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

Answer

C

Explanation

`T-mu_(k)m_(1)g=m_(1)a` (i)
`m_(2)g-T=m_(2)a` (ii)
`T-mu_(k)m_(1)g=(m_(1)(m_(2)g-T))/(m_(2))`
`T=m_(1)m_(2)g(mu_(k)+1)/(m_(1)+m_(2))` .