0
$\begingroup$

By Amonton's laws, we say that the force of kinetic friction takes the following form:

$$ F= \mu W$$

Where $ \mu$ is some constant and $ W$ is the amount of weight the body applies perpendicular to the surface which it moves on.

Now, many resources including Feynman's lectures say the mechanism of friction is related to the molecular vibrations which occur at the surfaces in contact between an object and the surface which it moves on(*). However, I think that this molecular depiction is completely unrelated to the actual physical interpretation of the equation!

For example, consider two blocks: Block-A and Block-B, and both blocks have rough surfaces and Block-B is much larger in dimensions than block-A. If block-A is given some velocity and moves on top of Block-B, then block-A moves on top of Block-B and this causes a friction action-reaction pair to trigger. Block-B moves on a 'smooth surface'.

Now, the friction retards block-A and accelerates block-B. Sometime after, when block-A has turned to rest relative to Block-B, then Block-B has gained some kinetic energy from block-A and is moving with velocity. However, notice that no energy was lost out of the system/ due to molecular interactions or sound!

So, does this suggest that the popular explanations are given for friction actually have no direct relation with the actual mathematical model we use for friction?

My argument that the model doesn't account for heating:

It is known that internal forces do no work. In this case, the frictional force does no work because it is internal. Hence, the kinetic energy of the block, in the beginning, must be equal to the final kinetic energy of the combined system, or in other words, no energy lost!


My question is different from this one because here my question is about understanding a certain nuance of the model used to explain friction rather than what the model is in itself.

Reference:

(*): Feynman lecture on physics

$\endgroup$
8
  • 1
    $\begingroup$ Your first equation only applies to horizontal surfaces where the normal force is the weight. I don’t think Feynman is saying molecular vibrations are the mechanism for friction but rather the consequences of kinetic friction, I.e., the generation of heat. $\endgroup$
    – Bob D
    Dec 10, 2020 at 10:26
  • $\begingroup$ Agreed with the first part, the second part I'm not so sure. Why does this heat generating mechanism turn off when the top block moves over the bottom one? $\endgroup$
    – Babu
    Dec 10, 2020 at 10:34
  • $\begingroup$ Possible duplicate of How is frictional force dependent on normal reaction? $\endgroup$ Dec 10, 2020 at 10:46
  • $\begingroup$ I've added an explanation as to why my question is different from that one @JohnRennie $\endgroup$
    – Babu
    Dec 10, 2020 at 10:52
  • $\begingroup$ @Bursian I don’t understand what you mean by the “heat generating mechanism turn off”. When the top block moves over the bottom friction work increases the temperatures of the surfaces. Then there’s heat transfer from the surfaces to the cooker surroundings. $\endgroup$
    – Bob D
    Dec 10, 2020 at 10:58

2 Answers 2

3
$\begingroup$

It is known that internal forces do no work. In this case, the frictional force does no work because it is internal. Hence, the kinetic energy of the block, in the beginning, must be equal to the final kinetic energy of the combined system, or in other words, no energy lost!

Total energy is (always) conserved, but part of the macroscopic kinetic energy of Block A is transferred to block B and the rest is converted to an increase in the microscopic internal kinetic energy of the two blocks due to internal kinetic friction work.

You didn't say how block A acquired its initial velocity on top of block B initially at rest, but let's just assume it was moving with horizontal velocity $V_1$ relative to the supporting surface when it gently landed on block B so that its initial velocity relative to block B and the supporting surface is $V_1$, and final velocity relative to block B is zero due to the friction force bringing it to a stop. Thereafter, both blocks move together at a final velocity $V_2$ on the supporting surface.

Now, looking at blocks A and B as a system, and considering the supporting surface to be frictionless, then there are no external forces (other than gravity) acting on the block AB system (the kinetic friction forces between the blocks being an internal forces of the system). Given that, the momentum of the system must be conserved.

The initial momentum of the system is that of block A, or $M_{A}V_1$, and the final momentum after block A comes to rest on block B is the momentum of the combined blocks, or $(M_{A}+M_{B})V_{2}$, where $V_1$ is the initial velocity of block A and $V_2$ is the final velocity of the combined blocks. For conservation of momentum we have

$$M_{A}V_{1}=(M_{A}+M_{B})V_2$$

$$V_{2}=\frac{M_A}{(M_{A}+M_{B})}V_1$$

The initial kinetic energy of the system is the initial kinetic energy of block A, or

$$KE_{1}=\frac{M_{A}V_{1}^2}{2}$$

The final kinetic energy of the system is

$$KE_{2}=\frac{(M_{A}+M_{B})V_{2}^2}{2}$$

Substituting the second equation for $V_2$

$$KE_{2}=\frac{M_{A}V_{1}^2}{2(1+\frac{M_B}{M_A})}$$

We see that $KE_{2}<KE_1$, i.e., the final kinetic energy of the system is less than the initial kinetic energy. The "lost" kinetic energy being due to the internal friction work between the blocks. If the masses of the two blocks are equal, the final kinetic energy is half the initial.

In effect, we have a perfectly inelastic collision between the blocks where momentum is conserved but macroscopic kinetic energy is not. It is partially converted into an increase in the microscopic internal kinetic energy of the blocks (increase in temperature at the interface).

Hope this helps.

$\endgroup$
4
  • $\begingroup$ Why does this method and the work done method by considering the action of friction force over a distance give different answers (The scheme highlighted in my post)? $\endgroup$
    – Babu
    Dec 10, 2020 at 22:16
  • $\begingroup$ Seems that internal forces makes conservation of momentum hold true but at the same time do work... $\endgroup$
    – Babu
    Dec 10, 2020 at 22:16
  • $\begingroup$ @Buraian I found your scheme a little hard to follow, but I'll look at it again and get back to you with an update to my answer which will include a response to your second comment as well.. $\endgroup$
    – Bob D
    Dec 10, 2020 at 22:51
  • $\begingroup$ I perhaps have solved the problem, see answer I posted. $\endgroup$
    – Babu
    Dec 11, 2020 at 8:09
3
$\begingroup$

"Block-B has gained some kinetic energy from block-A and is moving with velocity. However, notice that no energy was lost out of the system/ due to molecular interactions or sound!"

Block B hasn't gained as much kinetic energy as block A has lost. The sliding of A over B can be seen as an inelastic collision of long duration.

The blocks warm up slightly. Some energy is transferred from kinetic to internal (random thermal in the blocks).

$\endgroup$
10
  • $\begingroup$ How so? In this case, the friction is a force internal to the system and hence the total work done by it is zero $\endgroup$
    – Babu
    Dec 10, 2020 at 10:57
  • $\begingroup$ The blocks warm up slightly. Energy is transferred from kinetic to internal (random thermal of the blocks). $\endgroup$ Dec 10, 2020 at 11:14
  • $\begingroup$ How so? It is known that work done by internal forces is zero, so the total kinetic energy of block in start must be equal to total kinetic energy of combined block system at end. Hence no loss. $\endgroup$
    – Babu
    Dec 10, 2020 at 11:15
  • $\begingroup$ So the blocks don't warm up? And when two lead balls collide head-on they don't either? $\endgroup$ Dec 10, 2020 at 11:27
  • $\begingroup$ Good point, in that case I could calculate the kinetic energy change by considering conservation of momentum and I'd get some kinetic energy is lost. Not sure how to make a similar line of arguement would apply hsre $\endgroup$
    – Babu
    Dec 10, 2020 at 11:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.