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

a, c, d, e are all applying about success strategies.

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards