#### Answer

$a_A=9.66ft/s^2$, $a_B=15ft/s^2$

#### Work Step by Step

The acceleration of each block can be determined as follows:
We know that
$\Sigma F_y=0$
$\implies N-W=0$
$\implies N=W=8lb$
and $\Sigma F_x=ma_x$
$\implies kx-\mu_k N=m_A a_A$
We plug in the known values to obtain:
$20\times 0.2-(0.2\times 8)=\frac{8}{32.2}a_A$
This simplifies to:
$a_A=9.66ft/s^2$
Now for the second block
$\Sigma F_y=0$
$\implies N-W=0$
$\implies N=W=6lb$
and $\Sigma F_x=ma_x$
$\implies kx-\mu_k N=m_B a_B$
We plug in the known values to obtain:
$20\times 0.2-0.2\times 6=\frac{6}{32.2}a_B$
This simplifies to:
$a_B=15ft/s^2$