Two blocks are attached by a light inextensible string and the system is pulled by applying a force `F_(0)` as shown in the diagram. Find the acceleration of each block and tension in the string.

Explanation

Since the blocks travel same distance in equal time, their accelerations are equal (in this case, both magnitude and direction).
First make free body diagram of each block and apply.
Net force`=`mass`xx`acceleration in each direction

`x:T=ma` (i)
`Y:N_(1)=mg=0` (ii)

`x:F_(0)=T=2ma` (iii)
`y:N_(2)-2mg=0` (iv)
Solving `(i)` and we get
`a=(F_(0))/(3m)`
`T=(F_(0))/(3)`
Acceleration of each block is `F//3m` and tension in the string is `(F)/(3)`
Note`:` Since there is no acceleration in vertical direction, hence no need to write equations (ii) and (iv).
OR
We can find acceleration directly by taking two blocks together as a single system.

`F_(0)=(m+2m)a`
`rArr(F_(0))/(3m)`

`T=ma=(F_(0))/(3)`
Note: It is advised that in the beginning consider a single block as a system and solve. After sufficient practice try to solve problems quickly by taking two blocks or three blocks together as a system, see the next problems