A train of mass 19,000 kg has its speed changed from 45 km/h to 70 km/h in 32 secs. At what average rate does the engine work?

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txmedteach | High School Teacher | (Level 3) Associate Educator

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To answer this question, we need to connect the information to average power, which is done in terms of the rate of change of energy over time. This gives us the following equation:

Power = Joules/sec = (E2-E1)/32 sec

To find E2, the final energy level, and E1, the initial energy level, we need to find the final and initial total energies. Let's first assume that potential energy does not change (there is no information given about height or any other form of potential energy), so that all we need to worry about is kinetic energy.

To find the initil and final kinetic energies, though, we need to use the right units! We need to convert both velocities to meters per second:

V1: 45 km/h * 1 h/3600 sec * 1000 m/1 km = 25/2 m/s

V2: 70 km/h * 1 h/3600 sec * 1000 m/1 km = 175/9 m/s

Now that we have our velocities, we can find the initial and final kinetic energies:

E1 = 1/2 M V1^2 = 1/2*19000*(25/2)^2 = 1.484 * 10^6 J

E2 = 1/2 M V2^2 = 1/2*19000*(175/9)^2 = 3.592 * 10^6 J

To calculate the average rate at which the engine works, we now just find the difference in these two energies and divide it over the time spent working:

Power = (3.592 * 10^6 - 1.484 * 10^6)/32

Power = 65,875 Joules/sec = 65,875 Watts

There is your final answer! I hope that helps!


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