Answer
$62.63\dfrac {m}{s}$
Work Step by Step
Lets calculate magnitude of work done by the friction force:
The friction force is:
$F_{f}=\mu mg\cos \theta $
Also,
$L\sin \theta =H\Rightarrow \sin \theta =\dfrac {H}{L}=\dfrac {300}{500}=0.6\Rightarrow \theta =37.$
So the magnitude of the work done will be:
$W_{f}=F_{f}\times L=\mu {m}g\cos \theta \times L=0.25\times 520\times \cos 37\times 500\times 9.8\approx 5.1\times 10^{5}J$
Initial potential energy of the rock is:
$U_{top}=U_{buttom}+mg\Delta h=0+520\times 9.8\times 300\approx 1.53\times 10^{6}J$
So the kinetic energy at bottom will be:
$E_{k}=U_{top}-W_{f}\approx 1.02\times 10^{6}J$
So the speed will be:
$\dfrac {mv^{2}}{2}=E_{k}\Rightarrow v=\sqrt {\dfrac {2E_{k}}{m}}=\sqrt {\dfrac {2\times 1.02\times 10^{6}}{520}}\approx 62.63\dfrac {m}{s}$
