In the arrangement shown in figure, m_A = m_B = 2kg. String is massless and pulley is frictionless. Block B is resting on a smooth horizontal surface, while friction coefficient beteen block A and B is `mu=0.5` . the Maximum horizontal force F that can be applied so that block A does not slip over block B is.

Options

A. `25N`
B. `40N`
C. `30N`
D. `20N`

Answer

D

Explanation

Maximum value of friction on `A` , `f_(max)=mumg`
`=0.5xx2xx10=10N`
Hence, the block `B` is subject to this frictional force in the forward direction. Since the block `A` does not slip on `B` , both of them move with a common acceleration. The acceleration of `B` is the same as the acceleration of `A` .
To determmine acceleration of `'a'` , we find that `m_(B)a _(max)=f_(max)`
`a_(max)=(10)/(2)=m//s^(2)`
`F_(max)=(m_(A)+m_(B))a_(max)=4xx5=20N` .