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I am a student and I've been studying a physics book. There is one statement in the solution of one question (regarding calculation of Potential energy) which is confusing me. It states that:

If the particles are brought by some external agency without changing the kinetic energy, the work done by the external agency is equal to the change in potential energy = - 1.0 x 10 -11 J.

Now I want to know how is it possible to bring or move a particle without changing its kinetic energy because in bringing a particle there is some change in the velocity which should change the Kinetic energy?


marked as duplicate by sammy gerbil, Kyle Kanos, Jon Custer, Yashas, FGSUZ Sep 22 '18 at 20:21

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  • $\begingroup$ Could it be that the velocity changed direction, but kept the original magnitude? Hence, conserving kinetic energy? $\endgroup$ – Zirc Sep 19 '18 at 15:11
  • $\begingroup$ I don't think so,I am quoting the question "Find the work done in bringing three particles, each having a mass of 100 g, from large distances to the vertices of an equilateral triangle of side 20 cm." $\endgroup$ – user181463 Sep 19 '18 at 15:15
  • $\begingroup$ You have a particle far away. It isn't moving, so KE=0. Then you bring it towards you and place it on the corner of an equilateral triangle. You place it so it is still not moving. Thus KE=0 still. $\endgroup$ – Jahan Claes Sep 19 '18 at 15:22
  • $\begingroup$ Certainly while you are bringing it towards you it has kinetic energy, because it has velocity. But you're only comparing the initial and final points of the trajectory, not the whole trajectory. $\endgroup$ – Jahan Claes Sep 19 '18 at 15:22
  • $\begingroup$ @JahanClaes so you mean in the mean while while it was being brought the ke changed but final and initial values have same values? $\endgroup$ – user181463 Sep 19 '18 at 15:23

Yes it is possible. The key is the mass has to begin at rest and end at rest, that is, there is no net change in kinetic energy in going from 1 to 2. I’m not sure about the details of your 3 particle example, but consider the following simple example involving raising a mass $m$ from point 1 at rest to point 2 at rest where point 2 is height $h$ above point 1. The increase in potential energy is $mgh$.

Clearly, to accomplish this, the mass has to move, i.e., it must attain some velocity along the way. To get things going we need to apply an external upward force, $F_{ext}$, slightly greater than $mg$, to give it small upward acceleration $a$. Let’s say I do this for a brief time $dt$ and therefore over a short distance $dh$. The mass thereby attains a small velocity $v=adt$, a small increase in kinetic energy of $ ½ m(adt)^2 $ and a small increase in the potential energy of the mass m of $(mg) dh$.

We now immediately reduce our upward force so that it equals the downward gravitational force. The mass is now rising at constant velocity v and so there is no subsequent change in KE, however its potential energy keeps increasing since the mass continues to rise. The work to accomplish this increase in potential energy is due to my constant application of an upward force against gravity.

As we approach point 2 we are still left with the small kinetic energy. Therefore, prior to reaching point 2, we reduce the external upward force, $F_{ext}$, to slightly less than $mg$, to give it small negative acceleration. We do this for sufficient time to bring the mass to rest at point 2. During this period gravity now does a small amount of negative work resulting in a change in kinetic energy of $ -½ m(adt)^2 $. Consequently the total change in kinetic energy going from point 1 to point 2 is zero. Since there is no loss in height during this period, there is no loss in potential energy.

The end result in going from 1 to 2 is an increase in potential energy of

$$ \int_1^2 (mg)dh = (mg)h$$.

Hope this helps.

  • $\begingroup$ So the answer is acceptable? $\endgroup$ – Bob D Sep 19 '18 at 16:43
  • $\begingroup$ to me perfectly acceptable and Again thanks to all $\endgroup$ – user181463 Sep 19 '18 at 16:47

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