Let us consider the tension produced on both the sides of the rope is T.
We have been given that the rope is mass less and the rope is passing through a pulley which is stationary.
[tex]Let\ m_{1} and\ m_{2}\ are\ the\ masses\ of\ two\ blocks[/tex]
[tex]Let\ m_{1}\ is\ moving\ in\ vertical\ upward\ direction\ while\ m_{2}\ is\ in\ downward\ \ direction[/tex]
[tex]Hence\ m_{2} =4.5\ kg[/tex]
The body is moving downward with an acceleration of [tex]\frac{3}{4} g[/tex]
As the rope is the same one which is passing over a mass less pulley and connected by two masses,hence, acceleration of each block will be the same in magnitude.
[tex]For\ body\ m_{1}[/tex]
Here the tension is acting in vertical upward direction and the weight is acting in vertical downward direction. Here,the body is moving in vertical upward direction. Hence, the net force acting on it is-
[tex]T-m_{1} g=m_{1}a[/tex] [1]
[tex]For\ body\ m_{2}[/tex]
Here the tension is acting in vertical upward direction while weight is in vertical downward direction. The body is moving in downward direction. Hence the net force acting on it will be-
[tex]m_{2} g-T=m_{2} a[/tex] [2]
Combing 1 and 2 we get-
[tex]T-m_{1}g=m_{1}a[/tex]
[tex]m_{2} g-T=m_{2} a[/tex]
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[tex][m_{2} -m_{1} ]g=a[m_{1}+ m_{2}][/tex]
[tex][4.5-m_{1}]g =\frac{3}{4}g[4.5+ m_{1}][/tex]
[tex]4[4.5-m_{1}] =3[4.5+m_{1} ][/tex]
[tex]7m_{1} =4.5 kg[/tex]
[tex]m_{1} = 0.64286 kg[/tex] [ans]
