183_notes:examples:sledding

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183_notes:examples:sledding [2014/10/11 06:52] – 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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 $$\Delta U_{g} = W_{1} + W_{2}$$ $$\Delta U_{g} = W_{1} + W_{2}$$
 +
 +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 $f_{2}?$
  
-{{course_planning:course_notes:f1a.jpg?300|}}+{{course_planning:course_notes:f1a.jpg?200|}}
  
-$\sum{F_{x}} = f_{1} - mgsinθ = ma_{1} \longrightarrow$ don't need this because $f_{1}=μ_{k}N$+{{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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