1. **Problem statement:** A train of mass $2 \times 10^{5}$ kg moves at a constant speed of 72 km/hr up an incline. The frictional force opposing the motion is $1.28 \times 10^{4}$ N. The incline rises 1.0 m vertically for every 100 m along the incline. Calculate: a) The rate of increase per second of the potential energy of the train. b) The necessary power developed by the train. --- 2. **Convert speed to m/s:** $$72 \text{ km/hr} = 72 \times \frac{1000}{3600} = 20 \text{ m/s}$$ 3. **Calculate vertical velocity component:** The vertical rise per 100 m along incline is 1 m, so vertical speed $v_{vertical}$ is: $$v_{vertical} = 20 \times \frac{1}{100} = 0.2 \text{ m/s}$$ 4. **Calculate rate of increase of potential energy (a):** Potential energy $PE = mgh$, so rate of increase is: $$\frac{dPE}{dt} = mgv_{vertical}$$ Substitute values: $$\frac{dPE}{dt} = 2 \times 10^{5} \times 9.8 \times 0.2 = 392000 \text{ J/s} = 392 \text{ kW}$$ 5. **Calculate power to overcome friction:** Power to overcome friction $P_{fric} = F_{fric} \times v$: $$P_{fric} = 1.28 \times 10^{4} \times 20 = 256000 \text{ W} = 256 \text{ kW}$$ 6. **Calculate total power developed by the train (b):** Total power $P = \frac{dPE}{dt} + P_{fric}$: $$P = 392000 + 256000 = 648000 \text{ W} = 648 \text{ kW}$$ --- **Final answers:** a) Rate of increase of potential energy = $392$ kW b) Necessary power developed by the train = $648$ kW