How Does Jet Engine Thrust Affect Train Kinetic Energy?

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SUMMARY

The forum discussion centers on calculating the work done by a jet engine on an experimental train and its effect on the train's kinetic energy. The train, with a mass of 2.50 x 104 kg, is propelled by a thrust of 5.00 x 105 N over a distance of 509 m, resulting in a work done of 2.545 x 108 J. This work is equivalent to the change in kinetic energy, confirming that the work-energy principle applies. The final speed of the train can be determined using the equations of motion, specifically v(f)2 = v(i)2 + 2ad, where acceleration is derived from the thrust-to-mass ratio.

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  • Knowledge of kinematic equations
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Homework Statement


In the 1950s, an experimental train that had a mass of 2.50 x 10^4 kg was powered across the level track by a jet engine that produced a thrust of 5.00 x 10^5N for a distance of 509m.

a)Find the work done on the train
b)Find the change in kinetic energy
c)Find the final kinetic energy of the train if it started from rest
d)Find the final speed of the train if there were no friction.

Homework Equations





The Attempt at a Solution



a)
W=fd
W= (5.00 x10^5N)(509m) = 2.545 x 10^8

b)
I'd typically find the change in KE when I have differences in velocity b/c of the equation KE=(1/2)mv^2, but I don't have v in this equation and I don't know how to find it when I don't have time (t).

I know KE changes when there is change in height, but the train runs on a level track, does that mean there is none?

Then in my review book it says W= KE initial - KE final = change in KE
Does that mean the work I found done on the train is the same as the change in KE involved? I don't know...

c)
I used v(f)^2 = v(i)^2 = 2ad
v(f)^2= 2(9.8)(509)
=99.88

d)
I really don't know how to approach this one...
 
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Answer is the same for a,b and c.
For (d), find acceleration a = thrust / mass
then use the formula v(f)^2 = v(i)^2 + 2ad to find v(f)
 
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Welcome to PF.

Yes the work done will result in an increase in kinetic energy.

The problem generally wants you to ignore frictional forces.
 
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