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.

#Mechanics

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

A Former Brilliant Member - 5 years, 1 month ago

It must be " $\mu_k g s$ "

Kushal Patankar - 5 years, 1 month ago

Yeah. My bad.

A Former Brilliant Member - 5 years, 1 month 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?

Trevor Arashiro - 5 years, 1 month 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 ;)

A Former Brilliant Member - 5 years, 1 month ago

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

Kushal Patankar - 5 years, 1 month 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.

A Former Brilliant Member - 5 years, 1 month ago

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 - 5 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$