Answer
$F_{net}=877N$
Work Step by Step
We know that $F_{net}=\sqrt{f^2+N^2}$ where $f$ is the force of friction and $N$ is the normal reaction of the ground on the bicycle.
We also know that $N=mg$.
Putting the values of $m$ and $g$ into the formula $N=mg$ and solving:
$N=(85)(9.8)=833 N$
In addition, we had found from part (a) that $f=275 N$. Thus,
$F_{Net}=\sqrt{(275)^2+(833)^2}$
$F_{net}=877N$
