The diagram below, not drawn to scale, shows a vehicle with a mass of 1 500 kg starting from rest at point A at the bottom of a rough incline - English General - NSC Physical Sciences - Question 5 - 2018 - Paper 1

NO.

To calculate the average power, we use the formula:
$P = \frac{W}{\Delta t}$
where:

• W is the total work done, which is 4.80 x 10^6 J,
• \Delta t is the time taken, which is 90 s.
  So,
  $P = \frac{4.80 \times 10^6 \, J}{90 \, s} \approx 53333.33 \, W \approx 53.33 \, kW.$ Thus, the average power generated by force F is approximately 53.33 kW.

The speed of the vehicle at point B is given as 25 m·s⁻¹.

The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy. This means that the total work applied to an object results in a change in its motion.

To determine the work done by frictional forces, we first calculate the sum of potential and kinetic energy changes. The initial kinetic energy is zero since the vehicle starts from rest. The change in potential energy can be calculated as: $\Delta PE = mgh = (1500 \, kg)(9.81 \, m/s^2)(200 \, m) = 2943000 \, J.$
Then we use the work-energy principle: $W_{net} = \Delta KE + \Delta PE.$
Assuming the final total kinetic energy at point B: $W_{net} = \frac{1}{2}mv^2 - 0 + \Delta PE = \frac{1}{2}(1500)(25)^2 - 2943000.$
The total work done on the vehicle by friction, therefore, results in negative work being equal to the sum of energies calculated.