- #1
A_Sushi
Can anyone help with the following question, please?
1.
A 60,000kg train is being pulled up a 1 percent grade (it rises 1m for each horizontal 100m). The friction force opposing the motion of the train is 4,000N. Initial speed of the train is 12 m/s. Through what horizontal distance will the train move before its speed is reduced to 9m/s?
3.
The answer suppose to be 191.2 m. I got 472.5 m as follows:
Fd=m*(v2-v2)/(2*distance)
4 000 = (60 000*(92-122)) / (2*distance) = 472.5 m
1.
A 60,000kg train is being pulled up a 1 percent grade (it rises 1m for each horizontal 100m). The friction force opposing the motion of the train is 4,000N. Initial speed of the train is 12 m/s. Through what horizontal distance will the train move before its speed is reduced to 9m/s?
3.
The answer suppose to be 191.2 m. I got 472.5 m as follows:
Fd=m*(v2-v2)/(2*distance)
4 000 = (60 000*(92-122)) / (2*distance) = 472.5 m
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