Question #130972
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.
1
Expert's answer
2020-08-31T12:47:52-0400

The net force acting on a body

F=F1+F2+F3{\bf F}={\bf F}_1+{\bf F}_2+{\bf F}_3

F=2i^+3j^2k^5i^5j^+2k^+6i^+5j^3k^=3i^+3j^3k^{\bf F}=2\hat {\bf i}+3\hat {\bf j}-2\hat {\bf k}-5\hat {\bf i}-5\hat {\bf j}+2\hat {\bf k}+6\hat {\bf i}+5\hat {\bf j}-3\hat {\bf k}\\ =3\hat {\bf i}+3\hat {\bf j}-3\hat {\bf k}

The displacement of the body

d=r2r1=(4i^+5j^3k^)(2i^+5j^4k^)=6i^+0j^+k^{\bf d}={\bf r}_2-{\bf r}_1\\ =(-4\hat {\bf i}+5\hat {\bf j}-3\hat {\bf k})-(2\hat {\bf i}+5\hat {\bf j}-4\hat {\bf k})\\ =-6\hat {\bf i}+0\hat {\bf j}+\hat {\bf k}

The work done

W=Fd=3×(6)+3×0+(3)×1=21JW={\bf F}\cdot{\bf d}\\ =3\times (-6)+3\times 0+(-3)\times 1=-21 \:\rm J

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