# Work done on an object by the internal forces

How is the work done by the internal forces acting in a rigid body zero?

Actually I read in a book an example for the same.

Let me present that example here.

Consider a rigid body having two particles $A$ and $B$. Suppose, the particles move in such a way that the line $AB$ translates parallel to itself. The displacement $d\textbf{r}_A$ of the particle $A$ is equal to the displacement $d\textbf{r}_B$ of the particle $B$ in any short interval of time. The net work done by the internal forces ,i.e, the force that $A$ exerts on $B$ and the force that $B$ exerts on $A$, is zero.

How can it be analysed mathematically that the work done is zero?

This example is very unclear to me.

Your confusion might be coming from not clearly understanding that you need to define a system, and then all of your quantities are referenced to the system you have defined.

If your system is A and B, then the force that A applies to B is by definition an internal force, and no work is done by that force on the system. But you can choose your system however you would like. Nature doesn't care where you draw your boundary. If you take the system to be A, the the force of B on A is external, and work can be done on the system. Note that it's not the same system as the first case!

What's internal and what's external is a matter of bookkeeping. The setup of the equations will be different, but the final answer will always be the same.

• OK now i got it. Its all about the frame of reference. Apr 29, 2016 at 19:02

I disagree with both other answers.

• It's not enough that the sum of forces is zero
• This is not a vocabulary problem

The work-energy theorem applies also to systems of interacting bodies, where the total work of all forces (internal and external) equals the variation of the sum of the kinetic energies of all bodies.

Newton's third law gives you $\mathbf F_{A→B}=-\mathbf F_{B→A}$ (without condition, in any FoR). In your example, $\mathrm d\mathbf r_A=\mathrm d\mathbf r_B$, so you get $δW_{A→B}=\mathbf F_{A→B}·\mathrm d\mathbf r_B=-\mathbf F_{B→A}·\mathrm d\mathbf r_A=-δW_{B→A}$, hence the net work $δW_{A→B}+δW_{B→A}=0$.
• You've shown that the internal work done on the system A and B is zero. You've also shown that if you choose the system as B, the work done on that system is $-W$, and if you choose the system as A the work done on that system is $W$. The choice of system is essential in setting this up. Apr 29, 2016 at 23:18