Question

fr-rrce of F: 6i+3j-2kacts on a small bc.dy of mass 2 kg and undergoes

displacenrent X= 9i-.i+3p. Work done in joule is

Answer 1

As Work = Force x Displacement where Force F = 6i + 3j -2k and Displacement X = 9i-j+3k

Work is a scalar quantity...So we just calculate the Scalar product or Dot Product of F and X...

Thus Work = F x X
= (6i+3j-2k) . (9i-j+3k)
= 6x9 - 3x1 - 2x3 = 45 joules

Note that : a.b = b.a , i.i =j.j=k.k=1 and i.j=j.k=k.i=0