Weight of the block, `mg=(2)(10)=20N`
Work done by the applied force,
`W_(F) = Fs cos theta = Fh cos 0^(@)` (Here,s=h=2m and angle between force and displacement is `0^(@))`
or ` W_(F)=(40)(2)(1)=80 J` Similarly, work done by its weight ,
`W_(mg)=(mg)(h) cos180^(@)`
or `W_(mg)=(20)(2)(-1)=-40 J`