A 0.33-kg stone is held 1.1 m above the top edge of a water well and then droppe

Question

A 0.33-kg stone is held 1.1 m above the top edge of a water well and then dropped into it. The well has a depth of 5.3 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. J (b) Relative to the configuration with the stone at the top edge of the well, what is the gravitational potential energy of the stone-Earth 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) Total potential energy (gravitational energy) of the stone in this system.
using U=mgh
U1 = 0.33*9.81*1.1
U1 = 3.56 J

b) U2 = 0.33*9.81*(-5.3)
U2= -17.15 J

c) Change = U2 - U1
= -17.15 - 3.56
= - 20.71 J

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