Let the acceleration of the system be #a# and tension on the connecting string be #T#
Considering the forces on #W_B# we can write
#25xxgxxsin30-0.2xx25xxgxxcos30-T=25xxa.....(1)#
Considering the forces on #W_A# we can write
#15xxgxxsin30-0.4xx15xxgxxcos30+T=15xxa.....(2)#
Adding (2) and (1) we get
#(25+15)xxgxxsin30-(0.2xx25+0.4xx15)xxgxxcos30=(25+15)xxa#
#=>(40xxgxxsin30-(5+6)xxgxxcos30=40xxa#
#=>40xxgxxsin30-11xxgxxcos30=40xxa#
#=>a=(40xx32xx1/2-11xx32xxsqrt3/2)/40#
#=16-4.4sqrt3=8.38fts^-2#
Taking #"acceleration due to gravity"g=32fts^-2#
Inserting the value of #a# in (1)
#25xx32xxsin30-0.2xx25xx32xxcos30-T=25xx8.38#
#=>25xx32xx1/2-0.2xx25xx32xxsqrt3/2-T=25xx8
38#
#=>400-80xxsqrt3-T=209.5#
#=>261.44-T=209.5#
#=>T=261.44-209.5=51.94poundal#