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

To calculate the work done by gravitational force, we use the formula:

$W_g = m imes g imes h$

where:

• $m = 0.145 ext{ kg}$ (mass of the cellphone)
• $g = 9.81 ext{ m/s}^2$ (acceleration due to gravity)
• $h = 0.50 ext{ m}$ (height)

Calculating the work done:

W_g = 0.145 imes 9.81 imes 0.5 \= 0.710 \text{ J}

Power is defined as the rate at which work is done or energy is transferred. It can be expressed mathematically as:

P = rac{W}{t}

where:

• $P$ = power
• $W$ = work done
• $t$ = time taken.

Using the power formula:

P = rac{W}{t}

where:

• $W = W_{ ext{applied}}$ (the work done in lifting the cellphone)
• $t = 4 ext{ s}$

We have already calculated the work done as the force times the height:

$W = 4.9 imes 0.5 = 2.45 ext{ J}$

So,

P = rac{2.45}{4} = 0.613 ext{ W}

The principle of conservation of mechanical energy states that in an isolated system, the total mechanical energy remains constant. This means that the sum of potential energy and kinetic energy at any point in the system remains the same, provided that no external forces do work on the system.

Initially, the cellphone has gravitational potential energy given by:

$E_p = m imes g imes h = 0.145 imes 9.81 imes 0.5 \approx 0.710 \text{ J}$

As the cellphone falls, this potential energy converts to kinetic energy, which is given by:

E_k = rac{1}{2} m v^2

At the moment just before hitting the ground, all potential energy converts to kinetic energy:

0.710 = rac{1}{2} imes 0.145 imes v^2

Solving for $v$:

$0.710 = 0.0725 imes v^2$ v^2 = rac{0.710}{0.0725} \approx 9.79 $v \approx 3.13 ext{ m/s}$