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## Homework Statement

A particle is attracted towards the origin by a force proportional to the cube of its distance from the origin. How much work is done in moving the particle from the origin to the point (2,4) along the path y = x^2 assuming a coefficient of friction [tex]\mu[/tex] between the particle and the path?

## The Attempt at a Solution

For the path, set x = t. Thus, y = t^2

[tex]\vec{r}(t) = t\hat{i} + t^2\hat{j}[/tex]

[tex]\vec{r}'(t) = \hat{i} + 2t\hat{j}[/tex]

Because we're going from the origin to (2,4), t goes from 0 to 2.

For the force, we know that:

[tex]\left\|F\right\| = {k(x^2 + y^2)}^{3/2}[/tex]

I want this ideally in vector form, but the best I could come up with:

[tex]{\left\|F\right\|}^{1/3} = kx\hat{i} + ky\hat{j}[/tex]

My main problem is I can't really use [tex]{\left\|F\right\|}^{1/3}[/tex] to calculate [tex]\int_{C} {F\cdot dr}[/tex]. If I could just find a way to express F in vector form, I'm pretty much off to the races. (Because I can then calculate the work done by friction by finding the normal to [tex]\vec{r}(t)[/tex], multiplying it by [tex]\mu[/tex], and dotting it with [tex]\vec{r}'(t)[/tex] from [tex]0\leq t \leq 2[/tex], etc. etc.)

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