Answer
$2283.5\space J$
This rope has a lower value for stored potential energy than the ideal spring with the same spring constant.
Work Step by Step
Given that the $F= -Kx+bx^{3}$ where, F - Force, K - 244 N/m, $b = 3.24\space N/m^{3}$, x - Stretch of the rope.
This rope isn't a simple $F=-kx$ spring for which we already have a potential energy formula. Because the rope force varies with stretch, we'll have to integrate. We can use the following equation for stored potential energy.
$\Delta U=-\int_{x1}^{x2}F(x)\space dx$ ; Let's plug known values into this equation.
$U=-\int_0^x(-kx+bx^{3})dx=\frac{1}{2}kx^{2}-\frac{b}{4}x^{4}\Biggr |_{0}^{x} $
$U=\frac{1}{2}kx^{2}-\frac{1}{4}bx^{4}$
$U=\frac{1}{2}\times244\space N/m\times (4.68m)^{2}-\frac{3.24\space N/m^{3}}{4}(4.68m)^{4}$
$U=2672.1\space J -388.6\space J= 2283.5\space J$
If the rope was an ideal spring, the stored potential energy is $(U_{1})=\frac{1}{2}kx^{2}$
Let's plug known values into this equation.
$U_{1}= \frac{1}{2}\times 244\space N/m\times (4.68m)^{2}= 2672.1\space J$
$U\lt U_{1}$
This rope has a lower value for stored potential energy than the ideal spring with the same spring constant.
