Answer
a) $\Delta PE=-1086J$
b) The vertical height of the skater decreases by $2.02m$, so the skater is below the starting point.
Work Step by Step
a) The net work done on the skater is composed of conservative and nonconservative forces. In other words, $$\sum W=W_c+W_{nc}$$
The work done by nonconservative forces are already given by the exercise: $W_{nc}=+80-265=-185J$
The conservative forces here are gravitational forces, so $W_c=mgh_0-mgh_f=PE_0-PE_f=-\Delta PE$
The net work itself, according to the work-energy theorem, is $\sum W=\frac{1}{2}m(v_f^2-v_0^2)$
We know $v_0=1.8m/s$, $v_f=6m/s$ and $m=55kg$, so we calculate $\sum W=901J$
Therefore, $$901J=-\Delta PE-185J$$ $$\Delta PE=-901J-185J=-1086J$$
b) We have $\Delta PE=mg(h_f-h_0)=-1086J$ $$h_f-h_0=\Delta h=\frac{-1086}{mg}=\frac{-1086}{55\times9.8}=-2.02m$$
The vertical height of the skater decreases by $2.02m$, so the skater is below the starting point.
