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This question already has an answer here:

Say a train is moving in the positive $x$-direction at 100 meters per second with respect to the ground frame. Now let's say someone is pushing a large box in one of the cars of that train in the negative $x$-direction and is applying sufficient force to keep the box moving at a constant 5 meters per second in reference frame of the train car.

My intuition keeps telling me that this is like any other problem where the ground frame observer would witness the force being applied over a longer distance compared to the pusher/observer in the train frame, so work done is greater in ground frame, but isn't the energy dissipated as heat independent of reference frame?

Furthermore, the ground observer would see the box moving forward at 95 meters per second as the train moves forward at 100 meters per second. Correct? The person pushing is doing work and the friction is doing equal work in the opposite direction. From the ground frame we'd see much more work done but how do we account for the dissipated energy such as heat and vibration? I feel like from the ground frame you'd see enough heat to start a fire. I hope I have articulated my confusion clearly enough.

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marked as duplicate by sammy gerbil, John Rennie newtonian-mechanics Nov 22 '18 at 8:47

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Sliding friction is a bit odd. With most interactions you can use the power equation quite easily: $P=F\cdot v$. With sliding friction it is a little confusing because of the issue about which $v$ should be used since each surface has a different $v$. It turns out that you use both and the difference is the power that goes into thermal energy.

So assuming $F=10\:N$ then in the train’s frame the power on the box’s side of the interface is $P_{box}=F\cdot-v= -50 \: W$ and the power on the train’s side of the interface is $P_{train}=-F\cdot 0=0 \:W$. So 50 W is going into the interface and not coming out, this is the energy that increases the temperature.

In the ground frame the train’s velocity is $u=100 \: m/s$ so $P_{box}=F\cdot (u-v)=950 \: W$. Note, the sign of this power has changed, so this is energy that is going out of the interface and into the box, which is the opposite direction as previously. However, now $P_{train}=-F\cdot u=-1000 \: W$ meaning that 1000 W goes into the interface from the train side. Note again that 50 W goes into the interface and does not come out.

So the same 50 W of “heating” occurs in both frames. The explanation of where that power comes from is different, but all of the observable measurements are the same.

The nice thing about this approach is that there is no special reference frame. Any reference frame gives the same result with a simple systematic use of the standard equations.

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  • $\begingroup$ So you are essentially breaking the power delivered by friction into two parts where when you add them it just depends on the relative motion between the two surfaces. I'm unsure what $P_{train}$ physically is. $\endgroup$ – Aaron Stevens Nov 20 '18 at 18:34
  • $\begingroup$ $P_{train}$ is the power delivered from the train to the interface. $\endgroup$ – Dale Nov 20 '18 at 21:08
  • $\begingroup$ I agree that the numbers work out, but this doesn't make exact sense. To determine how much power is being delivered to an object by a force, you perform $\mathbf F\cdot\mathbf v$. Where $\mathbf F$ is the force in question, and $\mathbf v$ is the velocity of the object. So how can you justify using different velocities within the same frame? In each frame the box only has one velocity. $\endgroup$ – Aaron Stevens Nov 20 '18 at 22:26
  • $\begingroup$ Yes, $v$ is the velocity of the object. The box and the train have different velocities, hence two different velocities are not only justified but required. Because of the different velocities the power leaving one object is not the same as the power entering the other object. The difference is the power that warms the interface. $\endgroup$ – Dale Nov 20 '18 at 22:52
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    $\begingroup$ Right. Friction takes energy from the train and gives it to the box, but the transfer is not perfect. I was getting thrown off by your wording about talking about different sides of interfaces. Thanks for the clarification. $\endgroup$ – Aaron Stevens Nov 21 '18 at 1:08
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The heat produced by kinetic friction corresponds to the distance an object moves along a frictional surface, not the distance it moves in some arbitrary reference frame. In other words, the velocity used in heat from friction calculations is the velocity of the object with respect to the surface. In this example for the ground reference frame, the velocity of the box is 95 m/s and the velocity of the floor of the train is 100 m/s. Therefore, one would still find the sliding velocity between the box and the floor to be 95 - 100 = 5 m/s, so the heat produced by friction would be the same. Note that this is not the same as work done by friction.

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  • $\begingroup$ So it would be correct to say that all that matters with friction is that it is with respect to the relative velocity of the interface between sliding surfaces? so, since relative velocity is independent of reference frame, work by friction is also? $\endgroup$ – MattGeo Nov 20 '18 at 3:38
  • $\begingroup$ The static friction example has for some reason always made sense to me. I can't make sense of kinetic friction in relative motion. $\endgroup$ – MattGeo Nov 20 '18 at 3:44
  • $\begingroup$ @AaronStevens the question seems to be specifically about kinetic friction, which is only dependent on the relative velocity of the interface between the surfaces, so this is the case as described above. I understand your point about static friction though. $\endgroup$ – Sandejo Nov 20 '18 at 3:46
  • $\begingroup$ Could you give an example of what you mean by changing sign depending on frame? $\endgroup$ – MattGeo Nov 20 '18 at 3:56
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    $\begingroup$ @MattGeo Do not confuse "work done by friction" and "energy loss/heat dissipation by friction". The two are very different. $\endgroup$ – BowlOfRed Nov 20 '18 at 18:42
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My intuition keeps telling me that this is like any other problem where the ground frame observer would witness the force being applied over a longer distance compared to the pusher/observer in the train frame>

This is true. In each frame the net work is $0$, but the work done by each force depends on the frame. In fact, the work done by each force has different signs compared between your frames.

However, when it comes to how much energy is dissipated and lost due to heat, then only the relative motion between the box and the train matters (Dale's answer gives a proof of this, where it is shown that the power dissipated depends on the relative motion between the surfaces), and so it would be best to work in the frame of the train.

This is a good question, and causes us to really think about work and friction. We usually equate the work done by friction to the energy lost due to heat simply because most friction problems are done assuming we are at rest relative to the surface the object is sliding on. However, as your example has shown, this cannot always be the case.


A simple counter-example showing that the work done by friction does not always give the energy "lost to heat". Let's say I hold a box stationary while a conveyor belt moves underneath it. According to me, the box is not moving, so the work done by friction (and myself) must be $0$, yet I'm sure we would all agree that the box would start to heat up due to friction. To find how much energy is being lost to heat, we could move to a reference frame moving along with the belt so that the belt is at rest. Then the box would be moving relative to us, and any motion we see of the box is motion relative to the belt. Then the work done by friction would be the energy lost as heat.

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  • $\begingroup$ if you were in the ground frame observing the motion, could you calculate it based on seeing the train "sliding under" the box? being that the train is going 100 meters per second and the box 95 meters per second? That was the picture I had in my head but it didn't feel quite right. $\endgroup$ – MattGeo Nov 20 '18 at 3:50
  • $\begingroup$ @MattGeo I guess you can view it that way. This is a tricky scenario. Think about this too: if you let go of the box, then according to someone on the ground friction actually does positive work to get the box back up to the velocity of the train. $\endgroup$ – Aaron Stevens Nov 20 '18 at 3:57
  • $\begingroup$ @MattGeo I think I see what you are saying about the train sliding under the box. For example, if I hold a box stationary with a conveyor belt moving under the box, then I see friction doing no work, but things will be heating up. $\endgroup$ – Aaron Stevens Nov 20 '18 at 4:05

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