Three forces (5i^+4j^+2k^)N (−2i^−2j^+6k^)N,(i^−j−3k^)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
50J
25J
15J
-10J
Net force F→=F1→+F2→+F3→
F→=4i^+j^+5k^
Displacement
Δr→=r3→−r→1=3i^+j^+2k^−(i^−j^−k^)Δr→=2i^+2j^+3k^W=F→⋅Δr→=(4i^+j^+5k^)⋅(2i^+2j^+3k^)=8+2+15=25J