Calculating Work Using Dot Product: Constant Force and Particle Position

[PsychonautQQ]

Homework Statement

A constant force of 1i - 5j -8k moves (1,-4,2) (-3,2,-1), what is the work done on the particle?

Homework Equations

Avector*Bvector=ABsinθ
?? I think?

The Attempt at a Solution

I really am quite lost... but I found the coordinates for the position vector...

(-3-1)i + (2--4j) + (-1-2)k and the magnitude of the position vector is 7.81
so the unit vector is -.512i + .768j -.384k... am I on the right track here?

 

[Dick]

Homework Statement

A constant force of 1i - 5j -8k moves (1,-4,2) (-3,2,-1), what is the work done on the particle?

Homework Equations

Avector*Bvector=ABsinθ
?? I think?

The Attempt at a Solution

I really am quite lost... but I found the coordinates for the position vector...

(-3-1)i + (2--4j) + (-1-2)k and the magnitude of the position vector is 7.81
so the unit vector is -.512i + .768j -.384k... am I on the right track here?

For a constant force F, the work done is just the dot product of the force with the displacement. Use that.

 

[PsychonautQQ]

so the force is 1i - 5j -8k and the displacement is -4i +6j -3k? so dot product would be -4 -30 - 24 = -58?

 

[PsychonautQQ]

How do I find the angel between these vectors?

 

[Dick]

so the force is 1i - 5j -8k and the displacement is -4i +6j -3k? so dot product would be -4 -30 - 24 = -58?

Yes, if the force isn't constant you need to work harder and integrate, but if it's constant, it's that easy.

 

[Dick]

How do I find the angel between these vectors?

Why do you think you need the angle?

 

[HallsofIvy]

If you are required, separately, to find the angle between vectors, the dot product can be defined as
\vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta)

where \theta is the angle between the vectors.