A 0.23-kg stone is held 1.2 m above the top edge of a water well and then droppe

Question

A 0.23-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 5.2 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?
J

(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?
J

(c) What is the change in gravitational potential energy of the system from release to reaching the bottom of the well?
J

Explanation / Answer

a) E1 = 0.21*9.81*1.2 = 2.47 J b) E2 = 0.21*9.81*(-5.3) = -10.92 J c) Change = E2 - E1 = -10.92 - 2.47 = 13.39 J please dont forget to rate my answer :)

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