#### Answer

$v_2=9.22ft/s$, $\Sigma U_{1\rightarrow}2=3.04ft\cdot lb$

#### Work Step by Step

We can determine the required speed and work done as follows:
According to the principle of conservation of angular momentum
$m_B(v_B)r_1=m_B(v_{\theta})_2r_2$
$\implies 6\times 3=2(v_{\theta})_2$
$\implies (v_{\theta})_2=9ft/s$
Now we can calculate the speed as
$v_2=\sqrt{(v_{\theta})_2^2+(v_r)_2^2}$
We plug in the known values to obtain:
$v_2=9.22ft/s$
Now according to the principle of work and energy
$\frac{1}{2}m_Bv_1^2+\Sigma U_{1\rightarrow 2}=\frac{1}{2}m_Bv_2^2$
We plug in the known values to obtain:
$\frac{1}{2}(\frac{4}{32.2})(6)^2+\Sigma U_{1\rightarrow 2}=(\frac{4}{32.2})(9.22)^2$
This simplifies to:
$\Sigma U_{1\rightarrow}2=3.04ft\cdot lb$