Work done due to static friction in a boat-man system

[Nikhil Rajagopalan]

In a classic problem where a man walks along the length of a boat floating on the surface of water without friction, Should the net work done due to static friction between the man and the boat be zero. If the length of the boat is L, its mass being M, and the mass of the man being m, the displacements the boat and the man undergo could be calculated as mL / (M+m) and ML / (M+m) respectively. If F is the static friction between the boat and the man, it should be equal and opposite to the boat and the man. On calculating the net work done by the static friction on the boat - man system using F * displacement, even though the signs are opposite, it doesn't add up to zero.

Is the work done by static friction on a system of particles always zero,if the static friction acts only as internal force between the particles? If so, why is it not apparently verified mathematically.

 

[BvU]

it doesn't add up to zero

Oh ? Can you show that ?

 

[Nikhil Rajagopalan]

BvU , thank you for attending my query. On trying to prove it, i used the expressions for displacements of the boat and the man, +mL / (M+m) and -ML / (M+m) respectively ,calculated from an external point of reference. The static friction exerted by boat on the man and the man on the boat should be the same, equal and opposite. The product of force and displacement in both the cases here does not add up to zero. I am not certain if i am missing anything vital here.

 

[BvU]

I don't see a calculation that ends up not zero ?
What force does the man exercise on the boat to get moving ? What force to stop ?
After his walk, both boat and man are stationary. No kinetic energy is present, just like before.
Note that the friction force itself does no work: boat and shoes do not move wrt each other (assuming no slipping) wile the man exerts force on the boat to accelerate or decelerate.