A force of 202  is acting on a block of 1kg at  45∘ above the horizontal. Work done by all the forces in 2s is (g= 10ms−2)

Work done=F→.S→W=[Fcos45∘i^+(Fsin45∘-mg)j^].(Sxi^+Syj^)     here components of force F are ,   Fx=Fcos45 and Fy=Fsin45; components of displacement S are Sxand Syhere displacement Sx=ut+12axt2Sx=0+12Fxmt2=12×201×4Sy=ut+12ayt2=12×201×4W=[20i^+(20-10)j^].12axt2i^+12ayt2j^W=(20×12×201×4)+(10×12×20×4)Work=800+400Work done=1200J