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

( c) FBD of block A,

`T-m_(1)a=f_(k)`…….(i)
FBD of block B,

`m_(2)g-T=m_(2)a`…….(ii)
On adding Eqs.(i) and (ii), we get
`m_(2)g-m_(1)a=m_(2)a+f_(k)`
`implies m_(2)g-m_(1)a=m_(2)a+mu_(k)m_(1)g`
`implies a=((m_(2)-mu_(k)m_(1))g)/(m_(1)+m_(2))`
From Eq.(ii),
`T=m_(2)(g-a)=m_(2)[1-((m_(2)-mu_(k)m_(1)))/(m_(1)+m_(2))]g`
`T=(m_(1)m_(2)(1+mu_(k)))/(m_(1)+m_(2))g`