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?

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

How far will she travel along the flat?

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

No wind resistance.

System: Sled + Kid + Earth

Surroundings: Snow

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

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

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.

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 $f_{2}?$