Answer
$a=\frac{1}{2}(1-\mu_k)g$
Work Step by Step
We can determine the required acceleration as follows:
$\Sigma F_y=ma_y$
$\implies T-mg=-mg$
$\implies T=m(g-a)~~$ eq(1)
Now we apply Newton's second law
$\Sigma F_y=ma_y$
$\implies N-mg=0$
$\implies N=mg$
and $\Sigma F_x=ma_x$
$\implies T-\mu_k N=ma$
$\implies T-\mu_k mg=ma~~$ eq(2)
We plug in the value of $T$ from eq(1) into eq(2) to obtain:
$m(g-a)-\mu_k mg=ma$
This simplifies to:
$a=\frac{1}{2}(1-\mu_k)g$
