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

Answer: 2.55 joules, - 9.81 joules, - 12.36 joules

Explanation:

the parameters given from the question are:

mass (m) = 0.20 kg

height above water (h₁) = 1.3m

depth of the well (h₂) = 5m = - 5m (the negative sign is there because it is a depth below the surface)

constant value for acceleration due to gravity (g) = 9.8 m/s

potential energy (PE) before the stone is released = m x g x h₁

PE₁ = 0.20 x 9.8 x 1.3 = 2.55 joules

potential energy (PE) when it reaches the bottom of the well = m x g x h₂

PE₂ = 0.2 x 9.8 x (-5) = - 9.81 joules

change in potential energy = PE₂ - PE₁

= - 9.81 - 2.55 = - 12.36 joules