A 0.31-kg stone is held 1.2 m above the top edge of a water well and then dropped into it. The well has a depth of 5.6 m.
(a) Taking y = 0 at the top edge of the well, what is the gravitational potential energy of the stone–Earth system before the stone is released?
(b) Taking y = 0 at the top edge of the well, what is the gravitational potential energy of the stone–Earth system 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?
PE1 = 0
PE2 = -mgh = - 0.31•9.8•1.2 = - 3.65 J.
ΔPE = 3.65 J.
To calculate the gravitational potential energy of the stone-Earth system before the stone is released, we can use the formula:
Gravitational Potential Energy = m * g * h
where
m = mass of the stone (0.31 kg)
g = acceleration due to gravity (9.8 m/s^2)
h = height of the stone above the reference point (1.2 m)
Substituting the values into the formula, we can calculate:
Gravitational Potential Energy = 0.31 kg * 9.8 m/s^2 * 1.2 m
To calculate the gravitational potential energy of the stone-Earth system when it reaches the bottom of the well, we can use the same formula, but this time the height (h) will be the depth of the well (5.6 m):
Gravitational Potential Energy = 0.31 kg * 9.8 m/s^2 * 5.6 m
To calculate the change in gravitational potential energy of the system from release to reaching the bottom of the well, we need to subtract the initial potential energy from the final potential energy:
Change in Gravitational Potential Energy = Final Potential Energy - Initial Potential Energy
Substituting the calculated potential energy values into the equation, we can find the change:
Change in Gravitational Potential Energy = (0.31 kg * 9.8 m/s^2 * 5.6 m) - (0.31 kg * 9.8 m/s^2 * 1.2 m)
By evaluating the expression, you should be able to find the answers to parts (a), (b), and (c) of the question.