183_notes:examples:sliding_to_a_stop

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183_notes:examples:sliding_to_a_stop [2014/09/16 07:38] – pwirving183_notes:examples:sliding_to_a_stop [2018/02/03 23:24] (current) – [Example: Sliding to a Stop] hallstein
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 ===== Example: Sliding to a Stop ===== ===== 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 ⟨6,0,0⟩m/s. How long will it take for the block to come to a stop? How far does the block move?+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 ⟨6,0,0⟩m/s. How long will it take for the block to come to a stop? How far does the block move? 
  
 === Facts ==== === Facts ====
 +
 +Block is metal.
 +
 +Mass of metal block = 3 kg
 +
 +The coefficient of friction between floor and block = 0.4
 +
 +Initial velocity of block = ⟨6,0,0⟩m/s
 +
 +Final velocity of block = ⟨0,0,0⟩m/s
  
 === Lacking === === Lacking ===
 +
 +Time it takes for the block to come to a stop.
 +
 +The distance the block moves during this time.
  
 === Approximations & Assumptions === === Approximations & Assumptions ===
 +
 +Assume surface is made of the same material and so coefficient of friction is constant.
  
 === Representations === === Representations ===
 +
 +{{183_notes:friction_ground.jpg}}
 +
 +Δ→p=→FnetΔt
  
 === Solution === === Solution ===
  
-x:Δpx=−FNΔt+$ x: \Delta p_x = -\mu_k F_N\Delta t $
  
 y:Δpy=(FN−mg)Δt=0 y:Δpy=(FN−mg)Δt=0
  
-Combining these two equations and writing px=mvx, we have+Write equation of y direction in terms of FN to sub into x direction equation. 
 + 
 +(FN−mg)Δt=0  
 + 
 +Multiply out 
 + 
 +FNΔt−mgΔt=0 
 + 
 +Make equal to each other 
 + 
 +FNΔt=mgΔt 
 + 
 +Cancel Δt 
 + 
 +FN=mg 
 + 
 +Combining these two equations and substituting in mg for FN and writing $ p_x = \Delta(mv_x) $, we get the following equation: 
 + 
 +Δ(mvx)=−μkmgΔt 
 + 
 +Cancel the masses 
 + 
 +Δ(vx)=−μkgΔt   
 + 
 +Rearrange to solve for Δt and sub in 0 - vxi for Δ(vx) 
 + 
 +Δ(t)=0−vxi−μkg=vxiμkg 
 + 
 +Fill in values for variables and solve for Δt 
 + 
 +Δ(t)=6m/s0.4(9.8N/kg)=1.53s 
 + 
 +Since the net force was constant we can say the average velocity can be described as: vx,avg=(vxi+vxf)/2, so
  
-$ \Delta(mv_x) = -mg\Delta t $+$ \Delta x/\Delta t = ((6 + 0)/2) m/s = 3m/s $
  
-$ \Delta(v_x) = - g\Delta t $+Sub in for $\Delta tandsolvefor\Delta x$
  
-$ \Delta(t) = \dfrac{0 - v_{xi}}{-g} = \dfrac{v_{xi}}{g} $+$ \Delta x = (3 m/s)(1.53 s) = 4.5m $
  • 183_notes/examples/sliding_to_a_stop.1410853123.txt.gz
  • Last modified: 2014/09/16 07:38
  • by pwirving