You take a 3 kg metal block and slide it along the floor, where the coefficient of friction is only 0.4. You release the block with an initial velocity of $\langle 6, 0, 0\rangle m/s$. How long will it take for the block to come to a stop? How far does the block move?

Block is metal.

Mass of metal block = 3 kg

The coefficient of friction between floor and block = 0.4

Initial velocity of block = $\langle 6, 0, 0\rangle m/s$

Final velocity of block = $\langle 0, 0, 0\rangle m/s$

Time it takes for the block to come to a stop.

The distance the block moves during this time.

Assume surface is made of the same material and so coefficient of friction is constant.

$\Delta \vec{p} = \vec{F}_{net} \Delta t$

$x: \Delta p_x = -\mu_k F_N\Delta t$

$y: \Delta p_y = (F_N - mg)\Delta t = 0$

Write equation of y direction in terms of $F_N$ to sub into x direction equation.

$(F_N - mg) \Delta t = 0$

Multiply out

$F_N \Delta t - mg \Delta t = 0$

Make equal to each other

$F_N \Delta t = mg \Delta t$

Cancel $\Delta t$

$F_N = mg$

Combining these two equations and substituting in mg for $F_N$ and writing $p_x = \Delta(mv_x)$, we get the following equation:

$\Delta(mv_x) = - \mu_k mg\Delta t$

Cancel the masses

$\Delta(v_x) = - \mu_k g\Delta t$

Rearrange to solve for $\Delta t$ and sub in 0 - $v_{xi}$ for $\Delta(v_x)$

$\Delta(t) = \dfrac{0 - v_{xi}}{-\mu_k g} = \dfrac{v_{xi}}{\mu_k g}$

Fill in values for variables and solve for $\Delta t$

$\Delta(t) = \dfrac{6 m/s}{0.4 (9.8 N/kg)} = 1.53s$

Since the net force was constant we can say the average velocity can be described as: $v_{x,avg} = (v_{xi} + v_{xf})/2$, so

$\Delta x/\Delta t = ((6 + 0)/2) m/s = 3m/s$

Sub in for $\Delta t$ and solve for $\Delta x$

$\Delta x = (3 m/s)(1.53 s) = 4.5m$