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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?

Qmechanic
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user2800708
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    The energy would have come form the kinetic energy of the plane or, if the plane would stabilize its speed relative to the ground, from the engines. – CuriousOne Jan 17 '16 at 10:56
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    The formula for kinetic energy is $E=\frac12 mv^2$. – Steeven Jan 17 '16 at 13:17
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    If you're going to compare it to walking down the pavement, you should consider what would happen if you were to try walking on the outside of the plane... – Darrel Hoffman Jan 17 '16 at 17:48
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    You don't even need to consider a plane. You are walking on Earth, which rotates with speed up to 1670 km/h (depending on your latitude), orbiting the Sun at nearly 30 km/s (and orbiting the centre of galaxy, but I don't know what that speed is; probably even larger though). – Jan Hudec Jan 18 '16 at 05:54
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    "I needed 500 times the energy to do it." No, the plane's engines did that for you. You felt it as acceleration on takeoff pushed you a little firmer into your seat. – Lightness Races in Orbit Jan 18 '16 at 14:39
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    People walking down the aisle of a plane travelling at 250 m/s could, if the plane collided with a cliff, suddenly stop travelling at 250 m/s. That's when they're going to find walking down the aisle suddenly difficult. – Jon Hanna Jan 19 '16 at 03:53

5 Answers5

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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).

hmakholm left over Monica
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    I think the point about the static observer seeing the plane slow down is slightly misleading (though correct). Momentum is conserved so the velocity of the center of mass of the plane/you ensemble is constant. The appearance of slowing down is due to the "you" part of the ensemble's mass accelerating forward so the remaining mass has to accelerate backwards (the center of mass does not slow down). When you came to a stop, the plane would return to it's original speed. Perhaps a team of sprinters could cause a small plane to momentarily stall if they all ran up the aisle simultaneously! – ejrb Jan 18 '16 at 17:15
  • missed the point. – SkipBerne Jan 18 '16 at 17:51
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    @ejrb: The question here was "where does the passenger's increase in kinetic energy (as measured by the observer on ground) come from, given that it's not all supplied from chemical energy in his muscles?" My answer is that the same amount of energy is removed from the plane. Once the passenger stops (wrt the plane), this kinetic energy goes back into the plane. – hmakholm left over Monica Jan 18 '16 at 18:30
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    @ejrb Yes, but the point is that they would see the fuselage and wings slow down slightly (alternatively, the engine works harder, and more exhaust is accelerated out the back than usual). One assumes that the observer on the ground doesn't have a way to directly measure the position of the center of mass of the whole system of the plane+passengers. – Random832 Jan 18 '16 at 18:34
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    "as long as relativistic effects can be ignored" is misleading. Relativistic effects would not change this fact at all. That was the entire basis for SR. It seems you have forgotten the time dilation/length contraction terms in the Lorentz boost. – Aron Jan 19 '16 at 05:20
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    @Aron: That parenthesis was meant to forestall unproductive discussion about how one even counts the chemical energy released by the muscles in the relativistic case. There's at least an argument that the chemical energy is subject to relativistic mass increase by a factor of $\gamma$, such that in the ground frame we should say the energy released by the passenger's muscles is $\gamma$ times what it is in the comoving frame. – hmakholm left over Monica Jan 19 '16 at 09:49
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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.

John Rennie
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  • Very clear answer – innisfree Jan 17 '16 at 11:45
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    One could equally use the sun frame, for example, and ask the same question of us, on Earth, moving 100,000km/h around the sun. Or the galactic frame, etc. – J... Jan 17 '16 at 14:53
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    Would it be correct to say that for the ground observer, 25000 of those joules are work done by the plane and only the last 50 are done by the passenger? – Ixrec Jan 17 '16 at 21:17
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    If the ground observer has an unrealistically powerful electron microscope pointed at the plane, they will see your muscles performing chemical reactions to release only around 50 J of energy, correct? – user253751 Jan 17 '16 at 21:45
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    @immibis That sounds like an electron telescope. :-) – David Richerby Jan 17 '16 at 22:54
  • so $\Delta KE=W$ and not $KE=W$, that makes a great difference, thanks – user153330 Jan 17 '16 at 23:33
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    I don't think this gets at the question the user wanted answered. The user is wondering how, for a ground observer, you gain $25050$ J of kinetic energy when that energy is clearly not coming from your muscles. I think it's true that the energy you gain mostly comes from the plane as Makholm says below. – user35734 Jan 18 '16 at 00:19
  • Yes, its a good answer, but I'm still not sure how to calculate how much of that 25050J comes from my muscles, and how much from the plane? I presume, that it mostly comes from the plane. – user2800708 Jan 18 '16 at 09:19
  • @user2800708 It's the very first thing in the answer - the amount of energy required in the rest frame of the plane (though to be nitpicky, the amount of energy your muscles have to expend is much bigger, since muscles are rather inefficient). Since kinetic energy isn't linear, it's rather tricky to work with it directly in a more complicated scenario - using momentum (which is linear) is a lot simpler. When you figure out the momentum transfer, it's trivial to calculate the corresponding energy in a different rest frame. – Luaan Jan 18 '16 at 10:34
  • @user2800708: good point! I've updated my answer to address the point you raise. – John Rennie Jan 18 '16 at 11:22
  • you missed the obvious. – SkipBerne Jan 18 '16 at 17:51
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    @SkipBerne: do elaborate ... – John Rennie Jan 18 '16 at 17:56
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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.

Guill
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  • So, what if I change my velocity in increments of 1 m/s, from 0 to 250? By your reasoning the energy required would be proportional to (m * 1) * 250. When in fact the resultant kinetic energy would be proprtional to m * 250 * 250. I'm not convinced by your delta v equation. – user2800708 Jan 21 '16 at 11:31
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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.

wendy.krieger
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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.

SkipBerne
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    Could you explain how your observations about gravity and friction account for the change in KE the OP is asking about? – Brionius Jan 18 '16 at 17:54
  • that assumes that the kinetic system is relevant to a traction issue. which it is not... case in point the ISS. – SkipBerne Jan 18 '16 at 18:03