A small block is let slide down a curved surface as indicated in the picture from below. The curved surfaces have a negligible friction, however the horizontal bottom which is of $12\,m$ in length is rugged. The coefficient of kinetic friction of the bottom surface is $0.2$. The block departs from rest $8\,m$ above the rugged bottom. Find the position where the block stops.

The alternatives given are as follows:

$\begin{array}{ll}

1.&\textrm{At 4m from B}\\

2.&\textrm{At 3m from C}\\

3.&\textrm{At 4m from C}\\

4.&\textrm{At 3m from B}\\

\end{array}$

This problem involves the use of conservation of mechanical energy. And the way how I attempted to solve it was:

$E_u=E_k$

$mgh=\frac{1}{2}mv^2$

$v^2=2gh$

Then It mentions that the coefficient of friction is 0.2, this means that when the block moves in the rugged surface the frictional force will be F=ma.

$f_s=F$

$ma=\mu N = \mu mg$

Then:

$a=\mu g$

From this I can find the distance using the equation:

$v_f^2=v_o^2-2a\Delta x$

Since it indicates that the body will finish at rest $v_f=0$

Therefore:

$\Delta x= \frac{v_o^2}{2a}$

Replacing the given values results as follows:

$\Delta x= \frac{2gh}{2(\mu g)}=\frac{h}{\mu}=\frac{8}{0.2}=40\,m$

But this length is way too long and exceeds the given length of the bottom. Did I made any sort of mistake?. Can somebody help me here?. Supposedly the answer is at $4\,m$ from $C$.

Upon thinking more deeply into it. I think that this means the cube will rise to the other end of the curved surface. But there isn't any specific indication that the curved surface is a quarter of a circle.

If I assume that the cube rise to the top and only gets to $D$ and slides down back:

I did calculated the speed after traveling $12\,m$

$v_f^2=16g-\frac{4\times 12}{10}g=\frac{56}{5}g$

But after that what distance will be taken into consideration?. The vertical?.

Assuming it it is the vertical until traveling

$B-C-D-C=12+8+8=28\,m$

Then I thought that it really doesn't matter this distance because those curved surfaces are of negligible friction so the same speed will be preserved at the bottom as the energy is conserved.

In the second travel from $C-B$:

$0=\frac{56}{5}g-2(\mu g)\Delta x$

$0=\frac{56}{5}g-2(0.2 \times g)\Delta x$

$x= 28\,m$

Again this exceeds the $12\,m$.

Then I must calculate again the final speed after $12\,m$.

$v_f^2=v_o^2-2a\Delta x$

$v_f^2=\frac{56}{5}g-\frac{4\times 12}{10}g=\frac{32g}{5}$

Then again calculating for a third travel from $BC$.

$v_f^2=v_o^2-2a\Delta x$

$0=\frac{32g}{5}-2(0.2 g )\Delta x$

$x=16\,m$

Again this exceeds the $12\,m$ limit.

Then I'm forced to calculate the speed for $12\,m$ after that fourth travel from $BC$.

$v_f^2=\frac{32g}{5}-2(0.2g\times 12)=\frac{8g}{5}$

Then finally from $CB$:

$0=v_o^2-2a\Delta x$

$0=\frac{8g}{5} - 2(0.2g)\Delta x$

$x=4\,m$

Which accounts for $4\,m$ from $C$ and it checks with the alternatives. And this corresponds to the right answer. But this procedure was too extense. Does it exist a reasonable shortcut?.