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--- 183_notes:examples:sliding_to_a_stop [2014/09/22 04:35] – 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 $
  $y: \Delta p_y = (F_N - mg)\Delta t = 0$
@@ Line 56: / Line 56: @@
 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(mv_x) = - \mu_k mg\Delta t $
 Cancel the masses
- $\Delta(v_x) = - g\Delta t$
+$ \Delta(v_x) = - \mu_k g\Delta t $
 Rearrange to solve for  $\Delta t$  and sub in 0 -  $v_{xi}$  for  $\Delta(v_x)$
- $\Delta(t) = \dfrac{0 - v_{xi}}{-g} = \dfrac{v_{xi}}{g}$
+$ \Delta(t) = \dfrac{0 - v_{xi}}{-\mu_k g} = \dfrac{v_{xi}}{\mu_k g} $
 Fill in values for variables and solve for  $\Delta t$
@@ Line 70: / Line 70: @@
  $\Delta(t) = \dfrac{6 m/s}{0.4 (9.8 N/kg)} = 1.53s$
-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  $\Delta x$
  $\Delta x = (3 m/s)(1.53 s) = 4.5m$