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A block A of mass m(1) rests on a horizo...

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 kinetic friction between the block and the table is`mu_(k)`. When the block A is sliding on the table the tension in the string is

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)))`

Text Solution

Verified by Experts

The correct Answer is:
C
Frictional force acting on block A = `mu_(k)m_(1)g`
Hence resultant force on the blocks = `m_(2)g-mu_(k)m_(1)`g
Their acceleration, a = `(m_(2)g-mu_(k)m_(1)g)/(m_(1)+m_(2))`
If the tension in the string is T, then for block B,
`m_(2)g-T=m_(2)a`
`:. T=m_(2)(g-a) = m_(2)g(1-(m_(2)-mu_(k)m_(1))/(m_(1)+m_(2)))`
=`m_(2)g (m_(1)+mu_(k)m_(1))/(m_(1)+m_(2))`
`=(m_(1)m_(2)(1+mu_(k))g)/(m_(1)+m_(2))`
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