#### Answer

$v_A=5.8ft/s$

#### Work Step by Step

We can determine the required speed as follows:
$\Delta U_{A\rightarrow B}=W(3+0.25)(\frac{3}{5})-\mu_k N_B(3+0.25)+\frac{1}{2}k(x_1^2-x_2^2)$
We plug in the known values to obtain:
$\Sigma U_{A\rightarrow B}=4(3+0.25)(\frac{3}{5})-0.2\times 3.2(3+0.25)+\frac{1}{2}(50)(2-1.5)^2-(2-1.25)^2=-2.0925$
Now, according to the principle of work and energy
$\frac{1}{2}mv_A^2+\Sigma U_{A\rightarrow B}=\frac{1}{2}mv_B^2$
We plug in the known values to obtain:
$\frac{1}{2}(\frac{4}{32.2})v_A^2-2.0925=0$
This simplifies to:
$v_A=5.8ft/s$