Answer:
[tex]c=\frac{3}{2}\frac{k}{d}[/tex]
Explanation:
For the climbing rope, the force is given by,
[tex]F_c=-kx-cx^2[/tex]
Now, we know that for any conservative force,
[tex]F=-\frac{dV}{dx}[/tex] (one dimensional case)
where, V = potential energy
Therefore, the potential energy of the rope is given by,
[tex]V_c=\int {kx+cx^2} \, dx =\frac{kx^2}{2} +\frac{cx^3}{3} +a[/tex]
a is a constant, which can be taken to be zero depending on our choice of zero potential. Hence,
[tex]V_c=\frac{kx^2}{2} +\frac{cx^3}{3}[/tex]
Similarly, we can find the potential energy of an ideal spring,
[tex]V=\frac{1}{2}kx^2[/tex]
Now, it is given that at x = d,
[tex]V_c=2V[/tex]
or, [tex]\frac{kd^2}{2}+\frac{cd^3}{3}=kd^2[/tex]
or, [tex]c=\frac{3}{2}\frac{k}{d}[/tex]
