question_answer

For the arrangement shown in figure the coefficient of friction between the two blocks is\[\mu \]. If both the block are identical, then the acceleration of each block is-

A) \[\frac{F}{2m}-2\mu g\] 

B)   \[\frac{F}{2m}\]

C) \[\frac{F}{2m}-\mu g\]               

D) zero

Correct Answer: C

Solution :

In fig, direction of friction is shown, \[F-T-\mu mg=ma\] \[T-\mu mg=ma\] Solving the equations, \[a\,\,=\,\,\frac{F-2\mu mg}{2m}\]