Three blocks of masses `3kg ,2kg` and `1kg` are placed side by side on a smooth surface as shown in figure. A horizontal force of `12N` is applied on `3kg` block. Find the net force on `2kg` block.

Explanation

All the blocks wll move with same acceleration (say a) in horizontal direction. Let us take all the blocks as a system. Net external force on the system is 12 N in horizontal direction.

Using `sumF_(x)=ma_(x)`, we get
12=(3+2+1)a=6a or `a=(12)/(6)=2ms^(-1)`
Now, let F be the net force on 2 kg block in x-derction, then using `sumF_(x)=ma_(x)` for 2kg block, we get
F=(2)(2)=4N
Here, net force F on 2 kg block is the resultant of `N_(1)" and N_(2)(N_(1) gt N_(2))` where `N_(1)=` normal reaction between 3 kg and 2 kg block
and `N_(2)`= normal reaction between 2 kg and 1 kg block
Thus, `F=N_(1)-N_(2)`