Stopping force on wire rope from a falling object attached to it

Assuming the following properties for the rope:

• $L$: undeformed length of the rope
• $E$: young's modulus of the rope
• $A$: cross-section
• weight of the rope is zero (it greatly simplifies the calculation)

Then defining the x-axis as the vertical axis and

• setting 0 the location that the wire/rope is completely free,
• setting downward as positive axis

then at a distance x (again x is positive below 0), the tension on the rope will be equal to:

$$T(x) = frac{E A}{L} x$$

At that location on the object there are only two forces:

• tension on the rope T(x)
• Gravity $m*g$

Therefore the equation is: $$sum F_x = mcdot a_x$$ $$mg - T(x) = mcdot a_x$$ with some rearranging: $$ mcdot a_x + T(x) = mg $$

Because $a_x = ddot{x}$ $$ mcdot ddot{x} + T(x) = mg $$

$$ mcdot ddot{x} + frac{E A}{L} x = mg $$

You can substitute $frac{E A}{L}$ with $k$ and you get the very basic problem of the undamped harmonic oscillator (there are some caveats -e.g. there is only force when x is positive- but I won't go into that):

$$ mcdot ddot{x} + k x = mg $$

This equation along with the boundary conditions

• $x(t)=0$
• $dot{x} (t)=v_0$ : the initial velocity when the rope is at tension for the first time.

is a very basic Initial Value Problem (IVP). You can easily calculate the function for $x(t)$

$$x{left(t right)} = frac{g m}{k} + left(- frac{g m}{2 k} - frac{v_{0}}{2 sqrt{- frac{k}{m}}}right) e^{- t sqrt{- frac{k}{m}}} + left(- frac{g m}{2 k} + frac{v_{0}}{2 sqrt{- frac{k}{m}}}right) e^{t sqrt{- frac{k}{m}}}$$

and with that you can calculate the $T(x)$

Let's say you have a wire of steel with the cross-sectional area $A$ and length $L$ (for instance, say 20 km). Then, the strain energy of the stretched wire is equal to the kinetic energy of the falling mass once it has reached a depth beyong $L$ from the sky hook. Therefore, $frac{1}{2}mv^2+mgh=frac{1}{2}A E_{steel}epsilon_{steel}^2 $ since the work done by strain is a triangular area. We can then calculate for $epsilon = epsilon_{steel}$, $Delta x=epsilon cdot 20km.$ And as has been mentioned in the comment by @Jonathon R swift, the mass will bounce back up.