183_notes:examples:sledding [Projects & Practices in Physics]

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

Δ E_{s y s t e m} = W_{s u r r o u n d i n g s}

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

Δ K + Δ U_{g} = W_{f r i c t i o n}

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

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.

Δ E_{s y s t e m} = W_{s u r r o u n d i n g s}

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

The change in energy can be in the form of change of kinetic and change in gravitational potential energy.

Δ K + Δ U_{g} = W_{f r i c t i o n}

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

No change

Δ K = 0

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

Δ U_{g} = W_{f r i c t i o n} ⟶ W_{f r i c t i o n} ?

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

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

The frictional force is different in the two regions so we must consider the work they do separately.

Δ U_{g} = W_{1} + W_{2}

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

Breaking work down into force by change in distance.

Δ U_{g} = {\vec{f}}_{1} \cdot Δ {\vec{r}}_{1} + {\vec{f}}_{2} \cdot Δ {\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 are trying to solve for as this is the position change along flat part.

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

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 - m g c o s θ = 0

$\sum{F_{y}} = N - mgcosθ = 0$

m g c o s θ = N

$mgcosθ = N$

If $f_{1}=μ_{k}N$ then:

f_{1} = μ_{k} m g c o s θ

$f_{1} = μ_{k}mgcosθ$

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} = m a_{2} ⟶ f_{2} = μ_{k} N = μ_{k} m g

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

\sum F_{y} = N - m g = 0

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

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 $\vec{r}'s$ .

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

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

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