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

The work done by gravitational force is calculated using the formula:

$W_g = m imes g imes h$

where:

• $m = 0.145 ext{ kg}$ (mass converted from grams),
• $g = 9.81 ext{ m/s}^2$ (acceleration due to gravity),
• $h = 0.5 ext{ m}$ (height converted from cm).

Now substituting the values:

$W_g = 0.145 imes 9.81 imes 0.5 = 0.710 ext{ J}$

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:

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

Using the work done calculated earlier ( $W = 0.710 ext{ J}$) and the time taken ( $t = 4 ext{ s}$), we can find the power:

P = rac{W}{t} = rac{0.710}{4} = 0.178 ext{ W}.

The principle of the 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 is conserved.

When the cellphone is released, the potential energy at the height will convert into kinetic energy just before it hits the floor. The potential energy ($PE$) can be calculated as:

$PE = mgh = 0.145 imes 9.81 imes 0.5 = 0.710 ext{ J}$

At the moment just before it hits the floor, this potential energy will equal the kinetic energy ($KE$):

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

Setting $PE$ equal to $KE$:

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

Solving for $v$ gives:

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

v ext{ (velocity)} = ext{sqrt}igg(rac{1.42}{0.145}igg) = 3.33 ext{ m/s}.