Question

three forces 3(i+ 3j+k) ,5/6(-2i+9k) and 11(2i+j+6k) are acting on a particle calculate the work done in displacing the particle from point (4,-1,2) to point (11,6,8)​

Answer 1

Answer:

782.33 J

Explanation:

F1 = 3(i+ 3j+k)

F2= 5/6(-2i+9k)

F3= 11(2i+j+6k)

F = (3 - 5/3 +22)i + 20j + (69+7.5)k

  = (25-5/3)i + 20j + 76.5k

  = 70/3i + 20j + 76.5k

d = (11-4)i + (6+2)j + (8-2)k

  = 7i + 8j + 6k

F.d = 490/3 + 160 + 459

     = 163.33 + 619

     = 782.33 J

Answer 2

Answer:  The work done in displacing the particle from point (4,-1,2) to point (11,6,8)​ is 782.33 J .

Explanation:

F₁ = 3( i + 3j+k) = 3i + 9j + 3k

F₂= 5/6(-2i+9k) = -5/3i + 15/2k

F₃= 11(2i+j+6k) = 22i + 11j + 66k

∴Resultant Force F = (3 - 5/3 +22)i + 20j + (69+7.5)k

                              = (25-5/3)i + 20j + 76.5k
                              = 70/3i + 20j + 76.5k

Resultant force F acts on a particle and displaces it from point (4,-1,2) to point (11,6,8)​

Thus, Displacement  = (11-4)i + (6+2)j + (8-2)k

                                   = 7i + 8j + 6k

Work = Force x Displacement = F. d = 490/3 + 160 + 459 ( i.i + j.j+ k.k = 1)

                                                             = 163.33 + 619  = 782.33 J

Hence, the work done in displacing the particle from point (4,-1,2) to point (11,6,8)​ is 782.33 J .

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