An object of mass 2 kg slides down a frictionless track ABC - NSC Technical Sciences - Question 4 - 2022 - Paper 1

Gravitational potential energy (Ep) is calculated using the formula:

$Ep = mgh$

Substituting the values:

$Ep = (2 ext{ kg})(9.8 ext{ m/s}^2)(1.2 ext{ m}) = 23.52 ext{ J}$

The total mechanical energy (ME) at point B can be calculated as the sum of potential energy and kinetic energy:

$ME = Ep + Ek$ $Ek = \frac{1}{2}mv^2 = \frac{1}{2}(2)(0.88^2) = 0.7744 ext{ J}$

Thus,

$ME = 23.52 ext{ J} + 0.7744 ext{ J} = 24.29 ext{ J}$

At point C, all potential energy is converted into kinetic energy. Therefore, the mechanical energy at point C (ME) is equal to the mechanical energy calculated at point B:

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

From this, we can derive:

$v = \sqrt{\frac{2ME}{m}} = \sqrt{\frac{2(24.29)}{2}} = 4.93 ext{ m/s}$

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

$P = \frac{W}{t}$

The power delivered by the motor can be calculated using:

$P = F \cdot v$

Where:

• $P = 43000 \text{ W}$
• $v = 2 ext{ m/s}$

Rearranging gives us:

$T = \frac{P}{v} = \frac{43000}{2} = 21500 ext{ N}$