Question

1. set the simulation up like this: add a check to: pie chart, bar graph, grid, speed, slow motion is selected (at the bottom) 5. then, drag your skater onto the highest part of the u - shaped ramp and select the start arrow to begin. you may use the pause button to stop your skater and the slow motion button to slow the skater down. position of the skater on the u - shaped ramp which is more...kinetic or potential energy? does the total energy change? (yes or no?) 1 start the skateboarder at 6m stop the skateboarder at 4m (going down the ramp) 2 start the skateboarder at 6m. stop the skateboarder at 2m (going down the ramp).

Explanation:

Step1: Recall energy - conservation and formulas

The potential energy is $U = mgh$ and kinetic energy is $K=\frac{1}{2}mv^{2}$, and the total mechanical energy $E = K + U$. At the start, $K = 0$ and $U=mgh$. As the skater moves down, $U$ decreases and $K$ increases while $E$ is conserved (assuming no non - conservative forces like friction).

Step2: For the first case (start at 6m, stop at 4m)

The initial potential energy $U_1=mgh_1$ with $h_1 = 6m$. The potential energy at 4m is $U_2=mgh_2$ with $h_2 = 4m$. The change in potential energy $\Delta U=mg(h_1 - h_2)$ is converted into kinetic energy. Since $h_1>h_2$, potential energy $U$ is still greater than kinetic energy $K$ at 4m. And the total energy does not change (assuming no friction), so the answer is no.

Step3: For the second case (start at 6m, stop at 2m)

The initial potential energy is $U_{start}=mg\times6$. The potential energy at 2m is $U_{2m}=mg\times2$. A larger amount of potential energy has been converted into kinetic energy. At 2m, kinetic energy is greater than potential energy. And the total energy does not change (assuming no friction), so the answer is no.

Position of the Skater on the U - shaped Ramp	Which is more...kinetic or potential energy?	Does the total energy change? (yes or no)
Start the skateboarder at 6m, Stop the skateboarder at 2m (going down the ramp)	Kinetic energy	No

Answer:

Step1: Recall energy - conservation and formulas

The potential energy is $U = mgh$ and kinetic energy is $K=\frac{1}{2}mv^{2}$, and the total mechanical energy $E = K + U$. At the start, $K = 0$ and $U=mgh$. As the skater moves down, $U$ decreases and $K$ increases while $E$ is conserved (assuming no non - conservative forces like friction).

Step2: For the first case (start at 6m, stop at 4m)

The initial potential energy $U_1=mgh_1$ with $h_1 = 6m$. The potential energy at 4m is $U_2=mgh_2$ with $h_2 = 4m$. The change in potential energy $\Delta U=mg(h_1 - h_2)$ is converted into kinetic energy. Since $h_1>h_2$, potential energy $U$ is still greater than kinetic energy $K$ at 4m. And the total energy does not change (assuming no friction), so the answer is no.

Step3: For the second case (start at 6m, stop at 2m)

The initial potential energy is $U_{start}=mg\times6$. The potential energy at 2m is $U_{2m}=mg\times2$. A larger amount of potential energy has been converted into kinetic energy. At 2m, kinetic energy is greater than potential energy. And the total energy does not change (assuming no friction), so the answer is no.

Position of the Skater on the U - shaped Ramp	Which is more...kinetic or potential energy?	Does the total energy change? (yes or no)
Start the skateboarder at 6m, Stop the skateboarder at 2m (going down the ramp)	Kinetic energy	No