Question

Forces acting on a particle have magnitudes of 14, 7, and 7 N and act in the direction of vectors
6i+2j+3k, 3i-2j+6k, 2i-3j-6k respectively. The forces remain constant while the particle is
displaced from point A:(2,-1,-3) to B:(5,-1,1). Find the work done. The coordinates are specified in
meters.
(A) 75J
(B) 55 J
(C) 85 J
(D) 65 J
​

Answer 1

Dear Students,

● Answer -

W = -525 J

◆ Explaination -

Let F1, F2 & F3 be forces acting on a particle.

F1 = 14 × (6i+2j+3k) = 84i + 28j + 42 k

F2 = 7 × (3i-2j+6k) = 21i - 14j + 42k

F3 = 7 × (2i-3j-6k) = 14i - 21j - 42k

Total force acting on particle is calculated by -

F = F1 + F2 + F3

F = 84i + 28j + 42 k + 21i - 14j + 42k + 14i - 21j - 42k

F = 119i - 7j + 42k

Displacement s of the particle is calculated by -

s = (2,-1,-3) - (5,-1,1)

s = -3i - 4k

Work done in moving the particle is calculated by -

W = F.s

W = (119i - 7j + 42k).(-3i - 4k)

W = -357 - 168

W = -525 J

Thanks for asking..