A box is pushed so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is 0.15, and the push imparts an initial speed of 3.5 m/s?

When an object moves over a rough surface, it starts to decelerate due to the friction force, and after some time, it comes to rest.

Given data:

The initial speed of the box is $v_{\mathrm{i}}=3.5\frac{\mathrm{m}}{\mathrm{s}}$.

The final speed of the box is $v_{\mathrm{f}}=0$as at the end it stops.

The coefficient of kinetic friction is ${\mu }_{\mathrm{k}}=0.15$.

Assumption:

Let the mass of the box be m.

Let the box slide a distance of $\Delta x$before coming to rest.

Let a be the deceleration of the box due to the friction force acting on the box.

The normal force acting on the drawer is $N=mg$.

The kinetic friction force on the box is

$\begin{array}{c}{f}_{\mathrm{k}}={\mu }_{\mathrm{k}}N\\ ={\mu }_{\mathrm{k}}mg\end{array}$

Then, to find the magnitude of the deceleration of the box,

$\begin{array}{c}ma=f_{\mathrm{k}}\\ ma={\mu }_{\mathrm{k}}mg\\ a={\mu }_{\mathrm{k}}g\end{array}$

Now, to find the value of $\Delta x$,

$\begin{array}{c}{v}_{\mathrm{f}}^2=v_{\mathrm{i}}^2-2a\Delta x\\ v_{\mathrm{f}}^2=v_{\mathrm{i}}^2-2{\mu }_{\mathrm{k}}g\Delta x\\ 0^2={\left(3.5\frac{\mathrm{m}}{\mathrm{s}}\right)}^2-\left(2\times 0.15\times 9.80\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)\Delta x\\ \Delta x=4.17\mathrm{m}\end{array}$

Hence, the box slides 4.17 m across the floor.