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# All ramps are not equal

I have a very tiny block that slides down a ramp in the Cartesian plane from point $$a$$ to point $$b$$. This ramp can be modelled by the function $$f(x)$$ for $$x_a\leq x \leq x_b$$. For all values of c such that $$a<c \leq b$$, $$f(c)<f(a)$$. If the block manages to slide from point $$a$$ to $$b$$. Prove that the maximum velocity of the block at point $$b$$ is achieved when $$f(x)$$ is a straight line from $$a$$ to $$b$$ if friction is present.

I will post the proof for this if no one has within 48 hours.

Note by Trevor Arashiro
11 months, 3 weeks ago

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As the end points are fixed and are separated by a verical distance of $$h$$ (say), by work-kinetic energy theorem (and a little simplification), we have, $\frac{v_b^2}{2}=gh-\mu_kg s$ where $$\mu_k$$ is the coefficient of kinetic friction and $$s$$ is the length of the horizontal section of the ramp.

As the points $$a$$ and $$b$$ are fixed, so is $$s$$. Hence, speed of the block remains constant.

PS: This was a nice fallacy which our teacher had discussed.

@Trevor Arashiro · 11 months, 3 weeks ago

What if slope of the curve is less than or equal to $$\mu_{k}$$? · 11 months, 2 weeks ago

The question states that the body manages to slide to point $$b$$. So, kinetic friction will always act.

Plus, your objection makes sense if you said that $$\mu_s \geq \tan \theta$$. The inequality $$\mu_k \geq \tan \theta$$ makes no difference to the motion of the block. · 11 months, 2 weeks ago

That's the same start to my proof, but when you say horizontal section of the ramp, are you referring only to the length of the ramp in the x-direction or the total length of the ramp? And another thing to think about: at different angles, while u is constant, the frictional force is greater for flatter sections of the ramp, so does your equation still work for curved ramps? · 11 months, 3 weeks ago

I meant the length of the horizontal section only.

Yes, this is valid for curved ramps as well. Here's the proof:

Consider a point on the ramp. Let the tengent at this point make an angle $$\theta$$ with the horizontal. Then, if $$dW$$ denotes the work done by friction over a path length $$ds$$, $dW=-\mu_kmg\cos \theta ds$ But, note that $$ds\, \cos \theta=dx$$. Hence, $dW=-\mu_kmg dx\\\implies W=-\mu_kmg(x_b-x_a)$ Assuming $$x_b>x_a$$

So, all ramps are actually equal ;) · 11 months, 3 weeks ago

It must be "$$\mu_k g s$$" · 11 months, 3 weeks ago

Yeah. My bad. · 11 months, 3 weeks ago