Let’s say I can throw a baseball (0.142 kg) to a maximum speed of 10 mph (4.47 m/s). The kinetic energy of the baseball is thus 1.42 Joules.

Now let’s say I’m riding a train that’s traveling at 50 mph (22.35 m/s). Relative to the earth, the baseball has 35.47 Joules of kinetic energy. I again throw the baseball to a speed of 10 mph (relative to the train). Relative to the earth, the speed of the baseball is now 60 mph (26.82 m/s), and the kinetic energy of the baseball is now 51.07 Joules.

When I was on the ground, I was only able to give the baseball 1.42 Joules of energy. Likewise, it is assumed I can only give the baseball an additional 1.42 Joules of energy while on the train. Yet on the train, it would seem I was able to give the baseball an additional 15.61 Joules of energy.

I know I must be doing something wrong here. Can anyone help me out?

Your calculations are right. What’s happening is that the energy imparted to the baseball, in the second case, can be decomposed into two parts: one due to you, and one due to the motion of the train. In relativity (even classical “Galilean” relativity where objects are moving slowly) the total energy of a system obviously depends on the frame in which it is measured.

In this case, though, we seem to have gotten energy from nowhere. But consider the various forces acting: you are imparting a horizontal force on the ball, and the floor of the train is imparting an equal force on you (preventing you from sliding backwards across the floor when you throw the ball). The work done is the integral of force times displacement, so the work done on the ball is the 15.61 J you calculated. But the train does work on you, too; if you work this out you will find that it comes to exactly 15.61-1.42 J like you’d hope. (It’s useful to work out the integral explicitly – say, for the case of a baseball under constant acceleration a, with a train of constant velocity v[sub]0[/sub] – to see how these terms contribute to the total work done.) So the extra work is being done by the locomotive, pulling against the added force of your arm as you accelerate the baseball.

Would it be correct to say that which amount of energy is released on impact also depends on the frame of reference? That is, if the ball hits the back of a passenger’s head, it will only strike with the force imparted by the throw. But if it hits something outside of the train, it will strike with the force imparted by both the throw and the train.

Though your examples are right, I wouldn’t put it that way, because in the two cases the important thing is not your choice of frame of reference but your choice of the speed at which the passenger’s head is moving. I would just say that the energy released depends on the relative impact speed (or the speeds of the two bodies in the center-of-mass frame).

Quick shot without equations:
Basically, what’s happening is that in both cases, your arm muscle gives energy to the ball. But in the second case, the train’s locomotive is also adding energy.

Think about a very light train. Very light, like a skateboard, say. If you throw the ball while standing on the unmoving skateboard, the ball will go one way, the skateboard will roll back the other way (if it’s a perfectly frictionless skateboard). Just like a mini rocket.
In just the same way, if the skateboard is rolling forward and you throw the ball forward, the skateboard will slow down a little bit (like a rocket brake).

What just happened is the ball got more energy than just your arm muscles gave it: it also got a little energy from the skateboard (and your body) slowing down. In order to stay at the same speed, something would have to push the skateboard a little bit, adding energy to the system.

In the case of a train, the engine adds a little bit of energy as it brings the train back up to speed. With a real train, it’s so heavy compared to the ball that the speed changes are tiny, but that’s where the extra energy is coming from.

Does this make sense, or should I add numbers to this?

The essential point I think the OP is missing is that kinetic energy measurements are not required to agree in value between two different frames of reference. All energy measurements within a frame of reference (either the car or the earth) must obey conservation of energy laws. The fact that the value of energy added to the baseball in one frame of reference–within the train–does not equal the value of energy added as observed by someone outside the moving train is immaterial.

If v0 is the velocity of the baseball as thrown on the train and vt the velocity of the train, obviously the velocity of the baseball as observed from the track (assume Galileian approximation) is vt+v0. The kinetic energy of the baseball as measured by the train traveler is .5m(v0)^2, but the kinetic energy as observed from the track is .5m(v0+vt)^2 = .5m(v0)^2 +.5m(vt)^2 + m(v0)(vt).

It is tempting then to subtract out the kinetic energy of the train and say the amount of energy added to the baseball by the throw is .5m(v0)^2 + .5m(vt)^2, as observed from the track. This figure does not equal the measurement of .5m(v0)^2 as made by the on-train observer; it is larger by m(v0)(vt). But that doesn’t matter because the energy measurements are made from two different frames of reference, so they are not required to be equal.

To summarize, kinetic energy does not obey Galileian transformations like velocity, which allow us to convert velocities between frames of reference. They don’t have to, because all that’s required is that energy be conserved within any frame of reference.

After subtracting the ball energy .5m(vo)^2 from the ball-on-train energy .5m(vo+vt)^2 = .5m(v0)^2 +.5m(vt)^2+m(v0)(vt), you should get .5m(vt)^2+m(v0)(vt).

