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A block A of mass m₁ rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block B of mass m₂ is suspended. The coefficient of kinetic friction between the block and the table is μₖ. When the block A is sliding on the table, the tension in the string is

1. (m₂ − μₖm₁)g / (m₁ + m₂)
2. m₁m₂(1 + μₖ)g / (m₁ + m₂)
3. m₁m₂(1 − μₖ)g / (m₁ + m₂)
4. (m₁ + m₂)g / (m₁ + m₂)

Correct answer: (m₂ − μₖm₁)g / (m₁ + m₂)

Solution

The forces acting on block A are tension (T) and kinetic friction (μₖm₁g), while block B experiences tension (T) and gravitational force (m₂g). Using Newton's second law for both blocks and solving for T, we get T = (m₂ − μₖm₁)g / (m₁ + m₂).

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