Solving Potential Energy: Force & Potential Exam Ques.

[thenewbosco]

Just studying for an exam and the following question appeared on the sample exam:
Given the force: $$F=-c(x-y)^2(\hat{i}-\hat{j})$$ where i and j are the unit vectors.
a) Show the force is conservative
b) Show the potential energy is given by $$V(x,y) = \frac{c}{3(x-y)^3)}$$ assuming V(0,0) =0.


So for a) it is simple to show using the curl of F. but for b i am not sure how to get the potential energy function given the force. $$F=-\nabla V$$ will give the force easily if i have the potential function, but I am not sure how to go the other direction. perhaps for the exam to show it i could just take the negative gradient, (which appears to be wrong for the potential given in this question) but i would like to just know how to go the other way for knowledge. thanks

 

[tim_lou]

if a field is conservative, then the work done against the field on a object from point a to point b is conserved and equals to the change in potential. Well, basically, if you know the work done alone a path, you know the potential function.

 

[m2003]

a) To show that the given force F is conservative, we need to show that the curl of F is equal to zero. We can do this by taking the partial derivatives of each component of F with respect to the other variable:

∂F_x/∂y = -2c(x-y)

∂F_y/∂x = 2c(x-y)

Since these two partial derivatives are equal, we can conclude that the curl of F is equal to zero, and therefore F is a conservative force.

b) To find the potential energy function V(x,y) from the given force F, we can use the formula V(x,y) = -∫F⋅dr, where r is the position vector (x,y). Integrating the given force F along a path from the origin (0,0) to a point (x,y) will give us the potential energy at that point.

∫F⋅dr = ∫-c(x-y)^2dx + ∫c(x-y)^2dy

= -c∫(x-y)^2dx + c∫(x-y)^2dy

= -c∫(x^2-2xy+y^2)dx + c∫(x^2-2xy+y^2)dy

= -c(x^3/3 - xy^2 + x) + c(x^2y/2 - y^3/3 + y)

= -cx^3/3 + cxy^2 - cx + cx^2y/2 - cy^3/3 + cy

= -cx^3/3 + cxy^2 + cx^2y/2 - cy^3/3

= c(-x^3/3 + x^2y/2 + xy^2/2 - y^3/3)

= c(x^2y - xy^2)/2 - c(x^3 + y^3)/3

= c(x^2y - xy^2)/2 - c(x^3 + y^3)/3 + c(x^3 + y^3)/3

= c(x^2y - xy^2)/2 + c(x^3 + y^3)/3

= c(x^2y - xy^2)/2 + c(x^3 + y^3)/3 + C

= V(x,y