Aconstant force with vector representation f = 10i + 18j - 6k moves an object along a straight line from point (2,3,0) to the point (4,9,15). find the work done if the distance is measured in meters and the magnitude of the force is measured in newtons.

The work done is W = 38J.

Explanation:

We have a force verctor F=(10,18,-6) and two points P(2,3,0) and Q(4,9,15).

The position verctors of these points are P=(2,3,0) and Q=(4,9,15).

The vector that goes from P to Q will be:

R=Q-P=(2,6,15)

Now, considering that the force is constant and travels in the straight line represented by R, the work done will be:

W = F×R = (10,18,-6)×(2,6,15) = 10·2 + 18·6 + (-6)·15 = 38Nm = 38J

∴ W = 38J.

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