# Work done changes between reference frames?

(This is not homework; a friend shared with me this puzzler and neither of us can figure it out.) Suppose you are in a plane traveling at velocity $v_1$ relative to the ground. The flight attendent pushes a cart of mass, say, $m$, accelerating it from rest to $v_2$, relative to the plane (so relative to the ground, its velocity goes from $v_1$ to $v_1+v_2$). From your perspective, the work done is then $$W = \Delta E = \dfrac{1}{2}mv_2^2$$ but from the perspective of the ground, the work done is $$W = \Delta E = \dfrac{1}{2}m(v_1+v_2)^2 - \dfrac{1}{2}mv_1^2 = \dfrac{1}{2}mv_2^2 + mv_1v_2.$$ Why are these quantities unequal?

I've seen some similar problems, like a body colliding with a wall in different reference frames, but I don't have the physics experience to tell if that question is analogous to this one. So, my apologies if this is a duplicate. :)

• yes, both work and kinetic energy are dependent of the reference frame – user83548 Dec 12 '15 at 1:13
• What do the flight attendant's feet push on? – Daniel Griscom Dec 12 '15 at 1:28
• What Daniel says. Imagine a light boat on still water instead of a plane. To push an object and keep the boat still relative to the water, work must be done to compensate for the reaction on the boat. Same for keeping a boat/plane moving at constant velocity. – udrv Dec 12 '15 at 2:05
• This is a duplicate of this question if you just replace the waitress pushing a cart with a gun firing a bullet. – Brionius Dec 12 '15 at 5:31

There is a subtlety involved here. Let's phrase the problem this way: Say that the flight attendant has an amount of energy E to accelerate his cart to a speed $v_2$ with respect to the plane. If he does that while the plane is sitting on the ground, the total energy of the cart is $\frac{\text{mv}_2^2}{2}$. If he does that when the plane is in the air flying at speed $v_1$, then the total energy of the cart is $\frac{\text{mv}_2^2}{2}+\text{mv}_1 v_2$ which is greater than the total energy of the cart in the first case even though the attendant did the same amount of work in both cases. So we seem to have a paradox.
The key to unraveling the paradox is to recognize that when the flight attendant pushes the cart forward, he is pushing the rest of the plane backwards by some small velocity. That small decrease in the plane's velocity would seem to be so infinitesimal that it could be ignored, but when you consider the fact that the kinetic energy of the plane is proportional to its (very large) mass, even a small decrease in the plane's velocity can play a significant role here when considering the changes in energy. You'll find that the decrease in the kinetic energy of the plane cancels out the problematic $\text{mv}_1 v_2$ term, so that the total energy imparted to the entire system (plane+cart) by the attendant is the same in both cases.
Or, another way to look at it is if you require that the plane maintains the exact same velocity $v_1$ despite the fact that the flight attendant suddenly decides to push his cart forward at a velocity $v_2$, then the plane's engines are required to work a little harder briefly in order to maintain velocity $v_1$ against the force exerted by the flight attendant's feet in trying to push the rest of the plane slightly backwards when he pushes the cart forward. That additional energy provided by the plane's engines works out to be $\text{mv}_1 v_2$ which is the additional energy that you were wondering about in your originally stated problem.
Try working out the problem yourself in full and taking into account that the plane has some large (but finite) mass M and the cart has a smaller mass m. Assume that the plane has some high velocity $v_1$ and that an amount of energy is suddenly provided which propels the cart forward at a velocity $v_2$ with respect to the cart+plane center of mass. Using conservation of momentum, see how much the plane slows down as a result. Then calculate the total before and after energies and see how they compare.