Determine the gravitational potential energy, in kJ, of 3 m3 of liquid water at an elevation of 30 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 15 m.

A. 873 kJ

B. -436.5 kJ

Explanation:

Parameters given:

Volume of water = 3m³

Density of water = 1000 kg/m³

Height above the earth, h = 30m

The gravitational potential energy, PE, is given as:

PE = m*g*h

Where m is mass and g is acceleration due to gravity.

Mass, m, can be obtained:

Density = mass/volume

Mass = density * volume

Mass, m = 3 * 1000 = 3000kg

=> Potential ENERGY, PE = 3000 * 9.7 * 30

PE = 873000 J = 873 kJ

B. If the height is decreased by 15m,

h = 30 - 15 = 15m

PE = 3000 * 9.7 * 15

PE = 436500 J = 436.5 kJ

Change in PE will be:

ΔPE = PEfinal - PEinitial

ΔPE = 436.5 - 873

ΔPE = -436.5 kJ

answer; ///(c); ///bronchi i do believe this is the correct answer,; plz check first..good luck

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