4.1 Define work done - NSC Technical Sciences - Question 4 - 2021 - Paper 1

To calculate the work done:

1. Identify the values: The force $F = 60 ext{ N}$, the distance $d = 8 ext{ m}$, and the angle $\theta = 25°$.
2. Apply the formula for work done:

$W = F imes d imes ext{cos}( heta)$ 3. Perform the calculations:

$W = 60 imes 8 imes ext{cos}(25°)$ $W ≈ 60 imes 8 imes 0.9063 ≈ 435.03 ext{ J}$

Thus, the work done by the applied force is approximately 435 J.

The principle of conservation of mechanical energy states that the total mechanical energy of an isolated system remains constant, provided that no external forces do work on it. This means that the sum of kinetic energy (KE) and gravitational potential energy (PE) in the system does not change.

To calculate the potential energy gained:

1. Identify the values: The mass $m = 75 ext{ kg}$ and the height $h = 12 ext{ m}$. The gravitational acceleration $g = 9.8 ext{ m/s}^2$.
2. Use the formula for potential energy:

$PE = mgh$ 3. Perform the calculations:

$PE = 75 imes 9.8 imes 12$ $PE = 8820 ext{ J}$

The potential energy gained by the construction worker is 8820 J.

To calculate the kinetic energy:

1. Using the formula for kinetic energy:

KE = rac{1}{2} mv^2 2. Identify the values: As he is lifted at a constant speed $v = 3 ext{ m/s}$. 3. Perform the calculations:

KE = rac{1}{2} imes 75 imes (3)^2 KE = rac{1}{2} imes 75 imes 9 $KE = 337.5 ext{ J}$

Thus, his kinetic energy as he is lifted for 12 m is approximately 337.5 J.

To find the potential energy just before he was airlifted, we know the total mechanical energy at the highest point is 11,500 J, which is the sum of potential energy (PE) at that height and kinetic energy (KE) at that height.

1. Using the relationship:

$ME_{total} = PE + KE$ 2. Substituting the values:

$11500 = PE + 337.5$ 3. Solving for potential energy:

$PE = 11500 - 337.5 = 11162.5 ext{ J}$

The potential energy of the construction worker just before he was airlifted is approximately 11162.5 J.