===== 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?

=== Facts ===

Child on incline of θ.

The total mass of the sled and child = m.

There's a small bit of friction between the rails of the sled and the snow = (μ_k).

Slope length = L

Initial state: at rest, at height above horizontal

Final state: at rest on horizontal

=== Lacking ===

How far will she travel along the flat?

=== Approximations & Assumptions ===

Coefficient for kinetic friction for flat + incline is the same.

No wind resistance.

=== Representations ===

System: Sled + Kid + Earth

Surroundings: Snow

$$\Delta E_{system} = W_{surroundings}$$

$$\Delta K + \Delta U_{g} = W_{friction}$$

{{course_planning:course_notes:sledding_upload.jpg|}}

=== 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 $$ \Delta K = 0$$ as its initial and final state of the sled is at rest.

$$\Delta U_{g} = W_{friction} \longrightarrow W_{friction}?$$

Here, we pause because we have two different regions to consider.

{{course_planning:course_notes:region1c.jpg?300|}}

{{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.

$$\Delta U_{g} = W_{1} + W_{2}$$

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 are trying to solve for as this is the position change along flat part.

What's $f_{1}$ and $f_{2}?$

{{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.

$$\sum{F_{y}} = N - mgcosθ = 0$$

$$mgcosθ = N$$

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 $f_{2}$.

$$\sum{F_{x}} = f_{2} = ma_{2} \longrightarrow f_{2} = μ_{k}N = μ_{k}mg$$

$$\sum{F_{y}} = N-mg = 0$$

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 $\vec{r}'s$.

$$\Delta U_{g} = \vec{f}_{1}\cdot\Delta \vec{r}_{1} + \vec{f}_{2}\cdot\Delta \vec{r}_{2}$$

In the previous equation $\vec{f}_{1}\cdot\Delta \vec{r}_{1} \longrightarrow W_{1}<0$ and $\vec{f}_{2}\cdot\Delta \vec{r}_{2} \longrightarrow W_{2}<0$ because $\vec{f}$'s are opposite to $\Delta \vec{r}$'s

$$\Delta U_{g} = -(μ_{k}mgcosθ)d - (μ_{k}mg)x$$

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? 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.

$$-dsinθ = -μ_{k}(dcosθ + x)$$

$$dcosθ+x = \dfrac{d}{μ_{k}}sinθ$$

$$x = \dfrac{d}{μ_{k}}sinθ - dcosθ$$

$$x = d (\dfrac{sinθ-μ_{k}cosθ}{μ_{k}})$$

A check of the units reveals that:

[x]=m

[d]=m

Which makes sense as all the other quantities are unit less.


$E = γmc^2$