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As I am seeing this, the man and boat will have different displacement from ground frame, so there should be a net work done by friction . The books says, total work done by friction is zero when friction is static and internal to system. Please explain..

BioPhysicist
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Rahul Singh
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As I am seeing this, the man and boat will have different displacement from ground frame, so there should be a net work done by friction

Viewed from the ground frame (frame of the observer in picture) and considering the man alone as the system, the static friction force between the man and the boat does work on the man causing a horizontal displacement the man toward the observer. Per Newton's third law, the man exerts and equal and opposite external force on the boat that does work on the boat causing a horizontal displacement of the boat away from the observer.

The books says, total work done by friction is zero when friction is static and internal to system. Please explain..

When the man and the boat are considered to be a system, the static friction force is internal to the system. Given there is no friction between the boat and water (and assuming no air friction as well) then there are is no net external force acting on the system and therefore no net work is done on the system by the internal static friction force. To put it another way, there is no displacement of the center of mass of the man/boat system due to the static friction force.

Can you mathematically show friction work adding up to zero.

Keep in mind we need to differentiate between friction work that is done on the man and the boat individually in the reference frame of the observer vs the friction work that is done on the system of the man + boat in the reference frame of the observer.

In the first case we apply Newton's second law to the boat and man individually and calculate the work done on each by the friction force.

In the second case we need to apply Newton's second law to the combination of the boat and man. The acceleration of the boat/man system is the acceleration of the center of mass of the boat/man system. Since the boat and man will each accelerate with an acceleration inversely proportional their masses, the acceleration of the center of mass of the boat and man will be

$$a_{cm}=\frac{Ma_{M}+ma_{m}}{M+m}$$

Per Newton's third law, the friction force exerted on the man by the boat is equal and opposite to the friction force exerted by the man on the boat, or

$$Ma_{M}=-ma_{m}$$

Consequently, $a_{cm}=0$. Therefore there is no displacement of the center of mass of the boat/man system. The internal static friction force does no net work on the boat/man system.

Now, in response to your discussions with @Adrian Howard:

So am I correct to say that, in ground frame, the kinetic energy is due to internal friction forces. I am just trying to apply work energy theorem..

The net (static friction) force on the man equals its change in kinetic energy in the observers reference frame and equals the change in kinetic energy of the boat in the observers reference frame.

Also I read that, work done by non-conservative forces changes mechanical energy of system. So what I am thinking is that friction being internal nonconservative force, added kinetic energy to this system(in ground frame). Point me where I am wrong, or is it a correct explanation.

Static friction does add kinetic energy to the boat and man individually, but it does not add kinetic energy to the center of mass of the boat/man system, since there is no change in velocity of the center of mass of the system. For the boat/man system the total change in kinetic energy, at the macroscopic level, is the sum of the changes in its internal kinetic energy (man plus boat individually) and external kinetic energy (kinetic energy of the center of mass of boat/man system). The change in kinetic energy of the center of mass is zero.

In your example of the system of charges, the increase in kinetic energy of the charges accelerating towards each other involves an equal decrease in electrical potential energy. Total mechanical energy is conserved because the electrical force are conservative. For the boat/man system there is an increase in internal kinetic energy of the system with no change in potential energy. Mechanical energy is not conserved because friction forces are non-conservative. The increase in kinetic energy comes from the internal energy of the man (chemical energy).

But for both the boat/man system and system of charges there is no change in kinetic energy of the center of mass of either system, because there is no net external force acting on either system.

Hope this helps.

Bob D
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    Thanks for helping.But I am unable to grasp the second part. If I take an example of two attracting charge particles as a system, and no external force on system.The charges will gain kinetic energy and this will be due to work done by internal electrostatic forces.Then in my example why I can't say that, as the work done by friction is not adding up to zero from ground frame, the friction is responsible for gain in kinetic energy of boat and man.The book says, internal energy of man is being used up. Can you mathematically show friction work adding up to zero. – Rahul Singh Dec 21 '19 at 20:38
  • Yes I can show friction work adding up to zero mathematically when I get back to my computer (presently on mobile device) please stand by – Bob D Dec 21 '19 at 20:43
  • @RahulSingh Kinetic energy is not invariant and depends on reference frame – Adrian Howard Dec 21 '19 at 21:38
  • @Bob D.. Okay, waiting.. – Rahul Singh Dec 21 '19 at 22:01
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    @Adrian Howard - Yes, I understand that. So am I correct to say that, in ground frame, the kinetic energy is due to internal friction forces. I am just trying to apply work energy theorem. Also I read that, work done by non-conservative forces changes mechanical energy of system. So what I am thinking is that friction being internal nonconservative force, added kinetic energy to this system(in ground frame). Point me where I am wrong, or is it a correct explanation. – Rahul Singh Dec 21 '19 at 22:08
  • @RahulSingh I have updated my answer. Hope it helps. – Bob D Dec 21 '19 at 23:18
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    @BobD Isn't the kinetic energy of a collection of particles taken to be the sum of the kinetic energies of all the particles belonging to the collection? I ask this, because the final sentence of your answer reads "but it does not add kinetic energy to the boat/man system, since the change in kinetic energy of the center of mass of the system is zero," which seems to suggest that $\text{KE}{\text{system}}=\text{KE}{\text{CM}}$. – Ajay Mohan Dec 22 '19 at 03:55
  • @BobD (contd.) Would you also say that the electric force does no work on Rahul's example of two attracting charges coming closer together (with no external forces acting on the two-charge system), because the change in KE of the CM of the system is zero? – Ajay Mohan Dec 22 '19 at 04:02
  • @AjayMohan On your first comment, when I said add kinetic energy to the boat/man system I was referring to the kinetic energy of the center of mass of the system with respect to an external reference frame. It was not meant to say the total system kinetic energy is zero. However, I can see it might be interpreted that way so I added "center of mass". On the second comment, the electric field does work converting electrical potential energy to kinetic energy, while the change in KE of the CM is zero. I have added that as well. Thanks. – Bob D Dec 22 '19 at 12:16
  • @BobD " ...Therefore there is no displacement of the center of mass of the boat/man system. The internal static friction force does no net work on the boat/man system." Is my inference of this statement correct: If the displacement of the center of mass is zero, then there's no net work done by the force? This contradicts: Work done by a force on a system is equal to the sum of the work done by the force on all the particles belonging to the system. E.g: When I push two identical blocks towards each other by applying equal force on them, I'm doing work on the two-block system, right? – Ajay Mohan Dec 22 '19 at 19:06
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    (contd.) I have googled this question and as OP says, the answers seem to stress that the work done by static friction on the boat/man system is zero solely because the frictional force is static in the problem. I apologize for bothering you again. I have a feeling that OP is also stuck at this point and hence, your replies would help us both. – Ajay Mohan Dec 22 '19 at 19:15
  • It was my understanding that the total work done on a system of "particles" by both external and internal forces is given by the sum of the change in the kinetic energy of the CoM and the change in kinetic energy of the particles relative to the CoM. i.e. $$W_\text{ext}+W_\text{int}=\Delta K_\text{cm}+\Delta K_\text{rel}$$ In your answer you have shown that $\Delta K_\text{cm}=0$, and we have no external forces. Therefore, don't we have here $W_\text{int}=\Delta K_\text{rel}\neq0$? I think this is the confusion in the comments. – BioPhysicist Dec 26 '19 at 12:56