A train on a mountain railway is carrying 200 people of average mass 70 kg up a slope at an
angle of 30° to the horizontal and at a speed of 6 Oms. The train itself has a mass of 80 000 kg
The percentage of the power from the engine which is used to raise the passengers and the train
is 40%
What is the power of the engine?
Answers
Answer:
400kw
Explanation:
Due to increased in speed there will also increase in ke
Answer:
The power of engine is 6.9MW
Explanation:
Total mass in the system = Mass of passengers + Mass of train
Total mass in the system = 200(70) + 80000 = 94000kg
The weight acts vertically downwards. [Weight = mg] Since the slope is at an angle of 30o to the horizontal, the component of the weight acting (down) along the slope is
Component of weight along slope = (94000x9.81) sin30
Power required [output] = Work done / time = Force x velocity
To move at a constant speed of 6.0ms-1, the force provided by the engine should be equal (and opposite) to that component of the weight along the slope, such that the resultant acceleration is zero.
Force required by engine = (94000x9.81) sin30
Power output by engine = Fv = [(94000x9.81) sin30] x 6.0 = 2766420W
Efficiency = Power output / Power input = 40% = 0.4
Power input = 2766420 / 0.40 = 6.9MW