# Why only kinetic energy?

1. Jun 25, 2011

### modulus

I'm getting really confused about a specific application of the work-energy theorem, and I'm hoping you guys at PF could help out.
I'll start out by stating three concepts I've learnt, and I'll develop the apparent contradiction I run into.

Number one. the negative of the work done by the conservative internal forces on a system equals the change in potential energy of the system. Fine. It sounds good. It feels good. It makes sense. And I don't have problems applying it.

Number two. The work done by the external forces on a system equals the change in total (mechanical) energy. It makes perfect sense. Sounds perfectly natural. And it's easy to apply. But, this is exactly where my problems start. From what I know, and how I've been using this concept, I understand that the change can be in the kinetic or the potential energy of the system.

And number three. This is what I'm not getting. The work done by all forces (external and internal) equals the change in kinetic energy of the system. From the first concept, we get that the work done by the internal forces is indeed the change in kinetic energy (it's like the work-energy theorem). But the second part says the external forces change the kinetic energy too (or I should say 'only'). But that doesn't make sense, the external forces can cause changes in the kinetic or potential energy. So the sum of the internal and external forces' work cannot account for only the change in kinetic energy.

And before ending the post, I need to give you one example where the external forces cause a change in potential energy. This example really confirmed my doubts. There was a spring of spring constant k. It had two equal masses attached on each side, and was pulled by a distance x/2 on each side. as I evaluated, the total work done by the external forces equaled the negative of the work done by the spring, and, most importantly, the increase in potential energy of the spring. So, external forces do change potential energy.

2. Jun 25, 2011

### Petr Mugver

Number 1: it can't be correct, since a system can do work on another system, not on itself. And if you divide the system in two subsystems, and if the first does a certain amount of work to the second, then the second does minus the same amount of work to the first.

Number 2: correct, in absence of dissipative forces.

Number 3: same as 1.

3. Jun 25, 2011

### my_wan

I think maybe there is some confusion concerning Open and closed systems. For instance Number two states:
By having an "external force" do work on the system the system is by definition not enclosed. You added energy from an outside source. If you include that outside source as part of the system then no change in the total energy (kinetic and potential) has occurred.

If the kinetic energy came from external sources then it had to reduce the potential energy by the same amount, so the total energy did not change. However, if the work was done by an external source then the total energy of the system can increase, including both kinetic and potential. It does however cost that external source the same amount of energy gained by the system. So external energy can change the total energy of the system, internal energy can only trade kinetic and potential energy.

What is or is not internal or external to the system is purely up to you to define how you wish, but once defined any change in kinetic energy from internal sources alone exactly matches the loss of potential energy and visa versa. A pendulum is a system which merely oscillates between kinetic and potential energy.