Answer:
(a) Let ?a? be the acceleration of the blocks and T the tension in the string as shown in figure.
Taking the two blocks and the string as the system. Using\[\sum {{F}_{y}}=m{{a}_{y}}\]we get \[F-4g-2g=(4+2)a\]or \[120-40-20=6a\]or \[60=6a\] \[\therefore \,\,a=10\,\,m/{{s}^{2}}\]
(b) Free body diagram of 2 kg block is as shown in figure.
Using \[\sum {{F}_{y}}=m{{a}_{y}}\] we get \[T-2g=2a\,\,or\,\,T-20=(2)(10)\] \[\therefore \,\,\,T=40\,N\]
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