Average Power Output for 260000 kg Jet at 9200m Altitude

AI Thread Summary
The discussion centers on calculating the average power output of a 260,000 kg jet after it reaches a cruising altitude of 9,200 meters and a speed of 235 m/s. The initial kinetic energy is zero, and the final kinetic energy is calculated as 7,179,250,000 Joules. However, the calculation neglects the change in potential energy, which is crucial for determining the total work done. Participants suggest incorporating both the change in kinetic energy and potential energy into the work equation to find the correct power output. The conversation emphasizes the importance of considering all forms of energy in such calculations.
spockjones20
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Homework Statement



620 seconds after starting its take-off, a 260000 kg jet reaches its cruising altitude of 9200 meters and flies with a speed of 235 m/s. Neglecting friction, what has been the average power output of the engines?

Homework Equations



ΔKE = Work
Power = Work/time

The Attempt at a Solution



Initial KE = 0
Final KE = .5*260000*235^2 = 7179250000
Power = 7179250000/620 = 11579435.48 Watts

Can someone see an issue with what I am trying to do here? I tried to type in this answer but apparently it is wrong.
 
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spockjones20 said:

Homework Statement



620 seconds after starting its take-off, a 260000 kg jet reaches its cruising altitude of 9200 meters and flies with a speed of 235 m/s. Neglecting friction, what has been the average power output of the engines?

Homework Equations



ΔKE = Work
Power = Work/time

The Attempt at a Solution



Initial KE = 0
Final KE = .5*260000*235^2 = 7179250000
Power = 7179250000/620 = 11579435.48 Watts

Can someone see an issue with what I am trying to do here? I tried to type in this answer but apparently it is wrong.

I haven't checked your math, but I can see you left out the change in potential energy.
 
Ok. Would that go into the work equation?
 
spockjones20 said:
Ok. Would that go into the work equation?

I believe so.

Try this: ΔKE + ΔPE = Work

and see what happens.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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