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# Find the work done by the force?

A constant force F=<-1, 2, -5> acts on a particle as it moves from A(2, 4, 1) to B(-2, 1, 3). Find the work done by the force.

I have trouble for finding the dot products.

A constant force field is conservative, so you can assume a straight line path for simplicity.

The displacement vector is r = <-2 - 2, 1 - 4, 3 - 1> = <-4, -3, 2>

With a force of F = <-1, 2, -5>, you get

W = F·r = <-4, -3, 2>·<-1, 2, -5> = 4 - 6 - 10 = -12.

A negative number means that the particle is doing work against the field.

By the definition work W is

W = Fr cos θ

where θ is the angle between vectors F and r (F and r are modules of those vectors). If vectors A and B are given then the displacement r is defined as

r = B-A = (-2i +j +3k) -(2i +4j +k) = -4i -3j + 2k

Then

cos θ =( -4Fx + (-3)Fy +2Fz)/Fr

where Fx, Fy, Fz are x,y,z components of F: Fx = -1, Fy =2, Fz = -5. Thus, we have for W

W = (-4)(-1) + (-3)2 +2(-5) = 4=6 -10 = -12 units of work

Negative sign shows that the work is done by the component of force (F cos θ) which has a direction opposite to the direction of the displacement. This means that the particle willl slow down.