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.
  
 $$\Delta E_{system} = W_{surroundings}$$ $$\Delta E_{system} = W_{surroundings}$$
 +
 +The change in energy can be in the form of change of kinetic and change in gravitational potential energy.
  
 $$\Delta K + \Delta U_{g} = W_{friction}$$ $$\Delta K + \Delta U_{g} = W_{friction}$$
  
-no change $$ \Delta K = 0$$+No change $$ \Delta K = 0$$ as its initial and final state of the sled is at rest.
  
 $$\Delta U_{g} = W_{friction} \longrightarrow W_{friction}?$$ $$\Delta U_{g} = W_{friction} \longrightarrow W_{friction}?$$
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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.
  
 $$\Delta U_{g} = W_{1} + W_{2}$$ $$\Delta U_{g} = W_{1} + W_{2}$$
 +
 +Breaking work down into force by change in distance.
  
 $$\Delta U_{g} = \vec{f}_{1}\cdot\Delta \vec{r}_{1} + \vec{f}_{2}\cdot\Delta \vec{r}_{2}$$ $$\Delta U_{g} = \vec{f}_{1}\cdot\Delta \vec{r}_{1} + \vec{f}_{2}\cdot\Delta \vec{r}_{2}$$
  
-$\vec{r}_{2}$ is what we care about. (position change along flat part)+$\vec{r}_{2}$ is what we are trying to solve for as this is the position change along flat part.
  
 What's $f_{1}$ and $f_{2}?$ What's $f_{1}$ and $f_{2}?$
  
-$\sum{F_{x}} = f_{1} - mgsinθ = ma_{1} \longrightarrow$ don't need this because $f_{1}=μ_{k}N$+{{course_planning:course_notes:f1a.jpg?200|}} 
 + 
 +{{course_planning:projects:f2b.jpg?200|}} 
 + 
 +Need to find $f_{1}$ & $f_{2}$ 
 + 
 +To find $F_{1}$ 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 $f_{1}=μ_{k}N$. 
 + 
 +$\sum{F_{x}} = f_{1} - mgsinθ = ma_{1}$  
 + 
 +The sum of the forces in the y direction we do need because this allows us to express N.
  
 $$\sum{F_{y}} = N - mgcosθ = 0$$ $$\sum{F_{y}} = N - mgcosθ = 0$$
  
 $$mgcosθ = N$$ $$mgcosθ = N$$
 +
 +If $f_{1}=μ_{k}N$ then:
  
 $$f_{1} = μ_{k}mgcosθ$$ $$f_{1} = μ_{k}mgcosθ$$
 +
 +To find $f_{2}$ 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 $f_{2}$.
  
 $$\sum{F_{x}} = f_{2} = ma_{2} \longrightarrow f_{2} = μ_{k}N = μ_{k}mg$$ $$\sum{F_{x}} = f_{2} = ma_{2} \longrightarrow f_{2} = μ_{k}N = μ_{k}mg$$
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 $$\sum{F_{y}} = N-mg = 0$$ $$\sum{F_{y}} = N-mg = 0$$
  
-Again because not using kinematics we don't need accelerations.+We substitute in for $f_{1}$, $f_{2}$ and d the distance down the slope into the previous equation for gravitational potential energy with minuses on the $\vec{f}'s$ as they are in opposition of the $\vec{r}'s$.
  
 $$\Delta U_{g} = \vec{f}_{1}\cdot\Delta \vec{r}_{1} + \vec{f}_{2}\cdot\Delta \vec{r}_{2}$$ $$\Delta U_{g} = \vec{f}_{1}\cdot\Delta \vec{r}_{1} + \vec{f}_{2}\cdot\Delta \vec{r}_{2}$$
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 $$\Delta U_{g} = -(μ_{k}mgcosθ)d - (μ_{k}mg)x$$ $$\Delta U_{g} = -(μ_{k}mgcosθ)d - (μ_{k}mg)x$$
 +
 +Substitute in the equation for gravitational potential energy for $\Delta U_{g}$
  
 $$+mg(y_f - y_i) = -μ_{k}mgdcosθ - μ_{k}mgx$$ $$+mg(y_f - y_i) = -μ_{k}mgdcosθ - μ_{k}mgx$$
 +
 +Rearrange to get the following expression.
  
 $$y_f - y_i = -μ_{k}(dcosθ + x)$$ $$y_f - y_i = -μ_{k}(dcosθ + x)$$
  
-What is $y_f-y_i$ in terms of what we know?+What is $y_f-y_i$ 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:
  
 $$y_f-y_i =  -dsinθ$$ $$y_f-y_i =  -dsinθ$$
 +
 +Substitue $-dsinθ$ for $y_f-y_i$ and then rearrange to express x in terms of known variables.
  
 $$-dsinθ = -μ_{k}(dcosθ + x)$$ $$-dsinθ = -μ_{k}(dcosθ + x)$$
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 $$x = d (\dfrac{sinθ-μ_{k}cosθ}{μ_{k}})$$ $$x = d (\dfrac{sinθ-μ_{k}cosθ}{μ_{k}})$$
 +
 +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.
  
  
 +$E = γmc^2$
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