Answer
$f_k=1.9 \times 10^2 N$
Work Step by Step
Force of Kinetic friction is given as
$f_k=\mu_kN$
where $N$ is the normal reaction and $N=mg$
Therefore, $f_k=\mu_kmg$
Putting the values of $\mu_k$,$m$ and $g$ into the formula and solving:
$f_k=(0.35)(55)(9.8)$
$f_k=188.65 N$
$f_k=1.89\times 10^2 N$
After rounding off,
$f_k\approx1.9 \times 10^2 N$
