A man (weighing $86\space Kg$) is climbing a mountain when he suddenly falls. His security rope works like a spring with spring constant $1.2 > \times \space 10^3 N/m$ and after a fall of $0.75 \space m$ the rope start to stretch. How much does the rope extend?
I don't have the solution, so I post here my answer.
The work done by the weight force is: $W = Fs = mgs$ where $m$ is the mass of the man, $g$ is the gravity acceleration and $s$ is the displacement (I don't know the exact term - $0.75 \space m$ anyway).
When the rope stops the fall the energy transmits on it and becomes elastic potential energy:
$U_e = \frac{1}{2}kx^2$ where $k$ is the spring constant and $x$ the extension of the spring (the rope in this case).
So we can equal them:
$$\begin{align*}
W &= U_e\\
mgs &= \frac{1}{2}kx^2\\
\frac{2mgs}{k} &= x^2\\
\sqrt{\frac{2mgs}{k}} &= x
\end{align*}$$
and plugging in the numbers it becomes:
$x = \sqrt{\frac{2 \times 86 \times 9.8 \times 0.75}{1.2 \times 10^3}} = 1.03 \space m$
Is my answer correct?
Thank you, rubik
P.S. Sorry for my English, I'm not a native speaker nor I study physics in this language!
