Differences

This shows you the differences between two versions of the page.

--- 183_notes:examples:sliding_to_a_stop [2014/09/16 07:29] – pwirving
+++ 183_notes:examples:sliding_to_a_stop [2014/09/22 04:24] – pwirving
@@ Line 5: / Line 5: @@
 === 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 = -F_N\Delta t $
+$ y: \Delta p_y = (F_N - mg)\Delta t = 0 $
+$ (F_N - mg) \Delta t = 0 $
+$ F_N \Delta t - mg \Delta t = 0  \,\,\,\,\,\,\,Multiply\, out.$
+$ F_N \Delta t = mg \Delta t  \,\,\,\,\,\,\,\,\,Make\, equal\, to\, each\, other.$
+$ F_N = mg  \,\,\,\,\,\,\,\,\,\,\,\,Cancel\, \Delta t. $
+Combining these two equations and substituting in mg for F_N and writing $ p_x = mv_x $, we get the following equation:
+$ \Delta(mv_x) = -mg\Delta t $
+$ \Delta(v_x) = - g\Delta t $  Cancel the masses
+$ \Delta(t) = \dfrac{0 - v_{xi}}{-g} = \dfrac{v_{xi}}{g} $
+$ \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
+$ \Delta x/\Delta t = ((6 + 0)/2) m/s = 3m/s $
+$ \Delta x = (3 m/s)(1.53 s) = 4.5m $