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

The gravitational potential energy (Ep) at point B can be calculated using the formula:

$E_p = mgh$

Where:

• m = 2 kg (mass)
• g = 9.8 m/s² (acceleration due to gravity)
• h = 1.2 m (height above ground)

Thus, substituting the values:

$E_p = (2)(9.8)(1.2) = 23.52 \, J$

The total mechanical energy (ME) at point B is the sum of the gravitational potential energy at point B and the kinetic energy at point B.

From part 4.2, we calculated that: $E_p = 23.52 \, J$

The kinetic energy (Ek) at point B can be calculated using the formula: $E_k = \frac{1}{2} mv^2$ Where:

• m = 2 kg (mass)
• v = 0.88 m/s (velocity)

Thus: $E_k = \frac{1}{2}(2)(0.88^2) = 0.7744 \, J$

Consequently: $ME = E_p + E_k = 23.52 + 0.7744 = 24.2944 \, J$

At point C, all potential energy is converted into kinetic energy. Therefore, we can set the mechanical energy at point B equal to the kinetic energy at point C.

Thus, using:

= \frac{1}{2} mv_x^2$$ We have: $$24.2944 = \frac{1}{2}(2)v_x^2$$ Simplifying gives: $$24.2944 = v_x^2$$ Taking the square root: $$v_x = \sqrt{24.2944} \approx 4.93 \, m/s$$

Power is defined as the rate at which work is done or the rate at which energy is transferred. It can be mathematically represented as: $P = \frac{W}{t}$ Where:

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

The power (P) delivered by the motor can be expressed as: $P = F_{avg}v$ Where:

• F_{avg} is the average force (or tension, T)
• v is the velocity (2 m/s)

Given: P = 43 kW = 43000 W

Using the equation: $43000 = T(2)$

We can rearrange to find T: $T = \frac{43000}{2} = 21500 \, N$