Answer :
Given data,
Initial speed of the box,
[tex]v=4.80\text{ m/s}[/tex]Length of the floor,
[tex]d=4.70\text{ m}[/tex]Coefficient of friction,
[tex]\mu_k=0.100[/tex]Calculate the acceleration,
[tex]\begin{gathered} a=\text{coefficient of friction}\times g \\ a=0.100\times9.8 \\ a=0.98m/s^2 \end{gathered}[/tex]Consider the kinematic equation of motion.
[tex]\begin{gathered} v^2=u^2-2ad \\ v^2=(4.80\text{ m/s})^2-2\times0.98m/s^2\times4.70\text{ m} \\ v^2=23.04-9.212^{} \\ v=3.717^{}\text{ m/s} \end{gathered}[/tex]Therefore, the final speed of the box is
[tex]v=3.717\text{ m/s}[/tex]