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I had in mind this physical situation of a point mass rolling down a hill of some arbitrary shape ($y=y(x)$, for instance a parabola about $x=0$), with a peculiar friction force whose "expended work" is proportional to the arclength since $t=t_0$. If $x=x(t)$ and $y=y(t)$, then $$W=k \int_{x=x_0}^x \sqrt{1+y'(x)^{2}} dx=k\int_{t=t_0}^t\sqrt{x'{t}^{2}+y'(t)^{2}}dt$$

I've tried to get equations of motion starting from the energy theorem: $$\frac{1}{2}m(x'(t)^{2}-v_0 ^{2})=-mg(y(t)-y_0) +k\int_{t=t_0}^t\sqrt{x'{t}^{2}+y'(t)^{2}}dt$$ and then differentiate everytime by $t$: $$m x'(t) x''(t)=-mgy'(t)+k\sqrt{x'{t}^{2}+y'(t)^{2}}$$ I thought I would bring in the landscape $y(x)$ by using $$y'(t)=\frac{dy}{dx}x'(t)$$ Then: $$mx''(t)=-mg\frac{dy}{dx} +k\sqrt{1+y'(x)^{2}}$$ But I don't know how to work the equations further. Is there a better method to go at it (maybe even further from energetics)?

EDIT:

As shown in the comments exchange, I had to modify the method by expanding on $W=\int \mu N ds$ with $\vec N $ being the normal force. I obtained the peculiar result of $$N=mg \frac{1}{\sqrt{1+y'(x)^2}} \rightarrow W=\int \mu mg dx$$ seemingly path-independent.

First method: By generalizing that for the inclined plane $N=mg \cos{\theta}$ for $\theta$ the "slope angle". $\cos{\theta}$ then becomes $$=\frac{1}{\sqrt{1+\tan{\theta}^{2}}}=\frac{1}{\sqrt{1+y'(x)^{2}}}$$

Second method: I went in more rigorously, by using the definition of $N$ as the projection of "$\overrightarrow{\text{weight}}$" onto the instantaneous normal axis: $N=|\overrightarrow{\text{weight}} \cdot \hat n|$ with $\hat n$ the unit normal vector for the parametrization $(x,y(x))$. As $$\hat n=\frac{\hat t'}{|\hat t'|}$$ and $$\hat t=\frac{\vec r'}{|\vec r'|}=\frac{\left(1,y'(x)\right)}{\sqrt{1+y'(x)^{2}}} \rightarrow \hat t'=\left(\frac{-y''(x)y'(x)}{\left(1+y'(x)^{2}\right)^{3/2}},\frac{y''(x)}{\left(1+y'(x)^{2}\right)^{3/2}}\right) \rightarrow \hat n=\left(\frac{-y'}{\sqrt{1+y'^{2}}},\frac{1}{\sqrt{1+y'^{2}}}\right)$$

Using $\overrightarrow{\text{weight}}=(0,-mg)$ I got the same result as from method 1, and thus cancellation (??)

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  • $\begingroup$ In a sense work done (that is energy 'lost') by a friction force is always a function of arclength. If we assume a simple model with friction force $F_f=\mu F_N$ with $F_N$ the Normal force, then $dW=F_f ds$. Your problem doesn't seem to correspond to a real world problem though. I suggest simplifying a bit by specifying $y(x)$ and using a simple model for $F_f$ and see if you can handle the equation of motion for that specific but simpler problem. $\endgroup$ – Gert Aug 27 '15 at 1:12
  • $\begingroup$ Oh, so simple friction models most often depend on $ds$. My general (qualitative) inspiration was if a body was let to roll down a hill $y(x)$ (let's say $y=x^{2}$, starting at $(x_0,y_0)=(-1,1)$), because of friction the body would reach a final height less than $y_0$. My main issue is how to get that height analytically. What is non-real world about my problem though? $\endgroup$ – N.E. Aug 27 '15 at 14:25
  • $\begingroup$ It's not necessarily a 'non-real world' problem but you don't seem to have defined a mode of friction, only energy lost due to it. I'm not sure the problem can be solved that way. I'll formulate an answer shortly. $\endgroup$ – Gert Aug 27 '15 at 14:45
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I propose redefining this problem as follows (because I'm not sure it has a solution the way the OP has defined it).

enter image description here

Let $y=f(x)$ be some symmetrical (around $y$) function like $x^2$. Let the point mass experience a friction force acc. to the usual simple model $F_f=\mu F_N$, with $F_N$ the Normal force acting on the point mass in the point $(x,y)$ ($N$ is the Normal line in $(x,y)$) and $\mu$ a coefficient of friction (constant).

Now, from the balances of forces in the $x$ and $y$ directions, equations of motion can be set up.

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  • $\begingroup$ For some reason, I thought the problem was more complex than it should be. To get the height, I should have just used the regular energy theorem and get $W_f=y_0 -\frac{W}{mg}$, and $W=F_N \int_{y_0} ^{y_f}ds$, and solve analytically for $y_f$ . $\endgroup$ – N.E. Aug 28 '15 at 15:37
  • $\begingroup$ I guess $F_N$ is a function of (x,y(x)) right (I couldn't edit the integral due to the 5min edit limit)? $\endgroup$ – N.E. Aug 28 '15 at 15:45
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    $\begingroup$ $F_N$ is perpendicular to the tangent at $(x,y)$ and the slope of the tangent is of course related to $\frac{dy}{dx}$. That should help. $\endgroup$ – Gert Aug 28 '15 at 15:48
  • $\begingroup$ Without any friction you'd find of course that the point mass rolls back up on the right hand side to the same starting height (and the rolls back down, oscillating ad infinitum). With friction, some of the potential energy is converted to friction work, so it won't roll back up to the same height. Working out that total work done in the $y$ direction is the key, as you noted. $\endgroup$ – Gert Aug 28 '15 at 16:04
  • $\begingroup$ I obtained that $f=\mu mg \frac{1}{\sqrt{1+y'(x)^{2}}}$ and so $W=\int f ds =\int_{0}^{f}\mu mg \frac{1}{\sqrt{1+y'(x)^{2}}} \times \sqrt{1+y'(x)^{2}} dx$. Is the cancellation coincidental, or does the path $y(x)$ matter after all? $\endgroup$ – N.E. Aug 28 '15 at 16:52

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