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?

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

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

Combining these two equations and writing $p_x = mv_x$, we have

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

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

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

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

Since the net force was constant, $v_{x,avg} = (v_{xi} + v_{xf})/2$, so

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

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