How am I able to stand up and walk down the aisle of a flying passenger jet? The energy of a moving object is $E = mv^2\;.$ That is it increases with velocity squared.
I walk at say 3 miles per hour, or lets round that down to 1 meter per second for a slow walk. I weigh less than $100~\mathrm{kg}\;,$ but lets just round that up to $100~\mathrm{ kg}$ for convenience (it is just after Christmas).
So when I walk along the pavement, I have $100~\mathrm{kg\; m^2 s^{-2}}$, 100 joules of kinetic energy.
Now I get on a passenger jet, which is cruising at around 500 knots, call that 250 meters per second.
In my seat I have $100\times 250^2 = 6250000$ joules of kinetic energy.
But when I walk down the aisle I have $100\times 251^2 = 6300100$ joules of kinetic energy.
The difference between these is: 50100 joules.
It feels the same to me, walking down the pavement as walking down the aisle of the plane. I didn't have to make some huge effort to get up to speed on the plane, yet I needed 500 times the energy to do it.
How is this possible, and where did the energy come from?
 A: Due to momentum being conserved, when you accelerate yourself forwards relative to the plane, the tangential force you're applying to the floor will accelerate the rest of the plane backwards. Since the plane has a lot more mass than you, its velocity will not change by very much.
Thus, an inertial observer who was initially at rest with respect to the plane (and you) will see both you and the plane gain kinetic energy (due to your muscle work). The vast majority of the additional kinetic energy goes into you, though.
However, an observer on the ground will see the rest-of-the-plane slow down slightly, which means that it loses quite a bit of kinetic energy due to its large mass and velocity. This loss of kinetic energy from the plane cancels out the additional kinetic energy the ground observer thinks you gained, so the ground observer's energy ledger still balances.
(Mathematically, to the ground observer $v$ is, to a first approximation, both the ratio between your your gained momentum and your gained kinetic energy, and the ratio between the plane's lost momentum and its lost kinetic energy. So conservation of momentum leads to conservation of total energy, to first order. The term that comes from your muscle work is a second-order effect).
Both observers agree on the amount of energy your muscles contribute (at least as long as relativistic effects can be ignored).
A: Kinetic energy is not invariant under Galilean transformations. To see this consider the following:
In the rest frame of the plane you apply a force $F$ of 100N for one second to accelerate yourself to 1 m/s. During this time you move a distance $d$ of 0.5m so the work done is:
$$ W = Fd = 100 \times 0.5 = 50\,\text{J} $$
This of course is equal to your kinetic energy of:
$$ E = \tfrac{1}{2}mv^2 = \tfrac{1}{2} \times 100 \times 1^2 = 50\,\text{J} $$
The observer on the ground sees you applying a force of 100N for one second, but because the plane is moving at 250m/s the ground observer sees you move a distance of 250.5m. Therefore the work done is:
$$ W = Fd = 100 \times 250.5 = 25050\,\text{J} $$
For the ground observer your initial KE, before you started walking, is:
$$ E = \tfrac{1}{2}mv^2 = \tfrac{1}{2} \times 100 \times 250^2 = 3125000\,\text{J} $$
And your kinetic energy after you've reached a speed of 1m/s is:
$$ E = \tfrac{1}{2}mv^2 = \tfrac{1}{2} \times 100 \times 251^2 = 3150050\,\text{J} $$
So the change in your kinetic energy is:
$$ \Delta KE = 3150050 - 3125000 = 25050\,\text{J} $$
And as before this is equal to the work done.
Response to comment:
user2800708 points out, quite reasonably, that your muscles only produced 50J so if the ground observer sees your kinetic energy change by 25050J where did the rest of the energy come from?
The answer is that when you propel yourself forwards with a force of 100N you propel the plane backwards with a force of 100N. So in order to keep its speed constant at 250 m/s the planes engines have to provide an extra 100N of thrust. In the one second we watch the plane it moves 250m, so the extra work done by the plane's engines is:
$$ W_\text{plane} = Fd = 100 \times 250 = 25000\,\text{J} $$
Add this to the 50J provided by your muscles and we get the 25050J that we calculated above.
A: Because gravity works on you based on your distance from other masses. That changes little when you are at 30K', but measurably slightly less. This still forces your feet against the floor and friction coef. allows for traction. 
a Smart Academic answer would be because the "Captain has turned off the seatbelt sign". and you have legs. 
the energy came from 2 or more Pratt and Whitney Turbo jet engines that run on money.
your velocity is the same as the plane because of the structural strength of the plane. otherwise you would be a bloody pulp in a debris field, like many unsuccessful attempts. 
Note you can't do that in orbit.. like on the ISS for the very same reason. 
any body who thinks otherwise is missing the big picture and has a tenuous understanding of Newtonian physics. to say the least.
A: You, and the air and everything inside the aircraft, are travelling at the speed of the aircraft, and your motion is relative to that.
Lest there be turbulance, you would certainly no longer be moving relative to the speed of the aircraft, and you would be accelerated by the difference.  That's why they have seat belts.
A: The reason for the "apparent" confusion, is that you ere, inadvertently,  changing the frame of reference! In addition, the formula you are using is not correct for the cases in question.  The energy being calculated, is the energy required to make a change in velocity $$E = m(\Delta v)^2 \  were\   \Delta v = v - v_o$$ For the case "walking on ground" $v_o \  is \  0, and \  v = 1  $, therefore $\Delta v = 1 - 0 = 1.$
For the case "walking on plane" $v_o \ is \ 250, and \  v = 251 $, therefore $\Delta v = 251 - 250 = 1 $.  
As you can see, the amount of energy required from your muscles, is the same in both cases (same mass and velocity).  That's why "it feels the same" to you, and no "huge effort" (additional energy) is required.
