A constant 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.

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mmoyano077 mmoyano077 · Answer 1 · 2019-09-13T23:26:28+02:00

Answer:

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.