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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 m<s

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 \mu_k. When the block A is sliding on the table, the tension in the string is:

Option: 1

\frac{(m_2+\mu\:k\: m_1)g}{m_1+m_2}


Option: 2

\frac{m_{1}m_{2}(1+\mu_{k})g}{m_{1}+m_{2}}


Option: 3

\frac{m_1m_2(1-\mu_k)g}{m_1+m_2}


Option: 4

\frac{(m_{2}+\mu_{k}m_{1})g}{m_{1}+m_{2}}


Answers (1)

best_answer

\\ \text{As it is clear from the diagram, force of friction} \ f=\mu_{k} R=\mu_{k} m_{1} g \\ \text{Equations of motion of two blocks are} \ \ m_{2} g-T=m_{2} a \\ T-\mu_{k} m_{1} g=m_{1} a \\ From \ \ m_{2} g-m_{2} a=T \\ \text{Putting the value of} \ \ T in m_{2} g-m_{2} a-\mu_{k} m_{1} g=m_{1} a \\ a=\frac{\left(m_{2}-\mu_{k} m_{1}\right) g}{m_{1}+m_{2}} \\ \ \ Putting \ \ in \ \ m_{2}\left[\frac{m_{2}-\mu_{k} m_{1}}{m_{1}+m_{2}}\right] g \\ T_{2}=m_{2} g-m_{2}\left[\frac{m_{2}-\mu_{k} m_{1}}{m_{1}+m_{2}}\right] \\ g =\frac{m_{2} m_{1} g+m_{2}^{2} g-m_{2}^{2} g-m_{2}^{2} g+\mu_{k} m_{1} m_{2} g}{m_{1}+m_{2}} \\ \therefore T=\frac{m_{1} m_{2}\left(1+\mu_{k}\right)(g)}{m_{1}+m_{2}}

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jitender.kumar

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