All ramps are not equal

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

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

Note by Trevor Arashiro
3 years, 5 months ago

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As the end points are fixed and are separated by a verical distance of hh (say), by work-kinetic energy theorem (and a little simplification), we have, vb22=ghμkgs\frac{v_b^2}{2}=gh-\mu_kg s where μk\mu_k is the coefficient of kinetic friction and ss is the length of the horizontal section of the ramp.

As the points aa and bb are fixed, so is ss. Hence, speed of the block remains constant.

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

@Trevor Arashiro

Deeparaj Bhat - 3 years, 5 months ago

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It must be "μkgs \mu_k g s"

Kushal Patankar - 3 years, 5 months ago

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Yeah. My bad.

Deeparaj Bhat - 3 years, 5 months ago

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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?

Trevor Arashiro - 3 years, 5 months ago

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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 θ\theta with the horizontal. Then, if dWdW denotes the work done by friction over a path length dsds, dW=μkmgcosθdsdW=-\mu_kmg\cos \theta ds But, note that dscosθ=dxds\, \cos \theta=dx. Hence, dW=μkmgdx    W=μkmg(xbxa)dW=-\mu_kmg dx\\\implies W=-\mu_kmg(x_b-x_a) Assuming xb>xax_b>x_a

So, all ramps are actually equal ;)

Deeparaj Bhat - 3 years, 5 months ago

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What if slope of the curve is less than or equal to μk\mu_{k}?

Kushal Patankar - 3 years, 5 months ago

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

Plus, your objection makes sense if you said that μstanθ\mu_s \geq \tan \theta. The inequality μktanθ\mu_k \geq \tan \theta makes no difference to the motion of the block.

Deeparaj Bhat - 3 years, 5 months ago

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Not too difficult of a proof but it does require calc (at least the way I did it... which I pray is right)

Trevor Arashiro - 3 years, 5 months ago

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