A 0.26-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 4.5 m. (a) 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 before the stone is released?

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okpalawalter8 okpalawalter8 · Answer 1 · 2020-07-03T04:06:08+02:00

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Answer:

a

[tex]E_r = 3.058 \ J[/tex]

b

[tex]E_b = -11.466 \ J[/tex]

c

[tex]\Delta E_n = -14.524 \ J[/tex]

Explanation:

From the question we are told that

The mass of the stone is [tex]m_s = 0.26 \ kg[/tex]

The height above the top of the water is [tex]h = 1.2 \ m[/tex]

The depth of the well is [tex]d = 4.5 \ m[/tex]

The gravitational potential of the stone before it was released is

[tex]E_r = mgh[/tex]

substituting values

[tex]E_r = 0.26 * 9.8 * 1.2[/tex]

[tex]E_r = 3.058 \ J[/tex]

The gravitation potential of the stone when it reaches the bottom of the well is

[tex]E_b = mg(- d)[/tex]

The negative shows that the potential energy of the stone as compared to the earth is reducing

substituting values

[tex]E_b = 0.26 * 9.8 *(- 4.5)[/tex]

[tex]E_b = -11.466 \ J[/tex]

The change in the systems gravitational potential is

[tex]\Delta E_n = E_b - E_r[/tex]

substituting values

[tex]\Delta E_n = -11.466 - 3.058[/tex]

[tex]\Delta E_n = -14.524 \ J[/tex]