Variable friction on an inclined plane and maximum velocity

[madafo3435]

I think I've got it. First off, the total height is $h$. Let the incline's length be $d$ and the angle between it and the vertical $\theta$. Clearly, $d \cos \theta =h$. Firstly, the acceleration on the particle $= g \cos \theta -kgx \sin \theta$. Now here, $a=dv/dt=(dv/dx) \cdot v$. Cross multiplying both sides with $dx$ and integrating from $0$ to $d$ gives $0=gd \cos \theta -(kd^2g \sin \theta) /2$ (LHS is 0 as velocity is 0 at the beginning point and at $d$). So, after some simple algebra $d = 2 \cot \theta /k= h /( \cos \theta)$, yielding $h=2 \cos \theta \cot \theta /k$. Then, adopting the same method over distance $l$, where $v$ is maximum, I get $v^2/2=gl \cos \theta + (kl^2 g \sin \theta)/2$. Now, differentiating both sides with respect to $l$ gives $0=g \cos \theta - klg \sin \theta$(v is max here, so its derivative is zero). I end up with $k \cot \theta = l$. Putting $l$'s value in the last equation, simplifying, finally substituting $k$'s value gives me:
$$v= \sqrt { \frac {gh} 2} $$

There might be some errors here and there because I've typed it VERY briefly and hastily, but I think this is it. Because, the given answer matches as well.

In good time, his reasoning is very reasonable, but I must clarify that the signing force is not equal to -kmgxsen ($\theta$), it is actually -kmgx. Do you agree with me? Still, considering friction in this way, you may notice that you get the same result.

 

[rockinwhiz]

In good time, his reasoning is very reasonable, but I must clarify that the signing force is not equal to -kmgxsen ($\theta$), it is actually -kmgx. Do you agree with me? Still, considering friction in this way, you may notice that you get the same result.

Oh yeah. I guess I'll check it out later. Thanks for everything man.

 

[madafo3435]

Oh yeah. I guess I'll check it out later. Thanks for everything man.

you're welcome, our discussion was interesting

 

[haruspex]

In good time, his reasoning is very reasonable, but I must clarify that the signing force is not equal to -kmgxsen ($\theta$), it is actually -kmgx. Do you agree with me? Still, considering friction in this way, you may notice that you get the same result.

$\theta$ and k are not specified in the problem statement, they are unknown constants created by the solver. So whether the friction coefficient is written as $kx$ or $kx\sin(\theta)$ is irrelevant, provided it is done consistently.