Confused regarding conservative and non conservative force....


by sankalpmittal
Tags: confused, conservative, force
sankalpmittal
sankalpmittal is offline
#1
Oct27-12, 12:03 PM
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P: 700
Ok , so here is my explanation :

Suppose I take a box at a height 5m from a reference point upwards. Work done by gravity on the block is -mg(5) J. Upwards is taken as positive and downwards as negative. Then I again take that box 5m downwards to that reference point. Work done by gravity now is -mg(-5) = 5mg J. So net work done by gravity is -5mg+5mg = 0 J. So we find that work done by gravity depends only on initial and final position. So gravitation is a "conservative" force.

Now , I analyze force of push and pull.

Assume that the system is free from any other forces except that of push and pull.

Right is taken as positive and left as negative. I push the box 5m rightwards. Work done by the force of push is 5F J , where F is force of push. The I again take the box 5m leftwards. Work done by force of push is now (-F)(-5) = 5F J. Net work done is 10F J , and not zero. Many say its zero. Why ?

Thus I investigate that forces of push are non conservative. Am I correct ?
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bp_psy
bp_psy is offline
#2
Oct27-12, 01:18 PM
P: 452
No. A constant force is still a conservative force because it is the gradient of a potential. In this U=-kx and F=-dU/dx=k.What you are actually doing wrong is that you turn off the rightward force at some point. When you are talking about path independence the force must always be on. In this case this would mean that :
1. We apply F_r=k to the right until we get at some point Dx, the work done by F_r is E_r=kDx.
2. We turn on a force to the left F_l=-2k. The total force is then F=-k this will make the mass stop at 2Dx the work done by F_r is E_r=2kDx.
3.Now the mass is left to return to the starting point, The work done by the rightward force on the way back is -2kDx therefore the the total work done by it is 0, therefore F_r is conservative.
Note, the rightward force was never turned off and the magnitude of the leftward force can doesn't need to be -2k, it can even be time dependent. If you don't ever turn F_r of the work done by it to go to some point and return to the origin is always 0.


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