183_notes:examples:sliding_to_a_stop [Projects & Practices in Physics]

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===== 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 ===

{{183_notes:friction_ground.jpg}}

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

=== 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) = \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 $