[asy]
size(5cm);
pen p=linewidth(3), dark_grey=gray(0.25), ll_grey=gray(0.90), light_grey=gray(0.75);

transform dishift(real x) {
    return shift(x,x);
}

// Draw the table of latch of table
path ell = ((0,0)--(0,-1)--(-0.1,-1)--(-0.1,-0.1)--(-1,-0.1)--(-1,0)--cycle); // the ell shape
path corner = dishift(-0.85)*ell; // define the path
path table = dishift(-1)*scale(5)*ell; // define the table by scaling the pulley
filldraw(corner, ll_grey, light_grey+p); // base of pulley
filldraw(table, ll_grey, grey+p); // table

real block_size = 1.6;
// template for block
path block = unitsquare;
pair block_center = (0.5,0.5);

/* Resting block */
transform rest = shift(-5, -0.9) * scale(block_size); // transformation for resting block
filldraw(rest * block, ll_grey, light_grey+p); // draw block
draw(rest*(1,0.5)--dir(110), light_grey+p); // rope fr0m midpoint of right block to wheel
label("$m$", rest * block_center, fontsize(16)); // label block

/* Hanging block */
transform hang = shift(0.2,-4.1) * scale(block_size); // transformation for hanging block
draw((1,0)--(1,-2.5), light_grey+p); // string of pulley
filldraw(hang * block, ll_grey, light_grey+p); // fill it
label("$M$",hang * block_center,fontsize(16)); // label the small m

// Draws the actual pulley
filldraw(unitcircle, grey, p); // outer boundary of pulley wheel
filldraw(scale(0.4)*unitcircle, light_grey, p); // inner boundary of pulley wheel
path pulley_body=arc((0,0),0.3,-40,130)--arc((-1,-1),0.5,130,320)--cycle; // defines "arm" of pulley
filldraw(pulley_body, ll_grey, dark_grey+p); // draws the arm
filldraw(scale(0.18)*unitcircle, ll_grey, dark_grey+p); // inner circle of pulley
[/asy] 
A pulley system of two blocks, shown above, is released from rest. The block on the table, which has mass $m=1.0 \, \text{kg}$ slides after the time of release and hits the pulley to come to a dead stop. There was originally a distance of $ 1.0 \, \text{m} $ between the block and the pulley, which the block fully covers during the slide. From the time of release to the time of hitting the pulley, the angle that the rope above the table makes with the horizontal axis is a, nearly constant, $10.0^\circ$. The hanging block has mass $ M = 2.0 \, \text{kg} $. The table has a coefficient of friction of $0.50$ with the block that sits on it. The pulley is frictionless. Also, assume that, during the entire slide, the block never leaves the ground. Let $t$ be the number of seconds in takes for the $1.0\text{-m}$ slide. Find $100t$, rounded to two significant figures. 

[i](Ahaan Rungta, 4 points)[/i]
1. **Free Body Diagram (FBD) Analysis:**
   - Let \( T \) be the tension in the string.
   - Let \( a \) be the acceleration of the hanging block \( M \) (downwards).
   - Let \( N \) be the normal force on the block \( m \).
   - Let \( \mu \) be the coefficient of friction.
   - Let \( \theta = 10^\circ \) be the angle that the string makes with the horizontal.

2. **Equations of Motion:**
   - For the hanging block \( M \):
     \[
     Mg - T = Ma \quad \text{(1)}
     \]
   - For the block \( m \) on the table:
     - Horizontal direction:
       \[
       T \cos \theta - \mu N = ma \cos \theta \quad \text{(2)}
       \]
     - Vertical direction:
       \[
       T \sin \theta + N = mg \quad \text{(3)}
       \]

3. **Solving for Normal Force \( N \):**
   - From equation (3):
     \[
     N = mg - T \sin \theta \quad \text{(4)}
     \]

4. **Substitute \( N \) into Equation (2):**
   - Using equation (4) in equation (2):
     \[
     T \cos \theta - \mu (mg - T \sin \theta) = ma \cos \theta
     \]
     \[
     T \cos \theta - \mu mg + \mu T \sin \theta = ma \cos \theta
     \]
     \[
     T (\cos \theta + \mu \sin \theta) = ma \cos \theta + \mu mg \quad \text{(5)}
     \]

5. **Solving for \( T \):**
   - From equation (1):
     \[
     T = Mg - Ma \quad \text{(6)}
     \]

6. **Substitute \( T \) from equation (6) into equation (5):**
   \[
   (Mg - Ma)(\cos \theta + \mu \sin \theta) = ma \cos \theta + \mu mg
   \]
   \[
   Mg \cos \theta + Mg \mu \sin \theta - Ma \cos \theta - Ma \mu \sin \theta = ma \cos \theta + \mu mg
   \]
   \[
   Mg \cos \theta + Mg \mu \sin \theta = Ma \cos \theta + Ma \mu \sin \theta + ma \cos \theta + \mu mg
   \]
   \[
   Mg (\cos \theta + \mu \sin \theta) = a (M \cos \theta + M \mu \sin \theta + m \cos \theta) + \mu mg
   \]
   \[
   a (M \cos \theta + M \mu \sin \theta + m \cos \theta) = Mg (\cos \theta + \mu \sin \theta) - \mu mg
   \]
   \[
   a = \frac{Mg (\cos \theta + \mu \sin \theta) - \mu mg}{M \cos \theta + M \mu \sin \theta + m \cos \theta}
   \]

7. **Substitute Given Values:**
   - \( M = 2.0 \, \text{kg} \)
   - \( m = 1.0 \, \text{kg} \)
   - \( g = 9.8 \, \text{m/s}^2 \)
   - \( \mu = 0.50 \)
   - \( \theta = 10^\circ \)
   - \( \cos 10^\circ \approx 0.9848 \)
   - \( \sin 10^\circ \approx 0.1736 \)

   \[
   a = \frac{2 \times 9.8 \times (0.9848 + 0.50 \times 0.1736) - 0.50 \times 1 \times 9.8}{2 \times 0.9848 + 2 \times 0.50 \times 0.1736 + 1 \times 0.9848}
   \]
   \[
   a = \frac{2 \times 9.8 \times (0.9848 + 0.0868) - 4.9}{2 \times 0.9848 + 2 \times 0.0868 + 0.9848}
   \]
   \[
   a = \frac{2 \times 9.8 \times 1.0716 - 4.9}{2 \times 0.9848 + 0.1736 + 0.9848}
   \]
   \[
   a = \frac{21.0176 - 4.9}{2.1432 + 0.1736 + 0.9848}
   \]
   \[
   a = \frac{16.1176}{3.3016}
   \]
   \[
   a \approx 4.88 \, \text{m/s}^2
   \]

8. **Calculate Time \( t \):**
   - Using the equation \( s = \frac{1}{2} a t^2 \):
     \[
     1 = \frac{1}{2} \times 4.88 \times t^2
     \]
     \[
     t^2 = \frac{2}{4.88}
     \]
     \[
     t^2 \approx 0.4098
     \]
     \[
     t \approx \sqrt{0.4098}
     \]
     \[
     t \approx 0.64 \, \text{seconds}
     \]

9. **Calculate \( 100t \):**
   \[
   100t \approx 100 \times 0.64 = 64
   \]

The final answer is \(\boxed{64}\).