question: how does the speed relate to the potential and kinetic energy of the skater?

Explanation:

Step1: Recall energy formulas

The kinetic - energy formula is $K = \frac{1}{2}mv^{2}$, where $K$ is kinetic energy, $m$ is the mass of the skater, and $v$ is the speed of the skater. The potential energy formula is $U = mgh$, where $U$ is potential energy, $m$ is mass, $g$ is the acceleration due to gravity, and $h$ is the height.

Step2: Analyze kinetic - energy relationship

From $K=\frac{1}{2}mv^{2}$, we can see that kinetic energy is directly proportional to the square of the speed ($v$). As the speed of the skater increases, the kinetic energy increases quadratically.

Step3: Analyze potential - energy relationship

Potential energy is related to height. As the skater moves, if the total mechanical energy ($E = K + U$) is conserved (ignoring friction and air - resistance), when the skater's speed increases, the height ($h$) decreases (since $U=mgh$ and $E$ is constant), and thus the potential energy decreases.

Answer:

The kinetic energy of the skater is directly proportional to the square of the speed ($K=\frac{1}{2}mv^{2}$). If total mechanical energy is conserved, as the speed increases, kinetic energy increases and potential energy decreases because the skater's height decreases.