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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 questions 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 $\frac{1}{2}m_{sys}v_{COM}^2$ 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?

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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} $.

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  • $\begingroup$ @safesphere As the man is moving forward from rest, can't we say that a net external force must be acting on him in that direction ?? If we take work done as change in KE, there is a positive work done on him. As muscles are a part of our body and any action by them would result in motion of only particular parts of our body, the velocity of CoM of man's body will remain zero. So how will the muscles do work on the body ?? $\endgroup$
    – Atharva Kulkarni
    Commented Nov 30, 2017 at 15:52
  • $\begingroup$ Please read the above comment :) $\endgroup$
    – Atharva Kulkarni
    Commented Nov 30, 2017 at 16:00
  • $\begingroup$ Yes, from Newton's laws -In order for the man to accelerate from rest, there must be an external force on him, which is (static) friction. But that friction does no work (as your foot doesn't slip), The energy comes from your muscles, not from the ground. That's why you feel tired and your legs ache after walking too much. Think about what actually happens when you walk, you put one foot forward, push the ground with it (the reaction force helps you move your COM ahead) , then lift your other leg with your muscles and move ahead.) Friction is necessary for walking, but it doesn't do work. $\endgroup$
    – Rick
    Commented Nov 30, 2017 at 16:44
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There was no work done by friction in your example. Work is force by displacement. Friction is between man's feet and the board they contact. During this contact, each foot does not move relative to the board. Therefore the displacement is zero and so is work by friction. Actual work here is done by man's muscles by converting the chemical energy of food to mechanical energy.

The kinetic energy is defined the way you stated. The sum of kinetic energies of all bodies is non-zero, but the kinetic energy of the center of mass is zero. There is no contradiction here.

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  • $\begingroup$ As the man is moving forward from rest, can't we say that a net external force must be acting on him in that direction ?? If we take work done as change in KE, there is a positive work done on him. As muscles are a part of our body and any action by them would result in motion of only particular parts of our body, the velocity of CoM of man's body will remain zero. So how will the muscles do work on the body ?? $\endgroup$
    – Atharva Kulkarni
    Commented Nov 30, 2017 at 15:59
  • $\begingroup$ Your case is equivalent to the man throwing the board in one direction and consequently pushing himself off the board with the same momentum in the opposite direction. The muscular energy is spent on giving the board and himself the kinetic energy of motion. Friction is irrelevant here, as you can replace it by steps or grooves on the frictionless board that the man pushes against. $\endgroup$
    – safesphere
    Commented Nov 30, 2017 at 16:12

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