Answer
$l=2.77ft$
Work Step by Step
The required length can be determined as follows:
First, we find the velocity:
$v_B^2=\frac{\rho Wcos45}{m}=\rho gcos45=(1.5)(32.2)cos45=34.15(m/s)^2$
Now, according to the conservation of energy equation
$\frac{1}{2}mv_A^2+\frac{1}{2}kx_A^2=\frac{1}{2}mv_B^2+\frac{1}{2}kx_B^2+Wrsin45$
We plug in the known values to obtain:
$0+\frac{1}{2}(2)(1.5\pi-l)^2=\frac{1}{2}(\frac{2}{32.2})(34.15)+(2)(1.5sin45)+\frac{1}{2}(2)(1.5\times 0.75\pi -l)^2$
This simplifies to:
$l=2.77ft$
