Tuesday, July 11, 2017

newtonian mechanics - 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?




Answer



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).


No comments:

Post a Comment

classical mechanics - Moment of a force about a given axis (Torque) - Scalar or vectorial?

I am studying Statics and saw that: The moment of a force about a given axis (or Torque) is defined by the equation: $M_X = (\vec r \times \...