1
$\begingroup$

Let's say we are talking about gravitational potential energy, it is defined as:

At a point $X$, inside the gravitational field of mass $M$, the work done by an external agent to bring a mass $m$ from infinity to that point $X$ without changing its kinetic energy,is called the potential energy of that system.

Why don't we define it this way? We know,for conservative forces..like gravitational force, $W_{c}=-\Delta{U}$

$\implies$ $\Delta{U}=-W_c$

My question is why do we always define potential energy in terms of external forces?Isn't defining it in terms of conservative forces much simpler?

$\endgroup$
1
  • 1
    $\begingroup$ Potential energy can also be defined the way you've stated. And both definitions are equivalent. If an external force that is opposite to the conservative force brings a mass m from infinity to a point X without accelerating it, it essentially means that the force is the negative of the conservative force. How you'd like PE to be defined just depends on you. Personally, it makes sense to me that an external agent does work, and that work is stored as potential energy. $\endgroup$
    – archie
    Commented Aug 9 at 13:21

1 Answer 1

2
$\begingroup$

Potential energy can be defined in both ways (internal or external forces).

If there were no external forces acting on $m$ (and no internal dissipative forces) then mechanical energy of the $m$-$M$ system would be conserved. That would mean the sum of the changes in potential energy and kinetic energy of the system would be zero. So if the change in kinetic energy of $m$ is zero (per the quoted definition), it would mean the force doing the work on $m$ would have to be external to the system, as there is a change in mechanical energy of the system (in this case the gravitational potential energy of the system.

On the other hand, if the work is being done by a conservative force internal to the system (in this case the $m$-$M$ system), any change in potential energy plus the change in kinetic energy of the system would have to be zero, for conservation of mechanical energy. An example is the work done by gravity on a falling object. The loss of gravitational potential energy of the $m$-$M$ system equals the increase in kinetic energy of the system.

Hope this helps.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.