A rope is specially designed and its modulus of elasticity is known from specifications. For the purpose of this problem, assume that the rope is stretched to twice its natural length by a person of 75kg hanging at rest from the free end. It is necessary to adjust the natural length of the rope in terms of the weight of the jumper.
For a person of mass $m$ kg, show that the depth ($d$) to which the would fall if a attached to the type of rope described above, with length ($l$) is given by the model:
$$ d = \frac{2ml \pm l\sqrt{4m^2+600m}}{150} + l $$
I tried converting the equation above into its quadratic form, but once I did that I didn't really know where that could lead. I don't know how to progress any further from here. Any help would great, thanks.
When the rope is pulled beyond its natural length, the force it exerts (by Hooke's law) is given by $F=kx$, where $x$ is the distance beyond the natural length that the rope is being stretched and $k$ is a constant particular to the rope.
A person of mass $m$ will hang stationary on the rope if the force from the rope pulling them up, $kx$, is equal to the force of gravity pulling them down, $mg$. So for a person to hang on this rope, we need: $$\boxed{kx=mg}$$ This is the key equation needed for all steps of the solution.
Use the following statement to wrote the constant $k$ in terms of $l$:
Then you can solve for $x$ in terms of $m$ and $l$ (assuming you know the value of $g$). Finally, add $l$ to get the expected $d$.
Please let me know if you have any questions.