Question

What is the speed of a skeleton rider at the bottom of a track, assuming no nonconservative forces?

Answer 1

To determine the speed of a skeleton rider at the bottom of a track, we can apply the principle of conservation of mechanical energy, assuming no nonconservative forces (like friction or air resistance) are acting on the rider.

1. **Initial Energy**: At the top of the track, the skeleton rider has gravitational potential energy (PE) given by:

\[
PE = mgh
\]

where:
- \( m \) is the mass of the rider,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( h \) is the height of the track above the bottom.

2. **Final Energy**: At the bottom of the track, all of the gravitational potential energy is converted into kinetic energy (KE), which is given by:

\[
KE = \frac{1}{2}mv^2
\]

where:
- \( v \) is the speed of the rider at the bottom of the track.

3. **Conservation of Energy**: According to the conservation of mechanical energy, the potential energy at the top will equal the kinetic energy at the bottom:

\[
mgh = \frac{1}{2}mv^2
\]

4. **Solving for \( v \)**: We can simplify this equation by canceling the mass \( m \) (assuming it's not zero):

\[
gh = \frac{1}{2}v^2
\]

Multiplying both sides by 2 gives:

\[
2gh = v^2
\]

Taking the square root of both sides, we find the speed at the bottom of the track:

\[
v = \sqrt{2gh}
\]

This equation gives us the speed of the skeleton rider at the bottom of the track in terms of the height \( h \) from which they descended. The actual speed will depend on the specific height of the track.