A0.20-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?

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

PE₁ = 0.20 x 9.8 x 1.3 = 2.55 joules

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

  = -9.81 - 2.55 = -12.36 joules

(a) The gravitational potential energy of the stone before it was released is 12.348 J

(b) The gravitational potential energy of the stone when it reaches the bottom of the well is zero.

(c) The change in gravitational  potential energy of the stone is 12.348 J.

The given mass of the stone, m = 0.2 kg

(a) The gravitational potential energy of the stone before it was released is calculated as;

before the stone was released, potential energy will be maximum.

P.E = mgh

P.E = 0.2 x 9.8 x (1.3 + 5)

P.E = 12.348 J

(b) The gravitational potential energy of the stone when it reaches the bottom of the well is calculated as;

when the stone reaches the bottom, the height will be zero and the potential energy will be zero.

P.E = mgh

P.E = 0.2 x 9.8 x (0)

P.E = 0

(c) The change in potential energy is calculated as;

ΔP.E = 12.348 - 0 =  12.348 J.

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