Answer
$10.8ft/s$, $11.3ft\cdot lb$
Work Step by Step
According to the principle of conservation of angular momentum
$r_Am_A(v_A)_t=r_Bm_B(v_B)_t$
We plug in the known values to obtain:
$2(\frac{8}{32.2})\times 5=1\times (\frac{8}{32.2})\times (v_B)_t$
This simplifies to:
$(v_B)_t=10ft/s$
Now, $v_B=\sqrt{v_t^2+v_{B}^2}$
$\implies v_B=\sqrt{(4)^2+(10)^2}$
$\implies v_B=10.8ft/s$
Now we apply the principle of work and energy to calculate the required work done
$\frac{1}{2}m_Av_A^2+\Sigma U_{A\rightarrow B}=\frac{1}{2}m_Bv_B^2$
We plug in the known values to obtain:
$\frac{1}{2}(\frac{8}{32.2})(5)^2+\Sigma U_{A\rightarrow B}=\frac{1}{2}(\frac{8}{32.2})(10.8)^2$
This simplifies to:
$\Sigma U_{A\rightarrow B}=11.3ft\cdot lb$
