Answer
The friction exerts a force of 46.7 N on the skater.
Work Step by Step
We can find the rate of deceleration.
$a = \frac{v-v_0}{t} = \frac{0-2.40~m/s}{3.52~s}$
$a = -0.682~m/s^2$
The magnitude of acceleration is $a = 0.682~m/s^2$.
$F_f = ma = (68.5~kg)(0.682~m/s^2)$
$F_f = 46.7~N$
The friction exerts a force of 46.7 N on the skater.
