A block `A` of mass `m_(1)` 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_(2)` is suspended. The coefficient of knetic friction between the block and table is `mu_(k)` . When the block `A` is sliding on the table, the tension in the string is.

Options

A. `((m_(2) - mu k m_(1)) g)/((m_(1) + m_(2)))`
B. `(m_(1) m_(2) (1 + mu_(k)) g)/((m_(1) + m_(2)))`
C. `(m_(1) m_(2) (1 - mu_(k)) g)/((m_(1) + m_(2)))`
D. `((m_(2) + mu_(k) m_(1)) g)/((m_(1) + m_(2)))`

Answer

B

Explanation

`T - mu_(k) m_(1) g = m_(1)a " " … (i)`
`m_(2) g - T = m_(2) a " " ……. (ii)`
By (i) and (ii)
`(m_(1) + m_(2)) a = m_(2) g - mu_(k) m_(1) g`
`implies a = (m_(2) g - mu_(s) m_(1) g)/(m_(1) + m_(2))`

`T = m_(2) g - m_(2) a = m_(2) g - m_(2) [(m_(2) g- mu_(k) m_(1) g)/(m_(1) + m_(2))]`
`implies T = (m_(1) m_(2) g + m_(2)^(2) g - m_(2)^(2) g + m_(1) m_(2) mu_(k) g)/(m_(1) + m_(2))`
`implies T = (m_(1) m_(2) g (1 + mu_(k)))/(m_(1) + m_(2))`