Energy related question

  • #1
Hi,

Can someone please check my answers and possibly help with part c; I don't know where to begin on solving it.

Homework Statement



A 31000kg jet airliner is flying at 11000m doing 910 km/h.

(a) What is the jet’s gravitational potential energy (GPE) at 11000m? (GPE=0 at sea level)
(b) What is the jet’s kinetic energy (KE)?
(c) If the jet fuel used contains 30.3MJ of energy per litre and the jet engines are 30% efficient at converting the fuel’s
energy, how much fuel is used just to raise the jet’s potential from sea level to 11000m


The Attempt at a Solution


What is known:

m= 31000kg
h= 11000m
g=9.8m/s
v = 910km/h
e = 30.3 MJ/L

Answer:
a)
Gravitational Potential Energy:
U = mgh
U = 31000 x 9.8 x 11000
U = 3341800000 J = 3.3 x 10^9 J

b)

K = 1/2 mv^2
K = 1/2 31000 x 910000^2
K = 12835550000000000 = 1 x 1016 J

c) I have no idea where to start with this one!

Thanks, any help is greatly appreciated!

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
251
Welcome to PF!

A 31000kg jet airliner is flying at 11000m doing 910 km/h.

(a) What is the jet’s gravitational potential energy (GPE) at 11000m? (GPE=0 at sea level)
(b) What is the jet’s kinetic energy (KE)?
(c) If the jet fuel used contains 30.3MJ of energy per litre and the jet engines are 30% efficient at converting the fuel’s
energy, how much fuel is used just to raise the jet’s potential from sea level to 11000m
Hi girlinterrupt! Welcome to PF! :smile:

You know from a) how many joules you need, so for c) just do the arithmetic, making allowance for the megajoules and the 30%! :smile:
 
  • #3
Hi tiny-tim,

Firstly, thanks for the reply!

I'm finding it difficult to work out what the question was actually asking... im so new to physics.

so basically if I think of the 30.3MJ per litre at 100% efficiency, I can work out the energy required at 30% efficiency... the thing that has thrown me in the question is "how much fuel is used"... but i suppose it is valid to equate it in terms of this...

30300000 J per litres of fuel * 3.3 (efficiency) = 99990000 J / Litre of Fuel

Sound good?
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
251
Hi girlinterrupt! :smile:
so basically if I think of the 30.3MJ per litre at 100% efficiency, I can work out the energy required at 30% efficiency... the thing that has thrown me in the question is "how much fuel is used"... but i suppose it is valid to equate it in terms of this...

30300000 J per litres of fuel * 3.3 (efficiency) = 99990000 J / Litre of Fuel

Sound good?
That's it!

Just two tips …

i] use 10^ or mega … if you write all those 0s, sooner or later you'll make a mistake

ii] why approximate, with 3.3 … the question deliberately makes the arithmetic easy, so why not use 10/3 exactly? :smile:
 
  • #5
Thanks for the help and tips! :smile:

I Appreciate it!
 

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