NEET 2015 Physics question
A block A of mass 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 is suspended. The coefficient of kinetic friction between the block and the table is . When the block A is sliding on the table, the tension in the string is :-
Answer: (2)
Solution
Let the common acceleration be and the tension . For block A (horizontal): . For block B (vertical): . Adding: . Then m_2(gm_2\dfrac{m_2 - μ_k m_1}{m_1+m_2}$] =
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