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 clearv 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 `m_(2) g - m_(2) a = T`
Putting the value of `T` in
`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
` 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-m_(2)^(2)g+mu_(k)m_(1)m_(2)g)/(m_(1) +m_(2))`
`:. T = (m_(1)m_(2) (1+mu_(k))(g))/(m_(1) + m_(2))`
.