Answer
$f_k=821.73N$
Work Step by Step
1) Find the fireman's acceleration
Initial speed $v_0=0$, final speed $v=1.4m/s$ and distance $y=4m$
$$v^2=v_0^2+2ay$$ $$a=\frac{v^2}{2y}=0.245m/s^2$$
2) The fireman's weight is opposed by kinetic friction $f_k$. From Newton's 2nd Law of Motion, $$mg-f_k=ma$$ $$f_k=mg-ma=m(g-a)$$ $$f_k=86(9.8-0.245)=821.73N$$
