Question

introduction
if you didn’t already, turn on the ‘energy’ bar graph, ‘grid,’ and ‘speed’ options.
set the skater 2 meters above the ground on the ramp and release them.

1. 1. what type of energy does the skater have at the 2-meter mark? 5 points

1. 2. how high does the skater get on the other end of the ramp? 5 points

1. 3. explain, in terms of the conservation of energy, why the skater will never go higher than your answer to question 2 at this point. (idk is not a valid answer) 5 points

Explanation:

Question 1

At the 2 - meter mark, the skater is at rest (just released) and has a height above the ground. Potential energy (specifically gravitational potential energy) depends on height (\(PE = mgh\), where \(m\) is mass, \(g\) is acceleration due to gravity, and \(h\) is height) and since the skater is not moving (speed is zero), kinetic energy (\(KE=\frac{1}{2}mv^{2}\)) is zero. So the skater has gravitational potential energy.

Assuming no non - conservative forces (like friction) are acting on the skater, by the law of conservation of mechanical energy, the total mechanical energy (sum of kinetic and potential energy) is conserved. At the start, the skater has potential energy (and zero kinetic energy as it's released from rest) at a height of 2 meters. When it moves to the other end of the ramp, at the maximum height, its kinetic energy will be zero (momentarily at rest at the peak), so all the initial potential energy is converted back to potential energy. So the height at the other end should be 2 meters (neglecting energy losses).

The law of conservation of energy states that the total mechanical energy (kinetic + potential) of a system remains constant if no non - conservative forces (like friction, air resistance) do work on the system. Initially, the skater has a certain amount of mechanical energy (mostly potential energy at 2 meters height, \(ME = PE_1+KE_1\), with \(KE_1 = 0\) as it's released from rest). As the skater moves, energy can convert between kinetic and potential, but the total remains the same (assuming no energy loss). To go higher than 2 meters, the skater would need more potential energy (\(PE = mgh\), so higher \(h\) means higher \(PE\)) than the initial total mechanical energy. Since energy is conserved and there's no external source of energy adding to the system, the skater can't have more mechanical energy than it started with, so it can't go higher than 2 meters.

Answer:

Gravitational Potential Energy

Question 2