5.1 A learner lifts her school bag straight from the ground to a height of 0.9 m above the ground - NSC Technical Sciences - Question 5 - 2020 - Paper 1

To determine the net work done on the bag, we first calculate the gravitational potential energy gained by the bag, which is given by:

$PE = mgh$

where:

• $m$ is the mass of the bag (2 kg),
• $g$ is the acceleration due to gravity (approximately $9.8 ext{ m/s}^2$), and
• $h$ is the height (0.9 m).

Calculating the potential energy:

$PE = 2 ext{ kg} imes 9.8 ext{ m/s}^2 imes 0.9 ext{ m} = 17.64 ext{ J}$

The net work done on the bag is equal to the work done by the learner minus the gravitational potential energy:

$W_{net} = W - PE$

Since the work done by the learner is 22.5 J:

$W_{net} = 22.5 ext{ J} - 17.64 ext{ J} = 4.86 ext{ J}$

Thus, the net work done on the bag is 4.86 Joules.

The principle of conservation of mechanical energy states that the total mechanical energy of an isolated system remains constant if only conservative forces are acting on it. This means that the sum of potential energy and kinetic energy remains the same throughout the motion of the system.

At point A, the block is at rest, so its kinetic energy (KE) is zero. The mechanical energy (ME) is purely gravitational potential energy (PE), calculated as:

$ME = PE = mgh$

where:

• $m = 6 ext{ kg}$,
• $g = 9.8 ext{ m/s}^2$, and
• $h = 5 ext{ m}$. Substituting the values:

$ME = 6 ext{ kg} imes 9.8 ext{ m/s}^2 imes 5 ext{ m} = 294 ext{ J}$

So, the mechanical energy of the block at point A is 294 Joules.

To find the speed of the block at point B, we can use the conservation of energy principle:

$ME_{A} = KE_{B} + PE_{B}$

At point B, the potential energy is lower as the height is less. If we take point B at ground level, $PE_{B} = 0$:

• $ME_A = KE_B$. Thus:

294 ext{ J} = rac{1}{2}mv^2

Substituting the mass (6 kg):

294 = rac{1}{2}(6)v^2

Solving for $v$ gives:

v^2 = rac{294 imes 2}{6} = 98 ext{ m}^2/ ext{s}^2

Thus, $v = ext{sqrt}(98) ext{ m/s} ightarrow v ext{ = 9.9 m/s}$

So, the speed of the block at point B is approximately 9.9 m/s.

At point C, we need to calculate the total mechanical energy:

$ME_{A} = KE_{C} + PE_{C}$

At point C, the height is 3 m. So:

$PE_C = mgh = 6 ext{ kg} imes 9.8 ext{ m/s}^2 imes 3 ext{ m} = 176.4 ext{ J}$

Thus:

$294 ext{ J} = KE_C + 176.4 ext{ J}$

Therefore:

$KE_C = 294 ext{ J} - 176.4 ext{ J} = 117.6 ext{ J}$

Now, we will use the kinetic energy formula to find the speed:

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

Substituting the mass:

117.6 = rac{1}{2}(6)v^2

Solving for $v$ gives:

v^2 = rac{117.6 imes 2}{6} = 39.2 ext{ m}^2/ ext{s}^2

Thus,