The discussion centers on the application of the work-energy theorem to analyze the motion of a block on an inclined plane with variable friction. The participants establish that the potential energy lost by the block, represented as mgh, is converted into kinetic energy and work done against friction. The coefficient of friction is defined as μ = kx, where k is a constant and x is the distance slid. The key conclusion is that the total work done by all forces acting on the block must equal the change in kinetic energy, emphasizing the need to account for both gravitational and frictional forces in the analysis.
PREREQUISITESStudents of physics, educators teaching mechanics, and anyone interested in understanding the dynamics of motion on inclined planes with frictional forces.
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.