Let’s say that the train is just coasting, without an engine (on a level, frictionless track, of course). If I throw that baseball forward as hard as I can, I’m going to slow down the train a little bit. Admittedly, I won’t slow it much, since we have to conserve momentum. But it’ll slow down a little, and it’s much more massive than the ball, so it’ll lose energy. In fact, it’ll lose just enough energy to make up for the extra energy being added to the ball.

Now, if instead of saying that there’s no locomotive, suppose we say that the train continues at a constant speed. In this case, the locomotive needs to add a little more energy to keep the train at speed, and the energy comes from burning a little more diesel.

Let me try again. The short answer is (as Terminus Est says, “the train.” If you accelerate forward on a moving train, the locomotive has to do more work than if you were standing still; if you accelerate backward, it does less. (Think, for example, of pulling a sled with bunch of kids standing on it. If the one in back jumps off backward, he pushes the sled forward, making it momentarily easier to pull: you’re doing less work (here I’m ignoring the fact that now there’s less weight on the sled and so less friction to overcome).)

The long answer involves writing down the equations to see who’s doing all the work. First, think of the frame in which you’re at rest. You throw a baseball of mass m horizontally, accelerating it from rest to a final speed v. For simplicity, let’s suppose that the baseball undergoes constant acceleration a for a time t=v/a, so that you’re imparting a constant force F=ma on the baseball. The work you’ve done on the baseball is W=[integral]F.ds = Fs, where s=at[sup]2[/sup]/2 is the distance over which the baseball is accelerated. So W=(ma)(at[sup]2[/sup]/2)=mv[sup]2[/sup]/2. This is the final kinetic energy of the baseball, just as it should be.

Now, consider the case in which you’re on a train moving with speed v[/sub]0[/sub]. You throw the same baseball, forward along the train, so that it has final speed v relative to you (and v[sub]0[/sub]+v relative to the ground). The work done on the baseball is the product of the force (the same force as above) and the distance, which is now the sum of your arm’s travel and the distance s[sub]0[/sub]=v[sub]0[/sub]t the train moves while you’re accelerating the baseball: W’=F(s[sub]0[/sub]+s) =(ma)(v[sub]0[/sub]t+at[sup]2[/sup]/2) =m(v[sub]0[/sub]v+v[sup]2[/sup]/2). This, again, is not surprising; it’s just the difference between the baseball’s initial kinetic energy (E[sub]0[/sub]=mv[sub]0[/sub][sup]2[/sup]/2) and its final kinetic energy (E[sub]0[/sub]=m(v[sub]0[/sub]+v)[sup]2[/sup]/2). Notice that the work W’ has two terms. The second one we recognize as the work W our arm did above. The first term, then, must be the work done by the train. Let’s check this. The train exerts the same horizontal force F on you that you exert on the baseball (as friction, preventing you from sliding across the floor). It exerts this force over the time t that you are accelerating the baseball, during which it travels a distance v[sub]0[/sub]t. So the train does work F(v[sub]0[/sub]t)=mav[sub]0[/sub]t=mv[sub]0[/sub]v, just right to add the extra energy to the baseball.

Well, that depends on (as Chronos said) whether the train has to travel at a constant speed or not. If not, then the train slows down; its new velocity is the square root of (28.36 / M), where M is the mass of the train.

One way or the other, the train loses 14.18 joules of energy.

No, that extra fuel was burned accelerating the ball from rest with respect to the ground to 50mph with respect to the ground. If you were to drop the ball from the train it would remain travelling at 50mph.

No, Crafter_Man is right. The energy used to accelerate the ball from rest was the 35.47 J of #3. The 14.18 J is, as he says, the train’s contribution to the added energy when the ball is accelerated.

Don’t forget that the Earth is rotating; in the case of your ground-based example, when you throw the ball toward the east, it is actually travelling at 10mph plus several hundred miles per hour, when you throw it to the west, it is travelling 10mph minus several hundred miles per hour - all of this is relative to the position of an alien observer hovering in his spacecraft directly between the sun and the Earth - it’s all relative - Earth, train, spacecraft, galaxy - the question ‘how fast is this object moving’ only becomes meaningful when the question ‘relative to what?’ is answered.

So what is kinetic energy? You could think of it as the potential for work to be done by two objects moving relative to one another; it makes no difference whether the Earth is stationary and the asteroid smashes into it at ten thousand miles per second or if the asteroid is stationary in the path of the Earth that happens to be hurtling along at ten thousand miles per second, smashes into the meteorite and (because of its large mass) continues on its way - the same amount of energy is released in either scenario. In fact without an external frame of reference, the two scenarios are utterly indistinguishable.