# Work done walking on moving train

But let's say I walk forward/accelerate on a moving train, train at a constant speed. And there is an observer sitting watching this outside. They would see friction doing positive work on me relative to the ground?

This is correct. The power, $P$, of a contact force, $\vec F$, is given by $P = \vec F \cdot \vec v$ where $\vec v$ is the velocity of the material at the point of contact. Since $\vec F$ is in the same direction as $\vec v$ the power is positive meaning that positive work is done on the person.

The train is moving constant. The energy the person gains comes from internal. So why does it look as if friction from the train to the person is doing work. I'm confused, what am I seeing. This makes no sense to me

Actually, not all of the energy that the person gains is coming from the internal energy in the ground frame. Assuming a perfectly efficient conversion of chemical potential energy into mechanical energy, the amount of mechanical energy gained (in the ground frame) by the person is greater than the amount of chemical energy lost (which is Galilean invariant). The difference is precisely the positive work done by the contact force.

By Newton's 3rd law there is an equal and opposite contact force operating on the train. This force is in the opposite direction of $\vec v$ so negative work is done on the train. Since the train is travelling at a constant speed that means that its engine (unsurprisingly) must be supplying power.

So the mechanical energy gained by the person comes both from the internal conversion of chemical energy to mechanical energy and also from the energy of the train through the work done by the friction force in this frame. To see this quantitatively it is useful to consider a simplified example, such as a spring or something similar.

The train is moving at constant velocity. The energy the person gains comes from internal energy, like chemical potential energy. So why does it look as if friction from the train to the person is doing work. I'm confused, what am I seeing. This makes no sense to me.

Let's imagine a person at rest with respect to the surface they're standing on, and they push off that surface to move forward at a speed of 1m/s forward (relative to the surface).

Using a ground reference, this is pretty simple. Assume the person is 50kg, then the minimum energy for the step is: $$\Delta E = E_f - E_i$$ $$\Delta E = 0.5 (50\text{kg})(1 \text{m/s})^2 - 0 = 25\text{J}$$

Now let's imagine this happening on a moving train. We suppose the train is already at a speed of 5m/s. What's the energy change now?

$$\Delta E = 0.5 (50\text{kg})(6 \text{m/s})^2 - 0.5 (50\text{kg}) (5 \text{m/s})^2 = 275\text{J}$$

So when looking at the interaction from a different reference frame, the person has added much more energy. We resolve this problem by saying that besides the work the legs are doing, the train is also doing work on the person.

If the person took 1 second to make the step, then the force off the surface must be $F = ma = (50\text{kg})(1\text{m/s}^2) = 50 \text{N}$.

In the ground frame, we would say the train's work (or the work from friction) was therefore $$W = Fd = Fvt = (50\text{N})(5\text{m/s})(1s) = 250\text{J}$$

And that 250J exactly accounts for the difference in energy between the two cases.

If we just looks at this happening, walking inside the train, why don't we have to account for this other 250J energy? Shouldn't we have too? That is my confusion, when we are on earth walking why dont we have to account for any of these affects?

You have to account for it if you care about the actual energy of the train. This 250J came from the train. Now the massive train has a huge amount of KE, so this 250J will be a tiny fraction of that amount. But the faster the train is going, the more energy this change in speed represents.

Since the earth is even larger, as long as we're in a frame where its speed is small, we can usually ignore the energy exchanged with it.