(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A particle in the [itex]xy[/itex] plane is attracted toward the origin by a force of magnitude [itex]F=k/y[/itex]

(a) Calculate the work done when the particle moves from point [itex](0,a)[/itex] to [itex](2a,0)[/itex] along a path, which follows two sides of a rectangle consisting of a segment parallel to the [itex]x[/itex] axis from [itex](0,a)[/itex] to [itex](2a,a)[/itex], and a vertical segment from the latter point to the [itex]x[/itex] axis.

(b) Calculate the work done by the same force when the particle moves along an ellipse of a semi-axes [itex]2a,a[/itex] Hint: Set [itex]x=2a\sin\theta, y=a\cos\theta[/itex]

2. Relevant equations

[tex]W=\int\mathbf{F}\cdot d\mathbf{r}=\int F\cos\theta dr[/tex]

3. The attempt at a solution

For the part that is parallel to the [itex]x[/itex] axis, I get

[tex]W_{\alpha}=\int_0^{2a}\frac{k}{y}\cdot\frac{y}{\sqrt{x^2+a^2}}dx=\left.k\sinh^{-1}\left[\frac{x}{a}\right]\right|_{x=0}^{x=2a}=k\sinh^{-1}\left[2\right][/tex]

Then the second path gives a similar answer:

[tex]W_{\beta}=\int_0^{2a}\frac{k}{y}\cdot\frac{y}{\sqrt{4a^2+y^2}}dy=\left.k\sinh^{-1}\left[\frac{y}{2a}\right]\right|_{y=a}^{y=0}=-k\sinh^{-1}\left[\frac{1}{2}\right][/tex]

Then the total work is the sum of the two paths. My problem comes in setting up the next part, I thought to use the hint in this way:

[tex]W_\gamma=\int\frac{k d\theta}{r}=k\int\frac{d\theta}{\sqrt{4a^2\sin^2\theta+a^2\cos^2\theta}}=\frac{k}{a}\int\frac{d\theta}{\cos\theta\sqrt{1+4\tan^2\theta}}[/tex]

but this doesn't look correct to me. Any suggestions??

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# Homework Help: Work done over two paths

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