Anthony Pitt
Mar 9
A train of mass 330 tonnes attains a speed of 30 km/h from rest is 11 km. Find the excess of tractive force over resistance, both assumed constant?
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ChatGPT
Mar 9
First, we need to convert the speed from km/h to m/s:
30 km/h = 30,000 m/3600 s = 8.33 m/s
Next, we calculate the kinetic energy of the train:
KE = 0.5 * mass * velocity^2
KE = 0.5 * 330,000 kg * (8.33 m/s)^2
KE = 11,041,725 J
Now, we calculate the work done by the tractive force:
Work = KE
Work = 11,041,725 J
Since the train started from rest, the work done by the tractive force is equal to the excess of tractive force over resistance. Therefore, the excess of tractive force over resistance is 11,041,725 J.
30 km/h = 30,000 m/3600 s = 8.33 m/s
Next, we calculate the kinetic energy of the train:
KE = 0.5 * mass * velocity^2
KE = 0.5 * 330,000 kg * (8.33 m/s)^2
KE = 11,041,725 J
Now, we calculate the work done by the tractive force:
Work = KE
Work = 11,041,725 J
Since the train started from rest, the work done by the tractive force is equal to the excess of tractive force over resistance. Therefore, the excess of tractive force over resistance is 11,041,725 J.
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