A learner demonstrates the effectiveness of her robotic arm - NSC Technical Sciences - Question 6 - 2021 - Paper 1

The gravitational force acting on the cellphone can be calculated as:

$F_g = m imes g$

where:

• $m = 0.145 ext{ kg}$ (conversion from grams),
• $g = 9.81 ext{ m/s}^2$ (acceleration due to gravity).

Thus,

$F_g = 0.145 imes 9.81 = 1.42145 ext{ N}$

The work done by the gravitational force while the cellphone is lifted to a height of 0.5 m is:

$W_g = F_g imes h = 1.42145 imes 0.5 = 0.710725 ext{ J}$.

Power is the rate at which work is done or energy is transferred. It denotes how fast energy is converted or transferred from one form to another, mathematically defined as:

P = rac{W}{t}

where:

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

Using the work done calculated previously for the lifting:

$W = 0.710725 ext{ J}$

And the time taken:

$t = 4 ext{ s}$

The power can be calculated as:

P = rac{W}{t} = rac{0.710725}{4} = 0.17768125 ext{ W}

Approximately, this is:

The principle of conservation of mechanical energy states that in an isolated system, the total mechanical energy remains constant if only conservative forces are doing work. This means that the sum of potential energy and kinetic energy is conserved throughout the motion.

When the cellphone is released from a height, all of its potential energy is converted to kinetic energy just before it hits the ground. The potential energy (PE) at the top can be calculated as:

$PE = mgh$

where:

• $m = 0.145 ext{ kg}$,
• $g = 9.81 ext{ m/s}^2$,
• $h = 0.5 ext{ m}$.

Substituting these values gives:

$PE = 0.145 imes 9.81 imes 0.5 = 0.710725 ext{ J}$

This potential energy becomes kinetic energy (KE) just before the cellphone hits the floor:

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

Setting the potential energy equal to the kinetic energy gives:

0.710725 = rac{1}{2} (0.145) v^2

Solving for $v$ will yield the velocity: