# A skier is gliding along at 3.6 m/s on horizontal, frictionless snow. He suddenly starts down a...

## Question:

A skier is gliding along at 3.6 m/s on horizontal, frictionless snow. He suddenly starts down a 10-degree incline. His speed at the bottom is 13 m/s.

What is the length of the incline? Express answer with appropriate units.

How long does it take him to reach the bottom? Express answer with appropriate units.

## Conservation of Energy

Consider a particle moving under the action of conservative forces, that is, forces that are non-dissipative. The theorem of conservation of energy states that the total energy of the system remains unaltered.

An example of conservative force is gravitational force; an example of dissipative force is friction.

There is no friction between the glides and the snow, and the skier is only subjected to the gravitational force, which is conservative. So we can consider conservation of energy between points {eq}A {/eq} and {eq}B {/eq}, at the begining and at the end of the incline, respectively (See figure).

At the top of the incline the skier has:

• Kinetic energy: {eq}K_{A} = \dfrac{1}{2} m \, v_{A}^{2} {/eq};
• Potential energy: {eq}U = m \, g \, h = m \, g \, d \sin \theta {/eq}.

At the botton, the skier has only:

• Kinetic energy: {eq}K_{B} = \dfrac{1}{2} m \, v_{B}^{2} {/eq}.

The conservation of energy gives

\begin{align} K_{A} + U &= K_{B} \\ \\ \dfrac{1}{2} m \, v_{A}^{2} + m \, g \, d \, \sin \theta &= \dfrac{1}{2} m \, v_{B}^{2} \\ \\ v_{A}^{2} + 2 \, g \, d \, \sin \theta &= v_{B}^{2} \\ \\ \end{align}

Now we solve for the distance {eq}d {/eq}:

$$d = \dfrac{ v_{B}^{2} - v_{A}^{2} }{ 2 \, g \, \sin \theta } \, \, .$$

Applying values,

$$d = \dfrac{ ( 13 \, \text{m} / \text{s} )^{2} - ( 3.6 \, \text{m} / \text{s} )^{2} }{ 2 \, ( 9.8 \, \text{m} / \text{s}^{2} ) \, \sin ( 10^{\circ} ) } \approx 46 \, \text{m} \, \, .$$

{eq}\\ {/eq}

$$\color{blue}{ \boxed{ d \approx 46 \, \text{m} } }$$

Now consider the following diagram

We show that the component of the gravitational force along the incline is given by {eq}m g \sin \theta {/eq}, so that we can consider the accelaration along the incline to be {eq}a = g \sin \theta {/eq}.

We can calculate the time for the skier to go from {eq}A {/eq} to {eq}B {/eq} to be

\begin{align} \Delta t &= \dfrac{ \Delta v }{ a } = \dfrac{ v_{B} - v_{A} }{ g \, \sin \theta } \, \\ \\ &= \dfrac{ ( 13 \, \text{m} / \text{s} ) - ( 3.6 \, \text{m} / \text{s} ) }{ 9.8 \, \text{m} / \text{s}^{2} \, \sin ( 10^{\circ} ) } = 5.5 \, \text{s} \, \, . \end{align}

{eq}\\ {/eq}

$$\color{blue}{ \boxed{ \Delta t = 5.5 \, \text{s} } }$$