# Homework Help: Show that the following force is conservative

1. Feb 15, 2012

### nbram87

1. The problem statement, all variables and given/known data

Fx = K(2x + y), Fy = K(x + 2y)

2. Relevant equations

3. The attempt at a solution
I think what is confusing me is that it is two different forces (Fx and Fy). I know that the curl has to be zero for it to be conservative, and I am assuming I will have to figure out a value for the constant K for that too happen.

Last edited: Feb 15, 2012
2. Feb 15, 2012

### Staff: Mentor

Fx and Fy represent two components of a vector, so they describe a vector field. It might be written as:

$\vec{F} = Fx\;\hat{i} + Fy\;\hat{j}$

How would you form the curl of that?

3. Feb 15, 2012

### nbram87

I think that is one thing that is confusing me. How else could you determine that the force is conservative? Would you have to determine the work done by both Fx and Fy are equal to 0?

4. Feb 15, 2012

### Staff: Mentor

You could show that the work done in moving a particle along any closed path is zero (start at point P, traverse all possible paths (!) ending again at point P). The curl looks like the easiest approach.

5. Feb 15, 2012

### nbram87

When you do the curl of Fx and Fy, I think the constant K becomes useless because it equals to zero. What is the meaning of K in the problem then?

6. Feb 15, 2012

### Staff: Mentor

I don't understand your meaning. How does K become zero? Can you show your curl calculation?

7. Feb 15, 2012

### nbram87

Curl = d/dx(Fy) i - d/dy (Fx) j
= d/dx [K(x + 2y)] + d/dy [K(2x + y)]
= K(1+ 0) - K (0+1)
= 0
So K - K = 0?

8. Feb 15, 2012

### Staff: Mentor

That tells you that the curl is zero no matter what value K has.

9. Feb 15, 2012

### nbram87

Is it correct? Is my calculation of the curl and the value of K being meaningless correct?

10. Feb 15, 2012

### Staff: Mentor

The curl calculation result is correct. K is not "meaningless" (it's a scaling constant for the magnitude of the force, and likely makes the force equation units balance). It simply turns out to be irrelevant to the question of conservation.

11. Feb 15, 2012

### nbram87

Ok thank you.