1. ## Conservative force

A particle P moves on the x-axis under the force $\displaystyle F(x)= x^2-4x+3$. Show that the F is conservative . I want to show the work done on the particle is zero when it moves around a closed path. So I solved F(x) = 0, got x=1,3 then integrated with these as the limits. Problem is I didn't get 0.

Second part of the question is 'show that there is a single stable equibrium for the motion of P'. Anyone any ideas on what that means?

2. Originally Posted by poirot
A particle P moves on the x-axis under the force $\displaystyle F(x)= x^2-4x+3$. Show that the F is conservative . I want to show the work done on the particle is zero when it moves around a closed path. So I solved F(x) = 0, got x=1,3 then integrated with these as the limits. Problem is I didn't get 0.

Second part of the question is 'show that there is a single stable equibrium for the motion of P'. Anyone any ideas on what that means?
Hint for the first part. See here. Specifically look at option 3 under the "Mathematical Description" heading. In 1-D terms this means we are looking for a function $\displaystyle \Phi$ that satisfies
$\displaystyle F = -\frac{d \Phi}{dx}$

Can you find such a function?

For the second part, an equilibruim point is where the force is 0 N. You will find one of those. To tell if it will be a stable or unstable point, imagine a small displacement from equiilibrium. Does the graph of F tell you if the object is going to move back to the equilibrium point, or move away from it?

-Dan

3. Originally Posted by topsquark
Hint for the first part. See here. Specifically look at option 3 under the "Mathematical Description" heading. In 1-D terms this means we are looking for a function $\displaystyle \Phi$ that satisfies
$\displaystyle F = -\frac{d \Phi}{dx}$

Can you find such a function?

For the second part, an equilibruim point is where the force is 0 N. You will find one of those. To tell if it will be a stable or unstable point, imagine a small displacement from equiilibrium. Does the graph of F tell you if the object is going to move back to the equilibrium point, or move away from it?

-Dan
Ok I integrated -F(x) and got $\displaystyle -x^3/3 +2x^2 -3x.$ What do I do now?

4. Originally Posted by poirot
Ok I integrated -F(x) and got $\displaystyle -x^3/3 +2x^2 -3x.$ What do I do now?
Well, you can verify that this gives 0 J of work over a closed path. The second part deals with equilibrium. A point is an (stable) equilibrium point if, when given a small displacement, an object will return to the equilibrium point. (The point is unstable if the object moves away from the equilibrium point.

So solve dF/dx to find an equilibrium point. Is this point stable or not?

-Dan