Answer
$v_f=8.64m/s$
Work Step by Step
According to the work-energy theorem, $$W_{nc}=E_f-E_0$$ $$\Big(\frac{1}{2}mv_f^2+mgh_f\Big)-\Big(\frac{1}{2}mv_0^2+mgh_0\Big)=W_{f}$$ $$\frac{1}{2}m(v_f^2-v_0^2)+mg(h_f-h_0)=W_{f}$$
First, we notice that the student's initial speed $v_0=0$. Also, since we take the initial point to be when the student starts to slide down, $h_0=0$. From these, we can simplify the equation $$\frac{1}{2}mv_f^2+mgh_f=W_f$$ $$v_f=\sqrt{\frac{2(W_f-mgh_f)}{m}}$$
We have $m=83kg$, $h_f=-11.8m$ (since the student slides down) and $W_f=-6.5\times10^3J$. Therefore, $$v_f=8.64m/s$$
