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

As is clear from
force of friction `f = mu_(k) R = mu_(k) m_(1) g`
Equations of motion of two blocks are
`m_(2) g - T = m_(2) a`
`T - mu_(k) m_(1) g = m_(1) a`
From (i) `m_(2) g - m_(2) a =T`
Putting the value of `T` in (ii)
`m_(2) g - m_(2) a - mu_(k) m_(1) g = m_(1) a`
`a = ((m_(2) -mu_(k) m_(1))g)/(m_(1) + m_(2))`
putting in (i)
`m_(2) g - T = m_(2) [[m_(2) -mu_(k)m_(1)]/(m_(1) + m_(2))]g`
`T_(2) =m_(2) g-m_(2)[[m_(2) -mu_(k) m_(1))/(m_(1) + m_(2))]g`
`= (m_(2) m_(1) g + m_(2)^(2) g -m_(2)^(2) g + mu_(k) m_(1) m_(2) g)/(m_(1) + m_
(2)`
.