First slide
Work
Question

Three forces (5i^+4j^+2k^)N   (2i^2j^+6k^)N,(i^j3k^)N  displaces a particle from (1m,-1m,-1m) to (1m,0,3m) and then to (3m,1m,2m). The total work done by all these forces is

Moderate
Solution

Net force  F=F1+F2+F3

                 F=4i^+j^+5k^

Displacement 

Δr=r3r1=3i^+j^+2k^(i^j^k^)Δr=2i^+2j^+3k^W=FΔr=(4i^+j^+5k^)(2i^+2j^+3k^)=8+2+15=25J

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