Differences

This shows you the differences between two versions of the page.

--- 183_notes:examples:sledding [2014/10/11 07:03] – pwirving
+++ 183_notes:examples:sledding [2014/10/22 04:06] (current) – pwirving
@@ Line 1: / Line 1: @@
-===== 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?
@@ Line 40: / Line 40: @@
 === 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.
  $\Delta E_{system} = W_{surroundings}$
+The change in energy can be in the form of change of kinetic and change in gravitational potential energy.
  $\Delta K + \Delta U_{g} = W_{friction}$
-no change  $\Delta K = 0$
+No change  $\Delta K = 0$  as its initial and final state of the sled is at rest.
  $\Delta U_{g} = W_{friction} \longrightarrow W_{friction}?$
@@ Line 58: / Line 69: @@
  $\Delta U_{g} = W_{1} + W_{2}$
+Breaking work down into force by change in distance.
  $\Delta U_{g} = \vec{f}_{1}\cdot\Delta \vec{r}_{1} + \vec{f}_{2}\cdot\Delta \vec{r}_{2}$
- $\vec{r}_{2}$  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  $f_{2}?$
@@ Line 69: / Line 82: @@
 {{course_planning:projects:f2b.jpg?200|}}
-$\sum{F_{x}} = f_{1} - mgsinθ = ma_{1} \longrightarrow $don't need this because$ f_{1}=μ_{k}N$
+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.
  $\sum{F_{y}} = N - mgcosθ = 0$
  $mgcosθ = N$
+If  $f_{1}=μ_{k}N$  then:
  $f_{1} = μ_{k}mgcosθ$
+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  $f_{2}$ .
  $\sum{F_{x}} = f_{2} = ma_{2} \longrightarrow f_{2} = μ_{k}N = μ_{k}mg$
@@ Line 81: / Line 104: @@
  $\sum{F_{y}} = N-mg = 0$
-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$.
  $\Delta U_{g} = \vec{f}_{1}\cdot\Delta \vec{r}_{1} + \vec{f}_{2}\cdot\Delta \vec{r}_{2}$
@@ Line 88: / Line 111: @@
  $\Delta U_{g} = -(μ_{k}mgcosθ)d - (μ_{k}mg)x$
+Substitute in the equation for gravitational potential energy for  $\Delta U_{g}$
  $+mg(y_f - y_i) = -μ_{k}mgdcosθ - μ_{k}mgx$
+Rearrange to get the following expression.
  $y_f - y_i = -μ_{k}(dcosθ + x)$
-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:
  $y_f-y_i =  -dsinθ$
+Substitue  $-dsinθ$  for  $y_f-y_i$  and then rearrange to express x in terms of known variables.
  $-dsinθ = -μ_{k}(dcosθ + x)$
@@ Line 106: / Line 137: @@
  $x = d (\dfrac{sinθ-μ_{k}cosθ}{μ_{k}})$
+A check of the units reveals that:
 [x]=m
@@ Line 111: / Line 144: @@
 [d]=m
-All other quantities are unitless.
+Which makes sense as all the other quantities are unit less.
+ $E = γmc^2$