How Does Jet Engine Thrust Affect Train Kinetic Energy?

AI Thread Summary
The discussion focuses on calculating the work done on an experimental train powered by a jet engine, which produces a thrust of 5.00 x 10^5 N over a distance of 509 m. The work done on the train is calculated as 2.545 x 10^8 J, which directly relates to the change in kinetic energy, confirming that the work done results in an increase in kinetic energy. The train's final kinetic energy can be determined using the work-energy principle, and the final speed can be calculated by first finding acceleration from the thrust and then applying kinematic equations. The conversation emphasizes ignoring frictional forces for these calculations. Overall, the relationship between thrust, work, and kinetic energy is central to solving the problem.
citypigeon
Messages
1
Reaction score
1

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...
 
  • Like
Likes salma-2022
Physics news on Phys.org
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)
 
  • Like
Likes salma-2022
Welcome to PF.

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

The problem generally wants you to ignore frictional forces.
 
  • Like
Likes salma-2022
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top