Example: Sliding to a Stop

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?

Facts

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 = $\langle 0,0,0\rangle m/s$

Lacking

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

The distance the block moves during this time.

Approximations & Assumptions

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

Representations

$\Delta\overset{\rightarrow}{p} = {\overset{\rightarrow}{F}}_{net}\Delta t$

Solution

$x:\Delta p_{x} = - \mu_{k}F_{N}\Delta t$

$y:\Delta p_{y} = (F_{N} - mg)\Delta t = 0$

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 $\Delta(v_{x})$

$\Delta(t) = \frac{0 - v_{xi}}{- \mu_{k}g} = \frac{v_{xi}}{\mu_{k}g}$

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

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

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 $\Delta x$

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