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.

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TopNewestAs 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 – Deeparaj Bhat · 11 months, 3 weeks ago

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– Kushal Patankar · 11 months, 2 weeks ago

What if slope of the curve is less than or equal to \(\mu_{k}\)?Log in to reply

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. – Deeparaj Bhat · 11 months, 2 weeks ago

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– Trevor Arashiro · 11 months, 3 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?Log in to reply

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 ;) – Deeparaj Bhat · 11 months, 3 weeks ago

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– Kushal Patankar · 11 months, 3 weeks ago

It must be "\( \mu_k g s\)"Log in to reply

– Deeparaj Bhat · 11 months, 3 weeks ago

Yeah. My bad.Log in to reply

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 · 11 months, 3 weeks ago

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