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Homework Help: Energy related question

  1. Jul 31, 2008 #1
    Hi,

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

    1. The problem statement, all variables and given/known data

    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


    3. 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!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 31, 2008 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    Welcome to PF!

    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:
     
  4. Jul 31, 2008 #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?
     
  5. Jul 31, 2008 #4

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi girlinterrupt! :smile:
    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:
     
  6. Jul 31, 2008 #5
    Thanks for the help and tips! :smile:

    I Appreciate it!
     
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