Question

What is the upper block's acceleration if the coefficient of kinetic friction between the block and the table is 0.13?

Answer 1

The force pulling the block forward will be equal to the weight of the hanging box.
F = mg = g

The resistance will be equal to μR, where R is the reaction force and μ is the coefficient of friction. 

Resultant force = g - μmg
ma = g - μmg
a = g(1 - 2μ) / 2
a = 3.63 m/s²

Answer 2

Let us assume that pulley is mass less.

Let the tension produced at both ends of the pulley is T.

We are asked to calculate the acceleration of the block.

Let the masses of two bodies are denoted as $m_{1} \ and\ m_{2}\ respectively$

$Let\ m_{1} =1 kg\ and\ m_{2} =2 kg$

As per this diagram, the body having mass 1 kg is moving downward and the body having mass 2 kg is moving on the surface of the table.

Let the acceleration of each block is a .

For body having mass 1 kg:

The net force acting on 1 kg body will be-

                             $m_{1} g-T=m_{1} a$        [1]

Here tension in the rope will be vertically upward and weight of the body will be in vertical downward direction.

For body having mass 2 kg:

The coefficient of kinetic friction $[\mu]=0.13$

$Hence\ the\ frictional\ force\ F=\mu N$

                                                     $F=\mu m_{2} g$

Hence the net force acting on the body having mass 2 kg-

                                  $T-\mu m_{2} g=m_{2} a$  [2]

Here the tension of the rope is towards right i.e along the direction of motion of the 2 kg block and frictional force is towards left.

Combining 1 and 2 we get-

                           $m_{1} g-T=m_{1}a$             [1]

                           $T-\mu m_{2}g= m_{2} a$   [2]

                           ---------------------------------------------------

                           $[m_{1} -\mu m_{2} ]g=[m_{1} +m_{2} ]a$

                           $a=\frac{m_{1}-\mu m_{2}} {m_{1}+ m_{2}}*g$

                           $a=\frac{1-[2*0.13]}{1+2} *9.8\ m/s^2$

                           $a=\frac{0.74}{3} *9.8\ m/s^2$

                           $a=2.417 m/s$         [ans]