The Rough Time...

In this problem, what is important to note is that the friction is varying and will depend upon the mass of the rod on the rough surface.
Let the distance traveled by the front end of the rod be $x$ i.e., $'x'$ is the length of rod over the rough surface. Hence friction will be corresponding to the normal reaction only due to this length which is $N = \lambda xg$ where $\lambda$ represents the linear mass density of the rod. Hence,
$\displaystyle f = f_{max} = \mu N = \mu \lambda xg$ . Here $f = f_{max}$ because there is relative slipping.
$\displaystyle ma = - \mu \lambda xg \Rightarrow \ddot{x} = - \frac{ \mu \lambda xg}{m}$

Since $m = \lambda l$ , therefore, $\displaystyle \ddot{x} + \frac{\mu g }{l} x = 0$
This is a standard differential equation with a standard solution of the form,
$\displaystyle x = A\sin(\omega t + \phi) \text{ where } \omega^2 = \frac{\mu g }{l}$ Here , $A$ and $\phi$ are unknowns. which can be found out using the following method.

Differentiating the above equation, we get,
$\displaystyle v = A \omega \cos(\omega t + \phi)$
At $t=0$ , $x= 0 \Rightarrow A \sin(\phi) = 0 \Rightarrow \phi = 0$ . Also, at $t=0$ , $\displaystyle v=v_{0} \Rightarrow v_{0} = A \omega \cos(\phi) \Rightarrow A = \frac{v_{0}}{\omega}$

Since , we are interested only when the velocity first gets zero,we can write,
$\displaystyle \Rightarrow cos(\omega t ) = 0$ and we just want when was the first such instance, we can write $\displaystyle \omega t = \frac{\pi}{2}$ . Hence, $\displaystyle t_{0} = \frac{\pi}{2 \omega} \Rightarrow t_{0} = \frac{\pi}{2} \sqrt{ \frac{l}{\mu g}}$ .
Substitute the given values and round off to the nearest integer to get the answer as $5$ .

Note: The equations obtained above are those of simple harmonic motion which is sustained. However the above motion behaves in a similar manner as the retarding force is directly proportional to the displacement. In such a motion it will represent simple harmonic motion only for the one-fourth the time period, which actually is the sought answer.In this case the rod will not go back and the above equations of motion not hold true for the rod once the velocity becomes zero. If we wanted the length of the rod that goes onto the rough floor we would have simply given the answer as $A$ , which would have required the value of $v_{0}$ . But finding $A$ was much easier and could have been found out without solving the differential equation.

Parallel Question: Find $A$ given $v_{0} = 3 ms^{-1}$ . Without solving the differential equation.