A crane was used to lift an object of mass 250 kg vertically upwards to point A, which is 50 m above the ground, at a CONSTANT VELOCITY, as shown in the diagram below - NSC Technical Sciences - Question 4 - 2023 - Paper 1

To calculate the work done, we use the formula:

$W = F imes d$

First, we calculate the force due to gravity acting on the object:

$F = m imes g = 250 ext{ kg} imes 9.8 ext{ m/s}^2 = 2450 ext{ N}$

Now, substituting the values into the work formula:

$W = 2450 ext{ N} imes 50 ext{ m} = 122500 ext{ J}$

Power is defined as the rate of doing work, calculated as:

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

where $t$ is the time taken to do the work. Given that the object was lifted at a constant velocity, the time taken can be calculated:

Assuming it took $t = \frac{d}{v} = \frac{50 ext{ m}}{25 ext{ m/s}} = 2 ext{ s}$.

Now substituting into the power formula:

$P = \frac{122500 ext{ J}}{2 ext{ s}} = 61250 ext{ W}$

To convert watts to horse power:

$1 ext{ hp} = 745.7 ext{ W}$

Thus,

$\text{Power in hp} = \frac{61250 ext{ W}}{745.7 ext{ W/hp}} \approx 82.16 ext{ hp}$

The principle of conservation of mechanical energy states that in an isolated system, the total mechanical energy (the sum of potential and kinetic energy) remains constant, provided that no external forces do work on the system.

GREATER THAN

At a height of 50 m, the object possesses gravitational potential energy due to its height, which is not present when it is on the ground. Hence, the total mechanical energy at 50 m is greater than on the ground.

Using conservation of energy, the potential energy at point A converts entirely into kinetic energy just before hitting the ground:

$U = K \Rightarrow mgh = \frac{1}{2} mv^2$

Substituting the values:

$250 ext{ kg} \times 9.8 ext{ m/s}^2 \times 50 ext{ m} = \frac{1}{2} \times 250 ext{ kg} \times v^2$

Solving for $v$:

$122500 ext{ J} = 125 \times v^2$

$v^2 = \frac{122500}{125} = 980$

$v = \sqrt{980} \approx 31.30 ext{ m/s}$