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if a train (mass of 2500 kg) exerts a force of 15,000 N forward on a level track

ID: 1335218 • Letter: I

Question

if a train (mass of 2500 kg) exerts a force of 15,000 N forward on a level track, however a coefficient of friction exists between the train and track of .26, find the time it would take the train to accelerate up to 27.6 m/s if a train (mass of 2500 kg) exerts a force of 15,000 N forward on a level track, however a coefficient of friction exists between the train and track of .26, find the time it would take the train to accelerate up to 27.6 m/s if a train (mass of 2500 kg) exerts a force of 15,000 N forward on a level track, however a coefficient of friction exists between the train and track of .26, find the time it would take the train to accelerate up to 27.6 m/s

Explanation / Answer

Here ,

mass , m = 2500 Kg

F = 15000 N

let the acceleration is a

Using second equation of motion

15000 - .26 * 2500 * 9.8 = 2500 * a

a = 3.452 m/s^2

Now , for final speed , v = 27.6 m/s

Using first equation of motion

27.6 = 3.452 * t

t = 8 s

the time taken for the train to accelerate is 8 s