Single conservative force acting on a particle

[Robb]

Homework Statement

Homework Equations

U(final)(x)= (-) Integral F dx + U(initial)
Integration from (x-initial to x-final)

The Attempt at a Solution

U(final)(x)= (-) integral (-Ax+Bx^2)dx

Not sure on limits of integration

 

[andrewkirk]

You have to start the integration at a place where U is known. What place is that?

If you want to give the value of U at location coordinate x, what is the upper limit of integration?

 

[Robb]

Not sure. Lower = 0. upper x=2.00 & x=3.00?

 

[andrewkirk]

I was referring to part (a). There is no 2.00 or 3.00 in part (a). Once you have solved part (a), part (b) is just a question of substitution and subtraction.
I think you may be confusing yourself by mislabelling your integration variables.
You wrote:

U(final)(x)= (-) Integral F dx + U(initial)

That can't be right because the LHS depends on X while the RHS does not. By using x as the integration variable you make it disappear once the integral is taken. Instead use y as your integration variable and x as one of your limits of integration. Which one?

 

[Robb]

So, my potential energy function is U(x) = (-) integral F dx = (-) integral (-Ax+Bx^2) dx = 1/2Ax^2 + 1/3 BX^3

U=0 @ x=0

Though I am seeing this as a lower limit