Answer
The box will slide a distance of 4.1 meters.
Work Step by Step
The force of kinetic friction will oppose the motion of the box and bring it to a stop.
Let's calculate the magnitude of the acceleration.
$ma = F_f$
$ma = mg~\mu_k$
$a = g ~\mu_k$
$a = (9.8 ~m/s^2)(0.15)$
$a = 1.5~m/s^2$
When the box is sliding, the force of kinetic friction opposes the motion of the box. Therefore, $a = -1.5 ~m/s^2$
$v^2 = v_0^2 + 2ax$
$x = \frac{v^2-v_0^2}{2a}$
$x = \frac{0 - (3.5 ~m/s)^2}{(2)(-1.5 ~m/s^2)}$
$x = 4.1 ~m$
The box will slide a distance of 4.1 meters.
