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--- 183_notes:examples:sliding_to_a_stop [2014/09/22 04:24] – pwirving
+++ 183_notes:examples:sliding_to_a_stop [2014/09/22 04:39] – pwirving
@@ Line 37: / Line 37: @@
 $ 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  $
+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) = -mg\Delta t $
-$ \Delta(v_x) = - g\Delta t $  Cancel the masses
+Cancel the masses
+$ \Delta(v_x) = - 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}}{-g} = \dfrac{v_{xi}}{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 $