question_answer

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

Answer:

Since, all the blocks will 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 \[\sum {{F}_{x}}=m{{a}_{x}},\]we get \[12=(3+2+1)a=6a\,\,or\,\,a=\frac{12}{6}=2m/{{s}^{2}}\] Now, let F be the net force on 2 kg block in x-direction, then using\[\sum {{F}_{x}}=m{{a}_{x}}\]for 2 kg block, we get \[F=(2)(2)=4N\] Note: Here net force F on 2 kg block is the resultant of\[{{N}_{1}}\]and \[{{N}_{2}}({{N}_{1}}>{{N}_{2}}).\] 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,\[F={{N}_{1}}-{{N}_{2}}\]