Friction acting as an internal force I was solving this problem in my assignment:

Assuming a frictional force  F acts on the block of mass m, a force -F will act on plank of mass M. Hence, the net work done by frictional force should be zero, as friction is an internal force , but option D is given incorrect. What's the error in my reasoning?
Thanks in advance.
 A: Work is given by $$W=\int\mathbf F \cdot \text d \mathbf x$$ which, in the case of constant forces in 1D reduces to $$W=F\Delta x$$This definition depends on the path over which the force is applied, so having equal forces is not a sufficient condition to have the total work done on both blocks by friction be $0$.
I would consider how far each object moves relative to the chosen reference frame, and, assuming the friction force is constant, you should see what the answer is fairly easily.
A: The inner actions have a zero resultant but the associated power can be non-zero when there is slipping.
We have in general $\overrightarrow{{{F}_{1\to 2}}}=-\overrightarrow{{{F}_{2\to 1}}}$ but $P=\overrightarrow{{{F}_{1\to 2}}}\centerdot \overrightarrow{{{v}_{2}}}+\overrightarrow{{{F}_{2\to 1}}}\centerdot \overrightarrow{{{v}_{1}}}=\overrightarrow{{{F}_{1\to 2}}}\centerdot \left( \overrightarrow{{{v}_{2}}}-\overrightarrow{{{v}_{1}}} \right)=\overrightarrow{{{F}_{1\to 2}}}\centerdot \overrightarrow{{{v}_{g}}}$ with $\overrightarrow{{{v}_{g}}}$ the sliding velocity.
Sorry for my english !
A: Assume the unit vector $\hat i$ points to the right in your diagram, the bottom block is on a frictionless surface, block $M$ starts from rest and block $m$ has an initial velocity $v_{\rm i} \,\hat i$.  
The magnitude of the frictional force on both blocks is $F$ and their directions are opposite (Newton's third law).  
The acceleration of block $m$ is $-\frac Fm \,\hat i$ and that of block $M$ is $+\frac FM \,\hat i$.  
As there are no horizontal external forces on the two block system then the momentum of the two block system is conserved.
The final velocity of the two blocks is $\frac{m}{M+m}v_{\rm i}\, \hat i$.  
The velocity-time graphs for the two blocks are shown below.  

The area under a velocity-time graph is the displacement and from the graph it is clear that the displacement of mass $m$ is greater than that of mass $M$.
So although both frictional forces have the same magnitude the frictional force acting on mass $m$ does more work, $\int \vec F \cdot d\vec x$, than the frictional force acting on mass $M$.  
The work done on mass $m$ is negative because the frictional force and the displacement are in opposite directions and the work done on mass $M$ is positive because the frictional force and the displacement are in the same direction.
A: For the free body system we can write this equations:
$$m\,\ddot{x}_m+F_\mu=0$$
$$\dot{x}_m=-\frac{F_\mu}{m}\,t+v_0\tag 1$$
$$M\,\ddot{x}_M-F_\mu=0$$
$$\dot{x}_M=\frac{F_\mu}{M}\,t$$
Where $F_\mu$ the friction force between the block and the plank.
the friction force work is:  
$W=\int \,F_\mu\, dx=\int_0^{t_s}\,\,F_\mu\,\frac{dx}{dt}\,dt$
$t_s$ is the time that take the mass $m$ to reach the end of the block.
$\Rightarrow$
$$W_m=F_\mu\int_0^{t_s}\left(\,-\frac{F_\mu}{m}\,t+v_0\right)\,dt=F_\mu\left(-\frac{1}{2}\frac{F_\mu\,t_s^2}{m}+v_0\,t_s\right)\tag 2$$
and
$$W_M=F_\mu\,\int_0^{t_s}\,\left(\frac{F_\mu}{M}\,t\right)\,dt=\frac{1}{2}\frac{F_\mu^2\,t_s^2}{M}\tag 3$$
with equation (1) and the velocity $v_s$ that the mass $m$ reach at time $t_s$ we can calculate the final time:
$v_s=-\frac{F_\mu}{m}\,t_s+v_0\quad \Rightarrow$
$$t_s=\frac{m}{F_\mu}\,\left(v_0-v_s\right)\tag 4$$
with equation (4) in (2) and (3) we obtain for $W_m$
$$W_m=\frac{1}{2}\,m\left(v_0^2-v_s^2\right)\quad, v_0 > v_s \Rightarrow\quad W_m > 0$$
and for $W_M$
$$W_M=\frac{1}{2}\,\frac{\left(v_0-v_s\right)^2\,m^2}{M}> 0$$
so both work done by the friction force are positive! 
