91Ó°ÊÓ

The static frictional coefficient acts when the blocks are not moving. The acceleration will be zero in this case.

However, the kinetic frictional coefficient acts when the blocks move with constant speed.

The acceleration of the blocks remains zero. So, the only difference is that the frictional force uses the kinetic frictional coefficient while moving at a constant speed.

The given data can be listed below as:

• The mass of block A is $m_A=5\mathrm{kg}$.
• The mass of block B is $m_B=2\mathrm{kg}$.
• The static coefficient of friction is ${\mathrm{μ}}_s=0.40$.
• The kinetic coefficient of friction is ${\mathrm{μ}}_k=0.20$.

The free body diagram of block B can be shown as:

Here, T is the tension in the cord, and g is the acceleration due to gravity.

The acceleration of the blocks is zero.

At the equilibrium condition, the forces along the vertical direction of block B can be expressed as:

$\begin{array}{c}∑F_y=0\\ m_{B}g-T=0\\ T=m_{B}g…\left(i\right)\end{array}$

The free body diagram of block A can be shown as:

Here, $F_N$is the normal reaction force on block A, and $F_{fr}$is the frictional force. The tension is constant along the cord. So, the same tension force acts to the right to block A.

At the equilibrium condition, the forces along the vertical direction of block A can be expressed as:

$\begin{array}{c}∑F_y=0\\ F_N-m_{A}g=0\\ F_N=m_{A}g…\left(ii\right)\end{array}$

At the equilibrium condition, the forces along the horizontal direction of block A can be expressed as:

$\begin{array}{c}∑F_x=0\\ T-F_{fr}=0\\ T=F_{fr}\end{array}$

Since the block is not moving, the coefficient of static friction acts between block A and the table.

The static frictional force can be expressed as:

$F_{fr}={\mathrm{μ}}_s{F}_N$

Substitute the values from equations (i) and (ii) in the above equation.

localid="1645528746449" $\begin{array}{rcl}{m}_{B}g& =& {\mathrm{μ}}_s×m_{A}g\\ m_A& =& \frac{m_B}{{\mathrm{μ}}_s}…\left(iii\right)\end{array}$

Substitute the values in the above expression.

$\begin{array}{c}{m}_A=\frac{2\mathrm{kg}}{0.40}\\ =5\mathrm{kg}\end{array}$

Thus, the minimum value of 5 kg of $m_A$will keep the system from moving.

The system is moving at a constant speed. Therefore, the coefficient of kinetic friction acts between block A and the table.

From equation (iii), the mass of block A can be expressed as:

$\begin{array}{c}{m}_{B}g={\mathrm{μ}}_k×m_{A}g\\ m_A=\frac{m_B}{{\mathrm{μ}}_k}\end{array}$

Substitute the values in the above equation.

$\begin{array}{c}{m}_A=\frac{2\mathrm{kg}}{0.20}\\ =10\mathrm{kg}\end{array}$

Thus, 10 kg of $m_A$will keep the system moving at a constant speed.