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 by the crane, we use the formula:

$W = F imes d$

The force applied by the crane is equal to the weight of the object, which is given by:

$F = m imes g$

where:

• $m = 250$ kg (mass of the object),
• $g = 9.8$ m·s⁻² (acceleration due to gravity).

Thus,

$F = 250 imes 9.8 = 2450 ext{ N}$

Now, the displacement $d = 50$ m (height lifted), therefore:

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

Power is defined as the rate of work done over time. To find the average power used by the crane, we use:

P = rac{W}{t}

Where:

• $P$ is the power,
• $W$ is the work done,
• $t$ is the time taken.

Since the object is lifted at a constant velocity of 25 m·s⁻¹, the time $t$ taken to lift it 50 m is:

t = rac{d}{v} = rac{50}{25} = 2 ext{ s}

Now we can find the power:

P = rac{122500}{2} = 61250 ext{ W}

To convert watts to horsepower, use:

$1 ext{ HP} = 746 ext{ W}$

Thus,

ext{Power in HP} = rac{61250}{746} ext{ HP} \approx 82.0 ext{ HP}

The principle of conservation of mechanical energy states that in an isolated system, the total mechanical energy remains constant, provided that only conservative forces are acting on it. This means that the sum of kinetic energy and potential energy at any two points in the system will be equal.

The total mechanical energy of the object at 50 m above the ground is GREATER THAN its mechanical energy on the ground.

At a height of 50 m, the object possesses gravitational potential energy due to its height, which adds to its total mechanical energy. On the ground, this potential energy is zero; thus, the total mechanical energy at 50 m is greater than at ground level.

To calculate the velocity of the object when it hits the ground, we can use the conservation of mechanical energy principle. The potential energy at the height will convert to kinetic energy just before hitting the ground:

At height $h = 50$ m,

$PE = mgh = 250 imes 9.8 imes 50 = 122500 ext{ J}$

The kinetic energy (KE) just before impact is given by:

KE = rac{1}{2} mv^2

Setting $PE$ equal to $KE$:

122500 = rac{1}{2} (250) v^2

Solving for $v$ gives:

v^2 = rac{122500 imes 2}{250} = 980

Thus:

$v = ext{sqrt}(980) \approx 31.3 ext{ m·s⁻¹}$

The velocity of the object when it hits the ground is approximately 31.3 m·s⁻¹.