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+//Pulley
+//refer fig. 20.6
+//Let a be the resulting acceleration and T be the tension in the rope
+//Angular acceleration of pulley
+//alpha=1.667*a rad/sec^2
+//Dynamic equilibrium condition for the block gives
+//T=(600-(600*a)/(9.81))
+//From kinetic equation for pulley
+T=(200*7.358)/(9.81) //N
+a=(600*9.81)/(800) //m/sec^2
+printf("\nT=%.3f N\na=%.3f m/sec^2",T,a)