A construction worker has to pump water from reservoir A to reservoir B over a hill using an electric water pump, as illustrated in the diagram below - NSC Technical Sciences - Question 4 - 2024 - Paper 1

The force applied by the pump on the water can be determined using the force balance:

$F_{pump} = F_g$

Substituting the value of the weight of water:

$F_{pump} = 850 ext{ kg} imes 9.8 ext{ m/s²} = 8330 ext{ N} \text{ upwards}$

We can find the velocity of the water using the power equation:

$P = F imes v$

Given that the power of the pump is 7200 W and the force is 8330 N, we rearrange the equation to find velocity:

$v = \frac{P}{F} = \frac{7200}{8330} = 0.86 ext{ m/s} \text{ upwards}$

The principle of conservation of mechanical energy states that in an isolated system, the total mechanical energy remains constant. This means that the sum of kinetic energy and potential energy remains unchanged if only conservative forces are acting on the system.

The work done by the gravitational force can be calculated as:

$W = F_g imes d imes cos(\theta)$

Since the displacement is horizontal and the force of gravity acts vertically, $\theta = 90°$ and thus:

$W = 0 ext{ J}$

The work done by the gravitational force is zero because the direction of displacement (horizontal) is perpendicular to the direction of the gravitational force (vertical). Thus, the work done can be expressed as:

$W = F_g imes d imes cos(90°) = 0 ext{ J}$

Where the force component in the direction of displacement is zero.

Gravitational potential energy (PE) is calculated using the formula:

$PE = mgh$

Substituting the values:

$PE = 274 ext{ kg} imes 9.8 ext{ m/s²} imes 12.58 ext{ m} = 33779.82 ext{ J}$

The mechanical energy at any point consists of both potential energy and kinetic energy. Since we only have potential energy in this scenario (as water is flowing, not stationary), we calculate:

At the height of 6 m:

$PE = mgh = 274 ext{ kg} imes 9.8 ext{ m/s²} imes 6 ext{ m} = 16057.2 ext{ J}$

This is the potential energy at 6 m.