# Energy related question

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

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

Thanks, any help is greatly appreciated!

## The Attempt at a Solution

Related Introductory Physics Homework Help News on Phys.org
tiny-tim
Homework Helper
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! You know from a) how many joules you need, so for c) just do the arithmetic, making allowance for the megajoules and the 30%! Hi tiny-tim,

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?

tiny-tim
Homework Helper
Hi girlinterrupt! 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? Thanks for the help and tips! I Appreciate it!