Answer:
(a) Let a be the acceleration of each block and \[{{T}_{1}}\]and \[{{T}_{2}}\]be the tensions, in the two strings as shown in figure.
Taking the three blocks and the two strings as the system. Using \[\sum {{F}_{x}}=m{{a}_{x}}\]or \[14=(4+2+1)a\,\,or\,\,a=\frac{14}{7}2m/{{s}^{2}}\]
(b) Free body diagram (showing the forces in x-direction only) of 4 kg block and 1 kg block are shown in figure.
Using \[\sum {{F}_{x}}=m{{a}_{x}}\]For 1 kg block, \[F-{{T}_{1}}=(1)(a)\] or \[14-{{T}_{1}}=(1)(2)=2\] \[\therefore {{T}_{1}}=14-2=12N\] For 4 kg block, \[{{T}_{2}}=(4)(a)\] \[\therefore {{T}_{2}}=(4)(2)=8N\]
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