# Work done walking on moving train

Sometimes a system's internal chemical energy is converted into mechanical energy and thermal energy with no work being done on the system by external forces. For example, at the beginning of this section we described the energy conversions that take place when you start running. In order to move forward, you push back on the floor and the floor pushes on you with a static frictional force. This force causes you to accelerate, but it does not do work. It does no work because the displacement of the point of application of the force is zero (assuming your shoes do not slip on the floor). Because no work is done, no energy is being transferred from the floor to your body. The kinetic energy increase of your body comes from the conversion of chemical energy derived from the food you eat.

So when we walk on the ground, friction does no work, only providing grip so as our chemical energy can become kinetic energy.

But let's say I walk forward/accelerate on a moving train, with the train moving 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?

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.

Edit/picture is to end question about "does friction do work walking". This questions is not about that. I am trying to understand how in one frame, lets say walking, friction does no work, but in another frame, it appears it does. Maybe I am just confusing reference frames... not sure

• Kevin, this feels like Deja Vu. Have we been down this path before?. Feb 7, 2020 at 14:18
• Kevin, please site your source for the quote in your answer. Also, images are not accessible to all users. I have typed out the text image you had this time, but in the future, please make sure to type out the text and site your sources. Feb 7, 2020 at 18:39

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.

• @KevinCSpeltz in the ground frame the work done by friction on a person walking on the ground is zero since $v=0$ at the point of contact implies $P=F\cdot v=0$. This case is different because $v\ne 0$. Yes, both the energy from the train and from the internal energy must be accounted for. Usually classes on Newtonian physics focus on fairly narrow scenarios, so this type of thing simply never comes up. They have only so much time in class, so they have to choose what to exclude, and static friction with moving objects is one that gets excluded in the interest of more pressing topics.
– Dale
Feb 7, 2020 at 16:00
• @Dale "The difference is precisely the positive work done by the contact force." Just so we understand, you are acknowledging that positive work is being done by the static friction contact force. Correct? Feb 7, 2020 at 17:20
• @Dale Thanks. I think we are on the same page now. Feb 7, 2020 at 20:10
• @Dale, I get your point, you cannot feel energy. I guess what gets me is the force is the same. Thought maybe there would be another force. But really that is why we still have the same acceleration and thru that same force, the train/engine is providing and transferring the energy simultaneously unknown to the walker. Correct? Feb 14, 2020 at 17:20
• Have a good weekend. Thank you Feb 14, 2020 at 17:39

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.

• Bowlofred. Thank you. One question, the last part you calculate the 250j, why did you use 5 m/s becuase isn't it technically speeding up? Feb 7, 2020 at 20:58
• We're talking about the force of friction. In the ground frame, the object applying that force (the walking surface) is moving at a constant speed of 5m/s. The person's body is speeding up, but that's not where the force is applied. Feb 7, 2020 at 22:34
• Good point. I wish I knew more. I feel like I'm jumping into the middle of physics that's deeper than I learned. Thank you for help Feb 7, 2020 at 22:46
• Bowlofred and Dale, if this is true, couldn't the same be said that this is happening to earth, like it is the train in this situation? If so, why don't we see the affect? I assume like most things on Earth, it is too large to notice and overall generally sums to zero in the grand scheme. Feb 8, 2020 at 13:54
• This example is highlighting more the difference in energy accounting between frames. Yes, you can pick any frame you want and the energy at particular steps are different, but the change in energy is always the same. The fact that one frame is the earth and one isn't doesn't matter much. The size of the earth doesn't come into play much other than its size makes it a good approximation for an inertial frame. Feb 8, 2020 at 17:10