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DochEvi [55]
3 years ago
11

constant force F=5i+5j−1kF=5i+5j−1k is applied to an object that is moving along a straight line from the point (−5,−3,−4)(−5,−3

,−4) to the point (2,5,0).(2,5,0). Find the work done if the distance is measured in meters and the force in newtons. Include units in your answer.
Physics
1 answer:
meriva3 years ago
6 0

Answer:

W = 71J

Explanation:

Given force F = (5i+5j−1k)N

d = Δr

r1 = (−5,−3,−4)m

r2 = (2,5,0)m

Δr = r2 – r1 = (2-(-5), 5-(-3), 0-(-4))

Δr = (2+5, 5+3, 0+4) = (7i+ 8j +4k)m

W = F•d = (5i+5j−1k)•(7i+ 8j +4k)

W = 5×7 + 5×8 +-1×4 = 35 + 40 - 4

W = 71J

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