A 0.20 kg stone is held 1.3 m above the top edge of a water well and then dropped intoit. The well has a depth of 5.0 m. Relative to the configuration with the stone at the topedge of the well, what is the gravitational potential energy of the stone–Earth system:(a) before the stone is released.(b) when the stone reaches the bottom of the well?(c) What is the change in gravitational potential energy of the system from release toreaching the bottom of the well?

(a) Before the stone is released, the gravitational potential energy of the stone-Earth system can be calculated using the formula:

Potential Energy = mass * gravity * height

Given that the mass of the stone is 0.20 kg, the acceleration due to gravity is approximately 9.8 m/s^2, and the height above the top edge of the well is 1.3 m, we can substitute these values into the formula:

Potential Energy = 0.20 kg * 9.8 m/s^2 * 1.3 m

Calculating this expression, we find that the gravitational potential energy of the stone-Earth system before the stone is released is approximately 2.548 Joules.

(b) When the stone reaches the bottom of the well, the gravitational potential energy is converted into kinetic energy. At the bottom of the well, the stone has no height above the ground, so its gravitational potential energy is zero.

(c) The change in gravitational potential energy of the system from release to reaching the bottom of the well can be calculated by subtracting the initial potential energy from the final potential energy. Since the final potential energy is zero, the change in gravitational potential energy is equal to the initial potential energy:

Change in Potential Energy = Final Potential Energy - Initial Potential Energy Change in Potential Energy = 0 - 2.548 Joules Change in Potential Energy = -2.548 Joules

Therefore, the change in gravitational potential energy of the system from release to reaching the bottom of the well is approximately -2.548 Joules.