Two forces F1=(i^+j+k^)N and F2=(i^+2j^+3k^)N act on a block and displace it from point A (2,3,4) m to the point B (5,4,3) m. Determine work done (in joule).
The net work done,
W=F→1⋅s→+F→2⋅s→=F→1+F→2⋅s→=(2i^+3j^+4k^)⋅rB−rA=(2i~+3j~+4k~)⋅{(5i~+4j~+3k~)−(2i~+3j~+4k~)}.=(2i^+3j^+4k^)⋅(3i^+j^−k^)=6+3−4=5J