Example of conservative and non conservative force

AI Thread Summary
Conservative forces, like gravity, allow energy to be converted between potential and kinetic forms without loss, as demonstrated by a ball dropped from a height. In contrast, non-conservative forces, such as friction, dissipate energy, resulting in less energy available for motion after each bounce. The discussion highlights that the earth-moon system is conservative because the net work done is zero when returning to the same point, maintaining energy levels. Non-conservative forces, exemplified by friction in a spring-mass system, lead to energy loss over time. Overall, the distinction between conservative and non-conservative forces is crucial for understanding energy conservation in different physical systems.
Ravi Mandavi
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Hello frnds, i understand what conservative and non conservative force are but i didn't get it properly with practical example. so any article is there which explain it properly with practical example and in easy way, i searched but didn't get any article that satisfy me.
 
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Gravity is conservative, friction is non-conservative. Drop a ball from a height of 1.00 meter. Gravity actsdownward and the energy change is simply mgh, where h is height, 1.00 meter. Upon impact the ball heads upward. Gravity is conservative, the energy stored in potential gets converted to kinetic, back to potential, etc.

But air friction is non-conservative. During the ball's downward motion, friction (air) is taking away a fraction of its KE. The ball bounces, and the restitution is less than 100%, another non-conservative action. On the way up, friction again takes a fraction of the energy away.

This is just 1 example, there are many others. Did this help?

Claude
 
I got somthing but not fully, if we talkd about gravitation force acting on moon due to Earth then how can we justify that it is consevative force?
 
What does "conservative" mean to you? I've always been taught that a conservative force integrated along a path between 2 points produces the same work independent of the path. Non-conservative forces are path dependent.

The earth-moon system is indeed conservative. The net work from point a to b then back to a is zero. That is conservative. The moon follows an elliptical orbit. The energy at some specific point in said orbit remains the same. As the distance changes so does the energy. But when the moon returns to the same point it was at previously, it has exactly the same energy as previously, no change having taken place.

With non-conservative forces, that is not the case. Take a spring-mass system. An object rests on the floor while attached to a horizontal spring. The spring is stretched and harmonic motion occurs. If the floor was frictionless, the Hooke's law spring force would move the object towards the spring eventually compressing said spring, then the object reverses direction. At any point on the floor the object has a specific potential plus kinetic energy. After several oscillations the energy at each point remains the same.

Now add friction. As the object oscillates, it loses energy due to friction. Spring force is conservative. Friction is not. Did I help?

Claude
 
Yes, now i got it
thanx
 
The following example illustrates the difference between a conservative and a non conservative force field. The two molecules illustrate the same binding site with slightly different orientations. The energy of the bound complex is shown as a function of the bond distance. The energy curve has more than one minimum and therefore the energy minimum is not unique. The maximum energy is measured at the distance of 1.9 Å. The energy curve of a non conservative force field has more than one energy minimum and therefore the lowest energy is not unique. The curve is not straight and the energy minimum is located at the distance of 1.5 Å. The difference in the energy of the bound complexes is small, but it is not meaningful.
 
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I am looking at pressure in liquids and I am testing my idea. The vertical tube is 100m, the contraption is filled with water. The vertical tube is very thin(maybe 1mm^2 cross section). The area of the base is ~100m^2. Will he top half be launched in the air if suddenly it cracked?- assuming its light enough. I want to test my idea that if I had a thin long ruber tube that I lifted up, then the pressure at "red lines" will be high and that the $force = pressure * area$ would be massive...
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