A crane is lifting an object of mass 600 kg from point A to point B at a CONSTANT SPEED to a vertical height of 25 m in two minutes, as shown in the diagram below - NSC Technical Sciences - Question 4 - 2023 - Paper 1

Power is defined as the work done per unit time. Using the formula:

$P = \frac{W}{t}$

Where:

• $W = 1.47 \times 10^5 ext{ J}$ (from the previous calculation),
• $t = 2 ext{ minutes} = 120 ext{ s}$.

Substituting in the values:

$P = \frac{1.47 \times 10^5}{120} = 1225 ext{ W}$

Thus, the power at which the crane operates is 1225 W.

Gravitational potential energy is the energy that an object possesses due to its position in a gravitational field, which is determined by its height above a reference point, typically the surface of the Earth. This energy depends on the mass of the object and the gravitational acceleration.

The kinetic energy (KE) of an object can be calculated using the formula:

$KE = \frac{1}{2} mv^2$

Where:

• $m$ is the mass of the brick (3 kg),
• $v$ is the speed just before it hits the ground (7 m/s).

Substituting in the values:

$KE = \frac{1}{2} (3) (7^2) = \frac{1}{2} (3) (49) = 73.5 ext{ J}$

Thus, the kinetic energy of the brick just before it hits the ground is 73.5 J.

To find the height from which the brick was dropped, we can use the principle of conservation of energy, where the potential energy at the height is converted to kinetic energy:

$E_p = E_k \\ mgh = \frac{1}{2} mv^2$

Plugging in the values:

• $m = 3 ext{ kg}$,
• $v = 7 ext{ m/s}$,
• $g = 9.8 ext{ m/s}^2$.

Setting the equations equal:

$3 imes 9.8 imes h = \frac{1}{2} (3)(7^2) = \frac{1}{2} (3)(49) = 73.5 \\ 29.4h = 73.5 \\ h = \frac{73.5}{29.4} = 2.5 ext{ m}$

Thus, the height from which the brick was dropped is 2.5 m.