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--- 183_notes:examples:sliding_to_a_stop [2014/09/22 04:24] – pwirving
+++ 183_notes:examples:sliding_to_a_stop [2018/02/03 23:24] (current) – [Example: Sliding to a Stop] hallstein
@@ Line 2: / Line 2: @@
 ===== 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 ====
@@ Line 34: / Line 34: @@
 === 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 $
+Write equation of y direction in terms of $F_N$ to sub into x direction equation.
 $ (F_N - mg) \Delta t = 0 $
-$ F_N \Delta t - mg \Delta t = 0  \,\,\,\,\,\,\,Multiply\, out.$
+Multiply out
-$ F_N \Delta t = mg \Delta t  \,\,\,\,\,\,\,\,\,Make\, equal\, to\, each\, other.$
+$ F_N \Delta t - mg \Delta t = 0  $
-$ F_N = mg  \,\,\,\,\,\,\,\,\,\,\,\,Cancel\, \Delta t. $
+Make equal to each other
-Combining these two equations and substituting in mg for F_N and writing $ p_x = mv_x $, we get the following equation:
+$ F_N \Delta t = mg \Delta t  $
-$ \Delta(mv_x) = -mg\Delta t $
+Cancel $\Delta t$
-$ \Delta(v_x) = - g\Delta t $  Cancel the masses
+$ F_N = mg   $
-$ \Delta(t) = \dfrac{0 - v_{xi}}{-g} = \dfrac{v_{xi}}{g} $
+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, $v_{x,avg} = (v_{xi} + v_{xf})/2$, 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 $
+Sub in for $\Delta t$ and solve for $\Delta x$
 $ \Delta x = (3 m/s)(1.53 s) = 4.5m $