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thua
Oct16-05, 07:35 PM
When the Crampton coal-fired train engine was built in 1852, its mass was 48.3 t (1.0 t = 1.0 x 10^3 kg) and its force capability was rated at 22.4 kN. Assuming it was pulling train cars whose total mass doubled its own mass and the total friction on the engine and cars was 10.1 kN, what was the magnitude of the acceleration of the train?
What I have: (not sure if it's right though)
m = 48.3 t = (48.3 x 10^3 kg) x 3 = 144.9 x 10^3 kg
net force = 22.4kN - 10.1 kN = 12.3 kN = 12300 N
acceleration = ?
acceleration = net force / mass
= 12300 N / 144.9 x 10^3 kg
acceleration = 0.0849 m/s^2
acceleration = 8.49 x 10^3 kg
Is this right? because 0.0849 m/s^2 doesn't seem very reasonable to me. If it's wrong, could someone please help?

whozum
Oct16-05, 07:38 PM
Your procedure looks perfectly fine to me.

HallsofIvy
Oct16-05, 07:44 PM
Were you expecting a train in 1852 to take off really fast?

Notice that in only 100 seconds (1 min 40 seconds), the train will be traveling over 8 m/s. Now that's a pretty good pace!

By the way:
"acceleration = 0.0849 m/s^2" is correct.

"acceleration = 8.49 x 10^3 kg" is nonsense. I presume you meant
"8.49 x 10^(-2) m/s^2".

Pengwuino
Oct16-05, 08:15 PM
Yah, trains back then could go... 40mph at the most probably back then.