The discussion revolves around a problem from the chapter on Work and Energy, specifically focusing on a block sliding down an inclined plane with variable friction. Participants are exploring the relationship between potential energy, kinetic energy, and the work done by friction.
The discussion is active, with participants providing guidance on identifying forces and calculating work done. Some participants suggest using the definition of work and integrating to find expressions for the work done by friction. There is recognition of the need to clarify the roles of different forces and the implications of the work-energy theorem.
Participants note that friction is a non-conservative force and express uncertainty about how to handle its variable nature in calculations. There is also mention of a specific textbook that discusses these concepts in detail.
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 said: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.
Oh yeah. I guess I'll check it out later. Thanks for everything man.madafo3435 said: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.
you're welcome, our discussion was interestingrockinwhiz said:Oh yeah. I guess I'll check it out later. Thanks for everything man.
##\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.madafo3435 said: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.