When we say that the mechanical energy of a system is conserved that means,when the kinetic energy(of the system) is increased its potential energy must decrease. Also that work done by a conservative force is negative of the potential energy. Here's my approach and tell me where I am getting the concepts wrong. If there's a body,and (frictionless everywhere) you push it,you have given a extra energy to it which goes in increasing the kinetic energy of the body,but since this is the push which caused it,(the external energy imparted) why in this case then the potential energy of the body decreases? Please tell me where I am going wrong. the external force applied is a conservative force

  • $\begingroup$ The extra energy is given by the push and converted to kinetic energy. The potential energy does not decrease. $\endgroup$ Commented Apr 8, 2018 at 12:58
  • $\begingroup$ Then the law of Conservation of energy? Inc in P.E is dec in K.E and vice versa!? $\endgroup$
    – 123IR
    Commented Apr 8, 2018 at 13:00
  • 1
    $\begingroup$ COE holds for a system with no external force. Since the system is the ball energy for the system is not conserved. On the other hand if you consider the ball and the one who pushed it as the system then energy will be conserved. $\endgroup$ Commented Apr 8, 2018 at 13:04
  • $\begingroup$ There's something like conservative force where the COM holds true.So.the conservative force isn't something external? $\endgroup$
    – 123IR
    Commented Apr 8, 2018 at 13:09
  • $\begingroup$ I meant external non-conservative force :P $\endgroup$ Commented Apr 8, 2018 at 13:11

1 Answer 1


The definition of a conservative force is one which can be written in the form $$\vec F = -\nabla U_F$$ for some function of position $U_F$. We then call $U_F$ the potential energy associated to the force $F$.

If you don't yet know calculus, then for all intents and purposes, a conservative force is one which has a potential energy "partner": $$\text{Near-Earth Gravity:} \ U_g = mgy \iff \vec F_g = -\nabla U_g = -mg \hat y$$ $$\text{Newtonian Gravity:} \ U_G = G\frac{mM}{r} \iff \vec F_G = -\nabla U_G = G\frac{mM}{r^2} \hat r$$ $$\text{Elastic Force/Hooke's Law:} \ U_E = \frac{1}{2}k x^2 \iff \vec F_E = -\nabla U_E = -kx \hat x $$

So on and so forth. The relationship between the force $F$ and its associated potential energy $U_F$ is such that as $F$ does work on a particle (thereby changing its kinetic energy), then $U_F$ changes as well in such a way that the combination $$E = \frac{1}{2} mv^2 + U_F$$ remains unchanged. Therefore, if you have only conservative forces acting on an object, then you can simply add all of the associated potential energies to the kinetic energy, call the result the "total mechanical energy," and then notice that this quantity is conserved.

A non-conservative force, on the other hand, is one which does not correspond to a potential energy. Frictional forces are non-conservative, but so are forces which you exert with your hands. Such forces are able to change the kinetic energy of a body, but there is no associated potential energy to compensate, so the mechanical energy of the object will typically change as well.

From what I can tell, this is the source of your confusion - you assume that "non-conservative" means "frictional" and this is not true. At this level, the conservative forces are gravitational, elastic, or (possibly) electrostatic; pretty much everything else needs to be treated as non-conservative until you learn more sophisticated ways of keeping track of energy.

  • $\begingroup$ Thanks alot, though i have lot other questions regarding it but i would rather not get confused, accept it and move on until i finally get the intuition. Thanks once again. $\endgroup$
    – 123IR
    Commented Apr 19, 2018 at 4:25

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