# Calculate for each field the work done in moving a particle

1. Feb 5, 2004

### jlmac2001

Problem:

Which, if either, of the two force fields F1=-yi+xj+zk, F2=yi+xj+zk is conservative? Calculate for each field the work done in moving a particle counterclockwise around the circle x=cos(t), y=sin(t) in the (x,y) plane.

I got the part in telling if the force fields are conservative. F1 is not conservative because d/dx(Fy)-d/dy(Fx) is 2. F2 is conservative because everything is equal to zero.

How do I calculate the work done in each field? I don't get it.

2. Feb 6, 2004

### HallsofIvy

Staff Emeritus
If a force field is conservative, then work done is the difference in potential energy at the two end points. In particular, the work done in moving around a closed path is 0. (The work done in lifting a mass against gravity can be regained by dropping it back down.)

If a force field is not conservative, then you will have to integrate the dot product of the force with ds around the path.

In this case, x= cos(t), y= sin(t), z= 0, so ds= dxi+ dyj+ dzk=
-sin(t)i+ cos(t)j+ 0k. F . ds= (-yi+xj+zk).(-sin(t)i+ cos(t)j+ 0k)
= (-sin(t)i+ cos(t)j+0k).(-sin(t)i+ cos(t)j+ 0k)= sin22(t)+ cos2(t)= 1. The work done is just the integral of the constant 1 around the circle and so is just the circumference of the circle.

If you did integrate F2 in the same way, F2.ds= (yi+xj+zk).(-sin(t)i+ cos(t)j+ 0k)= (sin(t)i+cos(t)j).(-sin(t)i+ cos(t)j+ 0k)= cos2(t)- sin2(t)= cos(2t) and the integral of that around the circle (t going from 0 to 2&pi;) is 0.