Two identical blocks A and B each of mass M are connected to each other through a light string. The system is placed on a smooth horizontal floor. When a constant force F is applied horizontally on the block A, find the tension in the string.

Explanation

The acceleration of the system of two blocks A and
`B=("Force")/("Total mass")`
`therefore a=(F)/(M+M)=(F)/(2M)`.
If we consider the free body diagram of A, the forces acting on it are (i) the applied force F and (ii) the tension T on the string as shown in the following fig.

The resultant force `= F-T , Ma = F-T`
`M((F)/(2M))=F-T " " (therefore a=(F)/(2M))`
`(F)/(2)=F-T`

`T=(F)/(2)` (or) From FBD for B
`T=Ma=M(F)/(2M)=(F)/(2)`
`T=(F)/(2)`