Three forces (2i^+3j^-2k^)N, (-5i^-5j^+2k^)N and (6i^+5j^-3k)N are working simultaneously on body which moves from the position (in meter) (2,5,-4) to the position (in meter) (-4,5,-3). Calculate the amount of work done.
The net force acting on a body
F=F1+F2+F3
F=2i^+3j^−2k^−5i^−5j^+2k^+6i^+5j^−3k^=3i^+3j^−3k^The displacement of the body
d=r2−r1=(−4i^+5j^−3k^)−(2i^+5j^−4k^)=−6i^+0j^+k^The work done
W=F⋅d=3×(−6)+3×0+(−3)×1=−21J