Answer
a) The velocity of the sled and the person is $+3.17m/s$
b) $\mu_k=0.017$
Work Step by Step
a) The person, whose mass $M=60kg$ and initial velocity $\vec{V}=+3.8m/s$, jumps on the sled, whose mass $m=12kg$ and initial velocity $\vec{v}=0$. After that, these two become a system, whose mass $M+m=72kg$ and which has velocity $V_f$
We assume the total linear momentum before and after the jump is conserved, so $$M\vec{V}+m\vec{v}=(M+m)\vec{V_f}$$ $$\vec{V_f}=\frac{M\vec{V}+0}{M+m}=+3.17m/s$$
b) Kinetic friction is the force affecting the motion of the person-sled system by opposing it. According to Newton's 2nd law, $$f_k=(M+m)a$$ $$\mu_kF_N=(M+m)a$$
Because there is no vertical movement, $F_N=(M+m)g$, so $$\mu_k(M+m)g=(M+m)a$$ $$\mu_k=\frac{|a|}{g}$$
We can find $a$ since we know $v_0=3.17m/s$, $v_f=0$ and the distance the sled has gone $s=30m$, so $$a=\frac{v_f^2-v_0^2}{2s}=-0.167m/s^2$$
Therefore, $$\mu_k=0.017$$
