Answer
$l_{\circ}=2.77ft$
Work Step by Step
The required length of the cord can be determined as follows:
$\Sigma U_{1\rightarrow 2}=\frac{1}{2}k(x_1^2-x_2^2)-W(1.5sin45)$
We plug in the known values to obtain:
$\implies \Sigma U_{1\rightarrow 2}=\frac{1}{2}[(1.5\pi -l_{\circ})^2-(\frac{3\pi}{4}(1.5)-l_{\circ})^2]-(2\times 1.5sin45)$
This simplifies to:
$\implies \Sigma U_{1\rightarrow 2}=7.5887-2.36l_{\circ}$
Now, according to the principle of work and energy
$\frac{1}{2}mv_1^2+\Sigma U_{1\rightarrow 2}=\frac{1}{2}mv_2^2$
We plug in the known values to obtain:
$0+7.5887-2.36l_{\circ}=\frac{1}{2}\frac{2}{32.2}(5.84)^2$
This simplifies to:
$l_{\circ}=2.77ft$
