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Oduvanchick [21]

3 years ago

10

The constant forces F1 = 8 + 29 + 32 N and F2 = 48 - 59 - 22 N act together on a particle during a displacement from the point A

(20, 15,0)m to the point B (0,07) m. What is the work done on the particle? The work done is given by F. f, where is the resultant force (here F = Fi + F2) and is the displacement.

1 answer:

marshall27 [118]3 years ago

5 0

Answer:

-600 J

Explanation:

F₁ = 8i +29 j + 32k

F₂ = 48 i - 59 j - 22 k

F = F₁ +F₂ = 8i +29 j + 32k +48 i - 59 j - 22 k

F = 56i - 30 j + 10 k

displacement d = ( 0 - 20 )i + ( 0 - 15 )j + ( 7 -0) k

d = - 20 i - 15 j + 7 k

Work Done = F dot product d

F . d = - 56 x 20 - 30 x - 15 + 10 x 7

= - 1120 +450 + 70

= -600 J

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