Question

Three forces (2ˆi + 3ˆj − 2

ˆk) N, (−5ˆi − 5ˆj + 2ˆk) N and (6ˆi + 5ˆj − 3

ˆk) 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 don​

Answer 1

Answer:

-21 Joule

Explanation:

Here the total Force is,

• F =(2-5+6)î + (3-5+5)j + (-2+2-3)k

= 3î+3j-3k

And displacement,

• S= (-4-2)î + (5-5)j + {-3-(-4)}k

= -6î+0j+1k

= -6î+k

Now we know, work is the dot product of vector F and displacement S

W = F.S

= (3î+3j-3k) . (-6î+k)

= -18+0-3

= -21 Joule (ans)