# 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, $f(c). 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
3 years, 8 months 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.

- 3 years, 8 months ago

It must be "$\mu_k g s$"

- 3 years, 8 months ago

- 3 years, 8 months 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?

- 3 years, 8 months 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 ;)

- 3 years, 8 months ago

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

- 3 years, 8 months 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.

- 3 years, 8 months ago

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

- 3 years, 8 months ago