Answer
$N_B=857lb$
$N_A=1393lb$
$t=2.72s$
Work Step by Step
We can determine the required time and the normal reactions as follows:
According to equation of motion in x-direction
$\Sigma F_x=ma_x$
$\implies \mu_k(2N_B)=ma_x$
We plug in the known values to obtain:
$0.3(2N_B)=\frac{4500}{32.2}a_x$
This simplifies to:
$N_B=232.916a_x$ [eq(1)]
Similarly $\Sigma M_A=\Sigma (M_K)_A$
$\implies -2N_B(6)+W(2)=-ma_x(2.5)$
$\implies -2N_B(6)+4500(2)=-\frac{4500}{32.2}(2.5a_x)$
This simplifies to:
$N_B=750+29.1149a_x$ [eq(2)]
We plug in the known values from eq(1) into eq(2) to obtain:
$232.916a_x=750+29.1149a_x$
This simplifies to:
$a_x=3.68 ft/s^2$
We plug in this value in eq(1) to obtian:
$N_B=232.916(3.68)$
$\implies N_B=857lb$
Now we apply the equation of motion in y-direction
$\Sigma F_y=ma_y$
$\implies 2N_B+2N_A-W=m(0)$
$\implies 2(857)+2N_A-4500=0$
This simplifies to:
$N_A=1393N$
We know that
$v=v_{\circ}+a_xt$
We plug in the known values to obtain:
$10=0+3.68(t)$
This simplifies to:
$t=2.72s$
