Step 1

We are given that the coefficient of static friction is ${\mu _s} = 0.4$ , the coefficient of kinetic friction is ${\mu _k} = 0.3$, the force applied to the blocks is $P = 6\;{\rm{lb}}$, the weight of block A is ${w_A} = 20\;{\rm{lb}}$ and the weight of block B is ${w_B} = 50\;{\rm{lb}}$.

We are asked to calculate the acceleration of each block.

 
Step 2

The free body diagram of the system is shown as:

Here, f is the frictional force and P is the force applied to the blocks.

 
Step 3

To calculate the mass of block A we use the formula:

\[{m_A} = \frac{{{w_A}}}{g}\]
 
Step 4

Substitute the known values in the equation:

\[\begin{array}{c} {m_A} = \frac{{\left( {20\;{\rm{lb}}} \right)}}{{\left( {32.2\;{\rm{ft/}}{{\rm{s}}^2}} \right)}}\left( {\frac{{1\;{\rm{slug}} \cdot {\rm{ft/}}{{\rm{s}}^2}}}{{1\;{\rm{lb}}}}} \right)\\ = 0.621\;{\rm{slug}} \end{array}\]
 
Step 5

To calculate the mass of block B we use the formula:

\[{m_B} = \frac{{{w_B}}}{g}\]
 
Step 6

Substitute the known values in the equation:

\[\begin{array}{c} {m_B} = \frac{{\left( {50\;{\rm{lb}}} \right)}}{{\left( {32.2\;{\rm{ft/}}{{\rm{s}}^2}} \right)}}\left( {\frac{{1\;{\rm{slug}} \cdot {\rm{ft/}}{{\rm{s}}^2}}}{{1\;{\rm{lb}}}}} \right)\\ = 1.55\;{\rm{slug}} \end{array}\] Step 7

According to free body-diagram the net force along x-axis is given as:

\[{F_x} = P\]
 
Step 8

According to Newton’s law of motion, the net force acting on block along x-axis is given as:

\[{F_x} = \left( {{m_A} + {m_B}} \right)a\]
 
Step 9

Substitute the known value in above equation:

\[\begin{array}{c} P = \left( {{m_A} + {m_B}} \right)a\\ a = \frac{P}{{{m_A} + {m_B}}} \end{array}\]
 
Step 10

Substitute the known values in the equation:

\[\begin{array}{c} a = \frac{{\left( {6\;{\rm{lb}}} \right)}}{{\left( {0.621\;{\rm{slug}} + 1.55\;{\rm{slug}}} \right)}}\\ = \frac{{\left( {6\;{\rm{lb}}} \right)}}{{\left( {2.171\;{\rm{slug}}} \right)}}\left( {\frac{{1\;{\rm{slug}} \cdot {\rm{ft/}}{{\rm{s}}^2}}}{{1\;{\rm{lb}}}}} \right)\\ = 2.76\;{\rm{ft/}}{{\rm{s}}^2} \end{array}\] Step 11

To calculate the frictional force the net force acting on the block A along x-axis is given as:

\[{F_A} = P - f\]
 
Step 12

According to Newton’s law of motion, the net force acting on block A along x-axis is given as:

\[{F_A} = {m_A}a\]
 
Step 13

Substitute the known value in the equation:

\[\begin{array}{c} P - f = {m_A}a\\ f = P - {m_A}a \end{array}\]
 
Step 14

Substitute the known values in the equation:

\[\begin{array}{c} f = 6\;{\rm{lb}} - \left( {{\rm{0}}{\rm{.621}}\;{\rm{slug}}} \right)\left( {2.76\;{\rm{ft/}}{{\rm{s}}^2}} \right)\left( {\frac{{1\;{\rm{lb}}}}{{1\;{\rm{slug}} \cdot {\rm{ft/}}{{\rm{s}}^2}}}} \right)\\ = 6\;{\rm{lb}} - 1.71\;{\rm{lb}}\\ = 4.29\;{\rm{lb}} \end{array}\]
 
Step 15

The normal force acting on block A is equal to the weight of block A which is given as:

\[N = {w_A}\]
 
Step 16

To calculate the static force of friction acting on block A we use the formula:

\[{f_s} = {\mu _s}N\]
 
Step 17

Substitute the known value in the formula:

\[{f_s} = {\mu _s}{w_A}\]
 
Step 18

Substitute the known values in the equation:

\[\begin{array}{c} {f_s} = \left( {0.4} \right)\left( {20\;{\rm{lb}}} \right)\\ = 8\;{\rm{lb}} \end{array}\]

Since the frictional force acting on block A is less than the static friction therefore, the acceleration of both the blocks will be the same which is $2.76\;{\rm{ft/}}{{\rm{s}}^2}$.