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

Answer

A

Explanation

`m_(2)g-T=m_(2)a`
`T-mum_(g)=m_(a)`
`((m_(2)g-mum_(1)g))/(m_(1)+m_(2))=a`
`T=m_(2)(g-a)`