I had this doubt while thinking through a question about centre of mass. Consider a system, consisting of a man standing on one end of a plank which rests on a frictionless surface. Now the man starts running towards the other end of the plank(friction is present between the man and the plank). Once he reaches the end of the plank he jumps down and both the man and the plank keep on sliding endlessly on the surface with equal and opposite momentum. Although the net momentum is still zero, both of them now have some velocities and thus the kinetic energy of the system has increased. Therefore work is done on the system by friction. This has prompted the following ques. in my mind :-
1)How is the kinetic energy of the system defined ? In this case if we add the individual kinetic energies of the 2 bodies, we get a net increase in the KE of the system. However, if we take it as $\frac{1}{2}m_{sys}v_{cm}^2$ the KE will still be zero as velocity of centre of mass is zero.
2)If friction is doing work on the system, which energy is being converted into mechanical energy ? As this is an isolated system(assuming no form of heat exchange is present between the bodies and the surrounding) the total energy should always remain conserved. I thought that it must be the tiny deformations caused in the bodies by friction resulting in change of potential energy which is converted into KE. Is this right ?? Or will the bodies get cooler to keep the energy conserved ??
Best Answer
Friction isn't the force doing work. Human bodies are much better than inanimate blocks, it's the man using his internal energy from his muscles to kick the plank behind, the friction will stick his leg to the plank while he kicks and prevent it from slipping behind while he is lifting the other leg, ideally work done by friction should be zero if the man is walking properly and his shoes don't undergo plastic deformation. It's the man's internal energy causing the system to gain kinetic energy.
Yes, Kinetic energy is defined as how you stated, however for 2 particle systems of mass $m_1$ and $m_2$, you could write the total kinetic energy as
$KE = \frac{1}{2}M_{sys}v_{cm}^{2} + \frac{1}{2}\mu v_{rel}^{2}$, where $\mu$ is the reduced mass $\frac{m_1m_2}{m_1 + m_2} $.