₁, a₂ and a₃ respectively. F₁ and F₂ are external forces of magnitudes 2mg and mg respectively. Then
**Image Description:**
The image shows three separate vertical pulley systems labeled A, B, and C. Each system consists of a single pulley fixed to a ceiling (indicated by horizontal lines). A string passes over each pulley, with blocks attached to the ends of the string.
* **System A:** A block of mass m is attached to the left end of the string and labeled 'A'. The string continues over the pulley. An external force F₁ = 2mg is applied downwards to the right end of the string. The mass of the block on the right is not explicitly labeled by text next to the block, but based on the setup it appears to be part of the external force application. Assuming it's the end of the string where the force is applied, there isn't a second block of mass m shown directly attached to the string on the right side in the same way as the block labeled 'A' is on the left. The force F₁ is shown acting downwards at the end of the string.
* **System B:** A block of mass m is attached to the left end of the string and labeled 'B'. A block of mass 2m is attached to the right end of the string. The string passes over the pulley.
* **System C:** A block of mass m is attached to the left end of the string and labeled 'C'. A block of mass m is attached to the right end of the string. The string passes over the pulley. An external force F₂ = mg is applied downwards to the block on the right side (the one with mass m).
**Options:**
(a) a₁ = a₂ = a₃
(b) a₁ > a₃ > a₂
(c) a₁ = a₂, a₂ > a₃
(d) a₁ > a₂, a₂ = a₃
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Let's analyze three different pulley systems. System A has a block of mass m with an external force F₁ equals 2mg applied downward. System B is an Atwood machine with masses m and 2m. System C has two blocks of mass m each, with an external force F₂ equals mg applied downward to the right block. We need to find and compare their accelerations.
Let's analyze System A step by step. Block A has mass m and is connected to a string over a pulley. An external force F₁ equals 2mg is applied downward to the right side of the string. Since the string is massless, the tension throughout is equal to the applied force, so T equals 2mg. The forces on block A are tension T upward and weight mg downward. Using Newton's second law: T minus mg equals m times a₁. Substituting T equals 2mg: 2mg minus mg equals m times a₁, which gives us a₁ equals g.
Now let's analyze System B, which is a classic Atwood machine. We have block B with mass m on the left and a block with mass 2m on the right. Since 2m is greater than m, the heavier block will accelerate downward and block B will accelerate upward. For block B, the equation is T minus mg equals m times a₂. For the 2m block, the equation is 2mg minus T equals 2m times a₂. Adding these equations eliminates T and gives us mg equals 3m times a₂. Therefore, a₂ equals g over 3.
Now let's analyze System C. We have two blocks of mass m each, with an external force F₂ equals mg applied downward to the right block. The total downward force on the right block is its weight mg plus the external force mg, which equals 2mg. For block C, the equation is T minus mg equals m times a₃. For the right block, the equation is 2mg minus T equals m times a₃. Adding these equations gives us mg equals 2m times a₃. Therefore, a₃ equals g over 2.
Now let's compare our results. From System A, we found a₁ equals g. From System B, we found a₂ equals g over 3. From System C, we found a₃ equals g over 2. Comparing these values: g is greater than g over 2, which is greater than g over 3. Therefore, a₁ is greater than a₃, which is greater than a₂. This matches option (b): a₁ > a₃ > a₂. The answer is (b).