Timeline for How can the energy of a system increase even if net work done on it is zero?
Current License: CC BY-SA 4.0
13 events
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Nov 26, 2023 at 12:26 | comment | added | Dheeraj Gujrathi | @Dale,Got it,Will use more Precise words from now on | |
Nov 26, 2023 at 12:12 | comment | added | Dale | @DheerajGujrathi the contact point is always directly beneath the axle and the axle is moving at the speed of the car, so the contact point moves at the speed of the car. The material at the contact point is momentarily stationary and as the contact point moves it covers a different piece of material. | |
Nov 26, 2023 at 7:59 | comment | added | Dheeraj Gujrathi | @Dale,"Contact point with road moves at speed of car",Can you elaborate,I don't think so,The contact point is also at rest,Because the contact point is on the wheel's material | |
Nov 26, 2023 at 2:55 | comment | added | Dheeraj Gujrathi | @Dale,But yes, Being more precise in our definitions always help | |
Nov 26, 2023 at 2:54 | comment | added | Dheeraj Gujrathi | @Dale,According to "web.mit.edu", The point of application is "the exact location at which a force is applied to a body",So as long as we consider rigid Ideal bodies into play,The displacement of point of application is itself the displacement of material at point of Application ,Even for non contact force,Like consider a point Charge at centre of one square side surface of rigid cube, for any electric field That will apply force on charge,Even for such non contact force, displacement of point of application equals displacement of Material at the point of application.. | |
Nov 25, 2023 at 22:41 | comment | added | Dale | @DheerajGujrathi I agree with almost all you wrote except “It is technically dot product of Force and Displacement of point of application of force”. It is the displacement of the material at the point of application of the force. So for example, in an automobile the contact point with the road moves at the speed of the car but the material at that point is always stationary. | |
Nov 25, 2023 at 21:54 | comment | added | Dheeraj Gujrathi | @ArjunSharma, The "whole work" you refer is not zero,Neither on constituent particle,Nor On complete body, The Definition of work as dot product of Force and Displacement is For single particle,In other worlds,It is technically dot product of Force and Displacement of point of application of force, For Multi particle system,Work is summation of all such $F•ds$ ,Notice here it is a scalar,So even if you apply Two opposite forces on body,As long as Displacement of Point of application in in direction of force,Works will add,Not subtract,Hence net work will be positive,Not zero | |
Nov 25, 2023 at 18:36 | comment | added | Dale | @ArjunSharma said “as a whole on the system, work is zero”. This statement is ambiguous. The “as a whole” could either refer to the “total work”, which is not zero, or the “net work” which is zero. The “total work” and the “net work” are very distinct concepts. The total work is all of the work, the net work is the portion of the total work that changes the KE. So the change in the internal energy is the total work minus the net work. | |
Nov 25, 2023 at 14:24 | comment | added | Arjun Sharma | But I still am a little confused, since energy can be transferred only through work in this case, and as a whole on the system, work is zero, but still its internal energy increases?(Even though I do understand that positive work is being done on both the components) | |
Nov 25, 2023 at 14:17 | vote | accept | Arjun Sharma | ||
Nov 25, 2023 at 12:52 | comment | added | Dale | @ArjunSharma not just “in a sense” but “in fact” | |
Nov 25, 2023 at 12:35 | comment | added | Arjun Sharma | So in a sense, even thought net force on as system may be zero, it can increase the internal energy of the system due to the positive work done on the individual constituents, but cannot cause a change in the kinetic energy of the system? | |
Nov 25, 2023 at 3:51 | history | answered | Dale | CC BY-SA 4.0 |