A block of mass $m_1$ 2 kg is lying on table attached to string which is connected to another mass $m_2$ 1 kg which is overhanging the table. Find the coefficient of kinetic friction if the speed of $m_2$ is found to be 0.5 m/s after it has descended by 1m.
The solution in the example solves the problem using Work-Kinetic energy theorem, i.e
Change in KE=Work done
$$
( \frac12mv_1^2+\frac12mv_2^2)-0= \mathrm{(workdone\,by\,frictional force)+work\,done\,by\,weight}\\
(\frac12mv_1^2+\frac12mv_2^2)-0= -μN*S+m_2g*1\mathrm{m}
$$
where S is displacement of $m_1$.
My question is why are we considering work done by frictional force on $m_1$? why not tension on the $m_1$?
Best Answer
The tension in the string is an internal force. It does not have to be considered because it does no net work on the system. If the string were extensible, this force would have to be taken into account, eg by including elastic potential energy stored in the string.
There are other internal forces holding the atoms of the 2 blocks together, but these forces also do no net work on the blocks because the blocks are rigid, not elastic, not deformable.
The system here consists of the blocks and the string. It is assumed to be a rigid body : the distance between any 2 points does not change. So the work done by internal forces is zero. Then the work done by external forces on the system equals the increase in kinetic energy.