Answer to Question #130972 in Physics for Deepika

Question

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.

Answer 1

The net force acting on a body

$${\bf F}={\bf F}_1+{\bf F}_2+{\bf F}_3$$

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

The displacement of the body

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

The work done

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