Let the mass of the block be $\color{#20A900}{M}$ and initial velocity be $\color{#20A900}{u}$

Find Velocity at time t and initial velocity.

Now, let at time $\color{#20A900}{t}$ , the velocity of block be $\color{#20A900}{v}$ , and $\color{#20A900}{x}$ part of block be at rough part. Let acceleration at that time be $\color{#20A900}{a}$

So,

$\text{frictional force } = Ma$ $\mu \frac{Mx}{l} g = Ma$ $\mu \frac{x}{l} g= a$ $\mu \frac{x}{l} g= \frac{-vdv}{dx}$ $\mu g\frac{x}{l}dx = -vdv$ $\frac{\mu}{l}g\int_0^xxdx = -\int_u^v vdv$ $\frac{\mu}{l}g x^2 = u^2 - v^2$ $\boxed{v = \sqrt{u^2 - \frac{\mu g}{l}x^2}}$ $\text{At } x = l, v = 0$ $\text{So, } \boxed{u = \sqrt{\mu gl}}$

Finding time

So $v = \sqrt{u^2 - \frac{\mu g}{l}x^2}$ $\frac{dx}{dt} = \sqrt{u^2 - \frac{\mu g}{l}x^2}$ $\frac{dx}{\sqrt{u^2 - \frac{\mu g}{l}x^2}} = dt$ $\int_0^l\frac{dx}{\sqrt{u^2 - \frac{\mu g}{l}x^2}} = \int_0^tdt$ $\sqrt{\frac{l}{\mu g}}\sin^{-1}\bigg(\frac{\sqrt{\mu gl}}{u}\bigg) = t$ $\text{Put } u = \sqrt{\mu gl}$ $t = \sqrt{\frac{l}{\mu g}} \sin^{-1}1$ $t = \sqrt{\frac{(10)}{(\frac{\pi^2}{100})(10)}} \frac{\pi}{2}$ $\color{#3D99F6}{\boxed{t = 5 \text{ sec}}}$

I didn't want to do any integrating or ODE solving; I used a numerical simulation method where I coded in the value of acceleration.

Firstly, let's take the length of the block that's on the rough side of the surface to be $x$ .

Therefore, the normal force acting on that rough side would be $\bold{N} = \displaystyle -\frac{x}{10} Mg$ .

Multiplying the normal force by the friction coefficient and simplifying, we get:

$\ddot{x} = -x \mu$

We know that the block will come to rest when $\dot{x} = 0$ ; when we code in an initial value of $\dot{x}$ as a random value of $v$ , by simulating till $\dot{x} = 0$ , we can get the displacement travelled before rest for the test value of $v$ .

This method is sort of a "virtual laboratory" approach to the problem. It's just trial and error. What I have done is as follows:

• Give a random value for $v$ (initial velocity) and check the displacement travelled before rest
• Knowing that the value for displacement will have to be $10 m$ , modify the value for the velocity to arrive at the correct displacement
• When the value of the velocity produces a displacement of exactly $10$ , acquire the time which was recorded when the block stopped

This time will be the time it will take for the block to come to rest.

Here's the program:

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import math

time = 0
deltaT = 10**-6



v = math.pi #Test value for v

x = 0 #initial 
xDot = v #initial velocity

mu = math.pi**2/100 #friction coefficient
g = 10 #gravitational acceleration





while xDot >= 0: #till the body stops

  xDotDot = -x*mu #Friction force

  #Euler numerical integration 
  xDot += xDotDot*deltaT
  x += xDot*deltaT

  #updating value of time
  time += deltaT


#Calculating error in values of displacement and expected displacement
if abs(10-x)<= 0.001:
  print("The time it takes to come to rest is: "+str(time))
else:
  print("Test value of initial velocity isn't accurate.")

The equation of motion is

$\ddot{x}= - g \mu \dfrac{x}{10} = - \mu x$

Thus the differential equation governing the movement of the block is

$\ddot{x} + \mu x = 0$

The solution of this differential equation is

$x(t) = a \cos( \sqrt{\mu} t ) + b \sin ( \sqrt{\mu} t ) = a \cos \left( \dfrac{\pi t }{10} \right) + b \sin \left( \dfrac{\pi t }{10} \right)$

Since x(0) = 0, then $a = 0$ , and the solution $x(t) = b \sin \left( \dfrac{\pi t }{10} \right)$

The velocity of the block is $\dot{x}(t) = \dfrac{ b \pi }{10} \cos \left( \dfrac{ \pi t }{10} \right)$

Thus the velocity is decreasing (as the $\cos$ function is decreasing ) and becomes zero when $\dfrac{ \pi t }{10} = \dfrac{\pi}{2}$

From which, the block will stop at $t = 5 \text{ sec}$ .

Since we are given that the block stops when it has travelled $10$ m, then $b = 10$ and $v = \pi$ . For a given initial velocity $v$ , $b = \dfrac{10 v}{\pi}$ , so that if $v < \pi$ then the block will stop before travelling $10$ m. However if $v > \pi$ , then the solution we obtained is valid for $0 \le t \le t^*$ , where $b \sin \left( \dfrac{\pi t^* }{10} \right) = 10$ , i.e. $\sin \left( \dfrac{\pi t^* }{10} \right) = \dfrac{ \pi}{v}$ ; and for $t \gt t^*$ , the equation of motion becomes

$\ddot{x} = -\mu g$ , with $x(t^*) = 10$ and $\dot{x}(t^*) = \dfrac{ b \pi }{10} \cos \left( \dfrac{ \pi t^* }{10} \right)$

Integrating once, to obtain the velocity, we obtain,

$\dot{x} = \dot{x}(t^*) - \mu g (t - t^*)$

Equating the above to zero, we can obtain $t_f$ , the final time of movement,

$t_f = t^* + \dfrac{ \dot{x}(t^*) }{\mu g}$