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--- 183_notes:examples:sliding_to_a_stop [2014/09/16 07:25] – created pwirving
+++ 183_notes:examples:sliding_to_a_stop [2014/09/22 04:19] – pwirving
@@ 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
+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 = $\langle 6, 0, 0\rangle m/s$
+=== 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 $
+Combining these two equations and writing $ p_x = mv_x $, we have
+$ \Delta(mv_x) = -mg\Delta t $
+$ \Delta(v_x) = - g\Delta t $
+$ \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 $