183_notes:examples:sliding_to_a_stop

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183_notes:examples:sliding_to_a_stop [2014/09/16 07:45] – pwirving183_notes:examples:sliding_to_a_stop [2018/02/03 23:24] (current) – [Example: Sliding to a Stop] hallstein
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 ===== Example: Sliding to a Stop ===== ===== 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?+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 ==== === 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 === === Lacking ===
 +
 +Time it takes for the block to come to a stop.
 +
 +The distance the block moves during this time.
  
 === Approximations & Assumptions === === Approximations & Assumptions ===
 +
 +Assume surface is made of the same material and so coefficient of friction is constant.
  
 === Representations === === Representations ===
 +
 +{{183_notes:friction_ground.jpg}}
 +
 +$\Delta \vec{p} = \vec{F}_{net} \Delta t$
  
 === Solution === === Solution ===
  
-$ x: \Delta p_x = -F_N\Delta t $+$ x: \Delta p_x = -\mu_k F_N\Delta t $
  
 $ y: \Delta p_y = (F_N - mg)\Delta t = 0 $ $ y: \Delta p_y = (F_N - mg)\Delta t = 0 $
  
-Combining these two equations and writing $ p_x = mv_x $, we have+Write equation of y direction in terms of $F_N$ to sub into x direction equation.
  
-$ \Delta(mv_x) = -mg\Delta t $+$ (F_N - mg) \Delta t = 0 $ 
  
-$ \Delta(v_x) = - g\Delta t $+Multiply out
  
-$ \Delta(t) = \dfrac{0 - v_{xi}}{-g} = \dfrac{v_{xi}}{g} $+$ 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 $ $ \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+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 $ $ \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 $ $ \Delta x = (3 m/s)(1.53 s) = 4.5m $
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