Physics Asked by Meep on August 12, 2020
Suppose I have a little mass m resting on a big slab mass M resting on a table. There is a frictional force between the little mass and the slab, but not between the slab and the table. Suppose I now apply a force F to the little mass such that the static friction between the little mass and slab is not exceeded, and they move together.
Now if I take the table view as my initial reference frame, I correctly get that the increase in kinetic energy of the mass/slab system is equal to the work done by the applied force (found by integrating the force wrt displacement). This is all well and good but…
There is something that is bothering me: I have been taught that work is done by a force when the point of application of the force moves through some distance. Indeed, when I consider the applied force F and I am sat in my inertial (table) reference frame, the point of application of the frictional force moves through some displacement too. However it does no work in this system (as verified by calculation that the gain in kinetic energy of the mass and slab equals the work done by the applied force). I do not understand where this discrepancy in my understanding is arising from!
Thinking through this, I considered a few cases. The first case was a single mass being dragged on a floor with friction. Then I would see that both the applied force and the frictional force have done work, and would obtain the correct amount of work done by each through integrating force wrt displacement from my table inertial reference frame. The same would apply if I now consider this single mass as my mass/slab system where there is no slipping and a frictional force with the floor. So thinking of it in this way, it seems like I can only account for work done by a frictional force when there is relative motion between the two objects between which the frictional force acts. This makes sense to me.
However I feel like there is an important principle underlying this. Why is there this fundamental difference between a frictional force and the applied force? What other forces exhibit this discrepancy?
Also, it seems to be that reference frames have something to do with this. If I transport myself into the reference frame of the slab on which the mass is now slipping, I indeed see that the frictional force between me and the mass does work, whether or not I am accelerating too.
I’m not quite sure where I am going with any of this and would be very grateful for clarification of:
In the first scenario, the frictional force does work on the small mass, and it also does work (in the opposite direction) on the large mass (as reckoned from the laboratory frame of reference). The applied force minus the frictional force is equal to the acceleration of the small mass. The applied force does not do work on the large mass, only the reaction frictional force does work on the large mass. And its acceleration is the same as that of the small mass. $$F-f=ma$$ $$f=Ma$$ $$F=(M+m)a$$ $$f=Ffrac{M}{(M+m)}$$ For the work, $$(F-f)d=mfrac{v^2}{2}$$ $$fd=Mfrac{v^2}{2}$$ $$Fd=(M+m)frac{v^2}{2}$$
Answered by Chet Miller on August 12, 2020
In the first case, it is true that the frictional force is doing work on the small mass. If you choose to consider a frame in which only consists of the small mass, integration of Applied force $-$ The frictional force . $dx$ will be equal to the change in the Kinetic Energy of the small mass only.
On the whole system, the work done by the frictional force on the small mass is equal to the work done by it on the larger mass ( as they act on opposite directions, of course ) . Hence, you don't take it into account while applying the Work-Energy Theorem.
Answered by Aaryan Dewan on August 12, 2020
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