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A constant force F=5i−4j−5k is applied to an object that is moving along a straight line from the point (3,−4,−5) to the point (−1,−2,4). Find the work done if the distance is measured in meters and the force in newtons. Include units in your answer.

Respuesta :

Answer:

Work done = -73 J

Explanation:

We have given the force F =5 i-4 j- 5 k

Initial point = (3,-4,-5)

Final point = (-1,-2,4)

We have to find the ds

So [tex]ds=final\ point-initial\ point=(-1-3)i+(-2+4)j+(4+5)=-4i+2j+9k[/tex]

Now we know that work done is given by [tex]W=F.ds=(5i-4j-5k).(-4i+2j+9k)=-20-8-45=-73N[/tex]

So work done = -73 J

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