Question

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.

Answer 1

Explanation:

We will calculate the gravitational potential energy as follows.

                 $P.E_{1} = mgz_{1}$

       $P.E_{1} = (\rho V)gz_{1}$    

                    = $1000 kg/m^{3} \times 3 m^{3} \times 9.7 \times 40 m$

                    = 1164000 J

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

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

             $\Delta P.E = mg(z_{2} - z_{1})$

                         = $\rho \times V \times g (z_{2} - z_{1})$

                         = $1000 \times 3 \times 9.7 (10 - 40)m$

                         = -873000 J

or,                      = -873 kJ

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