I know that if gravity is the only thing that does work on a particle, its change of kinetic energy will be equal to the work done by gravity $$\Delta K = -\Delta U_{\rm grav}. \tag{1}\label{energyCons}$$ But how is that true for an object consisting of a large number of particles? There gravity is not the only thing that does work on a single particle of the object, other particles do work on that particle as well. So how is Eq. \eqref{energyCons} true for a "whole object", when only gravity does work?
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2 Answers
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Particles indeed do work on other particles within this system. But any work done by one is fully absorbed by the other. That corresponds to one doing positive work and the other doing negative work. The net work is zero.
Meaning, all energy exchanges within the system stay within the system and thus don't change the net energy of the system. One particle transferring 10 Joules to another will not change the net energy overall since these 10 joules still are within the system.
Therefor only external forces are relevant since only these do work that isn't balanced out; only these do work that results in net energy added or removed. This is a consequence of Newton's 3rd law.
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$\begingroup$ The statement 'one doing positive work and the other doing negative work' is not clear to me as both of the particles may not have the same displacement. But they do have forces, equal in magnitude and opposite in direction, acting on them. So how is the net work zero? Again thanks. $\endgroup$ Commented Aug 5, 2021 at 7:51
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$\begingroup$ @AzmainAsif The work that one particle does is a name for the transferred amount of energy. Where is it transferred to? The energy conservation law must hold true always, so this amount of work must necessarily be absorbed by something else 1-to-1. If the other particle isn't moved equally far so that not the same amount of work is being absorbed as was being done, then some other energy absorbing mechanism must be taking place. Maybe heat is exchanged as well; maybe energy is transferred to other particles in the same go. Nevertheless... $\endgroup$– SteevenCommented Aug 5, 2021 at 7:54
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$\begingroup$ ... all energy supplied is also absorbed within the system, cancelling out and causing a net energy change of zero from any internal factors. Only external factors - external forces and influences - can cause net energy changes. $\endgroup$– SteevenCommented Aug 5, 2021 at 7:58
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I know that "∆K = -∆Ugrav" is true for a "particle" is not a correct statement.
The first thing to define is the system which you have done in your question but then not applied the notion correctly.
System- the one particle that you are considering.
There is a net external force on the system (the particle) and that force does work on the particle and increases its kinetic energy.
System - all the particles with no external forces acting.
So now all you have is internal forces acting and in your case those internal forces do work resulting in the reduction of the gravitational potential energy of the system and in a corresponding increase in the kinetic energy of the particle(s).
Note that one particle on its own cannot have gravitational potential energy.
Also note that often when one particle is considered the other particle(s) are assume to be fixed in position (have a mass much, much greater that the particle under consideration.
This is an approximation because if you have a ball falling to Earth (system - Earth and ball) then the ball is gaining downward momentum and so as there are no external forces acting on the system the Earth must be rising up towards the ball - law of conservation of momentum.
Since the mass of the Earth is so much greater then the ball one can write $\Delta U_{\text{Earth + ball}} = \Delta KE_{\rm ball}$ as $\Delta KE_{\rm Earth}\ll \Delta KE_{\rm ball}$.
