ANSWER
2.29 m/s²
EXPLANATION
The free body diagram of this situation, when block C is lifted off A is
Now we know that blocks A and B are moving, so the equations according with Newton's second law of motion for each block are:
Block A:
[tex]\begin{gathered} F_T-F_f=m_A\cdot a \\ F_N-F_{gA}=0 \end{gathered}[/tex]
Block B (there are not horizontal forces):
[tex]F_T-F_{gB}=-m_B\cdot a[/tex]
From the equation for block B, we have that the tension of the rope is:
[tex]\begin{gathered} F_T=F_{gB}-m_B\cdot a \\ F_T=m_B\cdot g-m_B\cdot a \\ F_T=m_B(g-a) \end{gathered}[/tex]
And mB is:
[tex]m_B=\frac{F_{gB}}{g}[/tex]
So the tension force is:
[tex]F_T=\frac{F_{gB}}{g}(g-a)[/tex]
The kinetic friction force, when the coefficient of kinetic friction betwen the surfaces is μk is:
[tex]F_f=\mu_k\cdot F_N[/tex]
The normal force, from the second equation of block A is:
[tex]F_N=F_{gA}=44N[/tex]
So the friction force between block A and the table is:
[tex]F_f=0.15\cdot44N=6.6N[/tex]
Now, using the first equation for block A, we can find its acceleration. But before doing that, we have to find the mass of block A - remember that we have the weight, not the mass. This we find with the equation of weight:
[tex]\begin{gathered} F_{gA}=m_A\cdot g \\ m_A=\frac{F_{gA}}{g} \end{gathered}[/tex]
Replacing Ft, Ff and mA into the first equation of block A:
[tex]F_T-F_f=m_A\cdot a[/tex][tex]\frac{F_{gB}}{g}(g-a)-6.6N=\frac{F_{gA}}{g}\cdot a[/tex]
We have an equation which we can solve for a. Apply distributive property in the first term:
[tex]\begin{gathered} \frac{F_{gB}\cdot g}{g}-\frac{F_{gB}\cdot a}{g}-6.6N=\frac{F_{gA}}{g}\cdot a \\ F_{gB}-\frac{F_{gB}\cdot a}{g}-6.6N=\frac{F_{gA}}{g}\cdot a \end{gathered}[/tex]
Then add the second term from both sides:
[tex]\begin{gathered} F_{gB}-\frac{F_{gB}\cdot a}{g}+\frac{F_{gB}\cdot a}{g}-6.6N=\frac{F_{gA}}{g}\cdot a+\frac{F_{gB}\cdot a}{g} \\ F_{gB}-6.6N=\frac{F_{gA}}{g}\cdot a+\frac{F_{gB}}{g}\cdot a \end{gathered}[/tex]
Take a as a common factor from the right side of the equation:
[tex]F_{gB}-6.6N=(\frac{F_{gA}}{g}+\frac{F_{gB}}{g})a[/tex]
And then divide by the coefficient of a:
[tex]a=\frac{F_{gB}-6.6N}{(\frac{F_{gA}}{g}+\frac{F_{gB}}{g})}[/tex]
Since g is a common denominator we can simplify the expression a little before replacing with values:
[tex]a=\frac{F_{gB}-6.6N_{}}{\frac{F_{gA}+F_{gB}}{g}}=\frac{g(F_{gB}-6.6N)}{F_{gA}+F_{gB}}[/tex]
Now we can replace and find a:
[tex]a=\frac{9.81m/s^2(22N-6.6N)}{44N+22N}=2.29m/s^2[/tex]
