Question

A 0.30-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 5.1 m. (a) relative to the configuration with the stone at the top edge of the well, what is the gravitational potential energy of the stoneâearth 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 stoneâearth 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

Answer 1

(a) The gravitational potential energy of the stone before it is released is:

`U=mg \Delta h`
where m=0.30 kg is the mass of the stone,
`g=9.81 m/s^2` and
`\Delta h` is the difference in height relative to the reference level, which is the top edge of the well. Therefore, since the stone in this situation is at h=1.2 m above the top edge of the well,
`\Delta h=1.2 m`, and the gravitational potential energy is

`U=(0.30 kg)(9.81 m/s^2)(1.2 m)=3.5 J`

(b) Similarly, the gravitational potential energy when the stone reaches the bottom of the well is

`U=mg \Delta h`
but this time the stone is at the bottom of the well, so at 5.1 m below the top edge of the well, therefore
`\Delta h=-5.1 m`. So, the gravitational potential energy in this situation is

`U=(0.30 kg)(9.81 m/s^2)(-5.1 m)=-15.0 J`

(c) The change in gravitational potential energy from release to the bottom of the well is:

`\Delta U= U_b-U_a=-15.0 J-(3.5 J)=-18.5 J`