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Determine the gravitational potential energy, in kJ, of 3 m3 of liquid water at an elevation of 40 m above the surface of Earth. The acceleration of gravity is constant at 9.7 m/s2 and the density of the water is uniform at 1000 kg/m3. Determine the change in gravitational potential energy if the elevation decreases by 10 m.

Respuesta :

Explanation:

We will calculate the gravitational potential energy as follows.

                 [tex]P.E_{1} = mgz_{1}[/tex]

       [tex]P.E_{1} = (\rho V)gz_{1}[/tex]    

                    = [tex]1000 kg/m^{3} \times 3 m^{3} \times 9.7 \times 40 m[/tex]

                    = 1164000 J

or,                = 1164 kJ         (as 1 kJ = 1000 J)

Now, we will calculate the change in potential energy as follows.

             [tex]\Delta P.E = mg(z_{2} - z_{1})[/tex]

                         = [tex]\rho \times V \times g (z_{2} - z_{1})[/tex]

                         = [tex]1000 \times 3 \times 9.7 (10 - 40)m[/tex]

                         = -873000 J

or,                      = -873 kJ

Thus, we can conclude that change in  gravitational potential energy is -873 kJ.