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

The net work done can be computed using the formula:

$W_{net} = W_{applied} - W_{gravity}$

First, we calculate the gravitational force acting on the bag:

$W_{gravity} = m \times g \times h$

Where:

• $m = 2\, \text{kg}$
• $g = 9.8\, \text{m/s}^2$ (acceleration due to gravity)
• $h = 0.9\, \text{m}$

Now substituting the values:

$W_{gravity} = 2 \times 9.8 \times 0.9 = 17.64 \text{ J}$

Now, we use the values to find the net work done:

$W_{net} = 22.5 - 17.64 = 4.86 \, \text{J}$

Therefore, the net work done on the bag is 4.86 J.

The principle of conservation of mechanical energy states that in the absence of non-conservative forces (like friction), the total mechanical energy of an isolated system remains constant. This means that the sum of potential energy and kinetic energy at any point in the system is equal to the total mechanical energy.

At point A, the block starts from rest, so its initial kinetic energy is zero. The mechanical energy (ME) at point A is entirely gravitational potential energy (PE):

$ME = PE = mgh$

Where:

• $m = 6\, \text{kg}$
• $g = 9.8\, \text{m/s}^2$
• $h = 5\, \text{m}$

Substituting the values, we find:

$ME = 6 \times 9.8 \times 5 = 294\, \text{J}$

Thus, the mechanical energy of the block at point A is 294 J.

At point B, the mechanical energy is conserved, so:

$ME_{total} = KE + PE$

Using the conservation of mechanical energy:

$ME = KE_{B} + PE_{B}$

At point B, the height is 0 m (so $PE_B = 0$), therefore:

$294 = KE_{B}$

Since kinetic energy is given by:

$KE = \frac{1}{2}mv^2$

We can rearrange this to find the speed:

$294 = \frac{1}{2}(6)v^2$

Solving for $v^2$ gives:

$v^2 = \frac{2 \times 294}{6} = \frac{588}{6} = 98$

Thus, the speed at point B is:

$v = \sqrt{98} = 9.9 \, \text{m/s}$

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

At point C, the block has both kinetic and potential energies. To find the speed at point C, we first calculate the potential energy at that height:

$PE_{C} = mgh = 6 \times 9.8 \times 3 = 176.4\, \text{J}$

Using the conservation of mechanical energy again:

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

Substituting known values:

$294 = KE_{C} + 176.4$

Rearranging gives:

$KE_{C} = 294 - 176.4 = 117.6\, ext{J}$

Since:

$KE_{C} = \frac{1}{2}mv^2$

Substituting in:

$117.6 = \frac{1}{2}(6)v^2$

Solving for $v^2$ gives:

$v^2 = \frac{2 \times 117.6}{6} = \frac{235.2}{6} = 39.2$

Thus, the speed at point C is:

$v = \sqrt{39.2} = 6.26 \, \text{m/s}$

Therefore, the speed of the block at point C is 6.26 m/s.