183_notes:examples:sledding

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183_notes:examples:sledding [2014/10/11 06:32] – pwirving183_notes:examples:sledding [2014/10/22 04:06] (current) – pwirving
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-===== Example: The Jumper =====+===== Example: Sledding =====
  
 A little girl is riding her sled on a hill. If she starts a distance d up the hill, which makes an angle θ with the horizontal, how far will she travel along the flat snowy ground? A little girl is riding her sled on a hill. If she starts a distance d up the hill, which makes an angle θ with the horizontal, how far will she travel along the flat snowy ground?
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 === Solution === === Solution ===
 +
 +We could solve this using forces of kinematics; but, let's apply the energy principle because we can avoid vector quantities in the calculation.
 +
 +First we must decide the system and surroundings.
 +
 +System: Sled+Kid+Earth
 +Surroundings: Snow
 +
 +Starting with the principle that change in energy in the system is equal to the work done by the surroundings.
  
 ΔEsystem=Wsurroundings ΔEsystem=Wsurroundings
 +
 +The change in energy can be in the form of change of kinetic and change in gravitational potential energy.
  
 ΔK+ΔUg=Wfriction ΔK+ΔUg=Wfriction
  
-no change ΔK=0+No change ΔK=0 as its initial and final state of the sled is at rest.
  
 ΔUg=Wfriction⟶Wfriction? ΔUg=Wfriction⟶Wfriction?
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 Here, we pause because we have two different regions to consider. Here, we pause because we have two different regions to consider.
  
-{{course_planning:course_notes:region1a.jpg?600|}}+{{course_planning:course_notes:region1c.jpg?300|}}
  
-{{course_planning:course_notes:region2b.jpg?600|}}+{{course_planning:course_notes:region2b.jpg?300|}}
  
 The frictional force is different in the two regions so we must consider the work they do separately. The frictional force is different in the two regions so we must consider the work they do separately.
  
 ΔUg=W1+W2 ΔUg=W1+W2
 +
 +Breaking work down into force by change in distance.
  
 ΔUg=→f1⋅Δ→r1+→f2⋅Δ→r2 ΔUg=→f1⋅Δ→r1+→f2⋅Δ→r2
  
-→r2 is what we care about. (position change along flat part)+→r2 is what we are trying to solve for as this is the position change along flat part.
  
 What's f1 and f2? What's f1 and f2?
  
-$\sum{F_{x}} = f_{1} - mgsinθ = ma_{1} \longrightarrowdon′tneedthisbecausef_{1}=μ_{k}N$+{{course_planning:course_notes:f1a.jpg?200|}} 
 + 
 +{{course_planning:projects:f2b.jpg?200|}} 
 + 
 +Need to find $f_{1}&f_{2}$ 
 + 
 +To find F1 we can say that the sum of the forces in the x direction are equal to $ma_{1}$ But we don't need this because we know that f1=μkN. 
 + 
 +∑Fx=f1−mgsinθ=ma1  
 + 
 +The sum of the forces in the y direction we do need because this allows us to express N.
  
 ∑Fy=N−mgcosθ=0 ∑Fy=N−mgcosθ=0
  
 mgcosθ=N mgcosθ=N
 +
 +If f1=μkN then:
  
 f1=μkmgcosθ f1=μkmgcosθ
 +
 +To find f2 we must do the same thing and add all the forces in the x and y directions. Again because not using kinematics we don't need accelerations and instead want an equation that expresses f2.
  
 ∑Fx=f2=ma2⟶f2=μkN=μkmg ∑Fx=f2=ma2⟶f2=μkN=μkmg
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 ∑Fy=N−mg=0 ∑Fy=N−mg=0
  
-Again because not using kinematics we don't need accelerations.+We substitute in for f1, f2 and d the distance down the slope into the previous equation for gravitational potential energy with minuses on the $\vec{f}'sastheyareinoppositionofthe\vec{r}'s$.
  
 ΔUg=→f1⋅Δ→r1+→f2⋅Δ→r2 ΔUg=→f1⋅Δ→r1+→f2⋅Δ→r2
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 ΔUg=−(μkmgcosθ)d−(μkmg)x ΔUg=−(μkmgcosθ)d−(μkmg)x
 +
 +Substitute in the equation for gravitational potential energy for ΔUg
  
 +mg(yf−yi)=−μkmgdcosθ−μkmgx +mg(yf−yi)=−μkmgdcosθ−μkmgx
 +
 +Rearrange to get the following expression.
  
 yf−yi=−μk(dcosθ+x) yf−yi=−μk(dcosθ+x)
  
-What is yf−yi in terms of what we know?+What is yf−yi in terms of what we know? Eventually we want to express x in terms of variables we know. 
 + 
 +{{course_planning:course_notes:final_sledding.jpg?200|}} 
 + 
 +From the diagram of the incline we get:
  
 yf−yi=−dsinθ yf−yi=−dsinθ
 +
 +Substitue −dsinθ for yf−yi and then rearrange to express x in terms of known variables.
  
 −dsinθ=−μk(dcosθ+x) −dsinθ=−μk(dcosθ+x)
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 x=d(sinθ−μkcosθμk) x=d(sinθ−μkcosθμk)
 +
 +A check of the units reveals that:
  
 [x]=m [x]=m
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 [d]=m [d]=m
  
-All other quantities are unitless. +Which makes sense as all the other quantities are unit less.
  
  
 +E=γmc2
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