A load of mass 75 kg is initially at rest on the ground - NSC Physical Sciences - Question 5 - 2018 - Paper 1

The non-conservative force acting on the load is the tension in the rope (T). This force is doing work on the system as it moves the load upward.

The work done (W_g) by the gravitational force is given by:

$W_g = F_g imes d imes ext{cos}( heta)$

Where:

• $F_g$ is the gravitational force (735 N)
• $d$ is the distance moved (12 m)
• $heta$ is the angle between the force and the direction of motion (180° for gravitational force when moved upward)

Thus,

$W_g = 735 imes 12 imes ext{cos}(180°) = -8820 ext{ J}$

The negative sign indicates that the gravitational force is doing work against the upward movement.

The work-energy theorem states that the work done on an object by the net force is equal to the change in the object's kinetic energy. This means that the energy transferred to the object is converted into its motion.

Using the work-energy theorem:

W = rac{1}{2} mv^2 - 0

Where:

• $W$ is the net work done on the object, which can be calculated by considering the non-conservative works done. Here, we find: $W = W_T + W_g$
• $W_T$ will be equal to the tension (T) work done while moving upward against gravity.

We previously calculated $W_g = -8820 J$, so to find the speed (v) after incorporating work done, rearranging:

egin{align*} ext{Let Tension force, } T = m(g + a) = 75 ext{ kg} (9.8 + 0.65) ext{ m/s²}
T ext{ can then be calculated.}
ext{Using}
T imes d - 735 imes 12 = rac{1}{2} m v^2
ext{Calculating yields: } v ≈ 3.95 ext{ m/s} ext{(Result verified by computation.)} ext{Final speed is } 3.95 ext{ m/s.} \end{align*}