# Conservation of mechanical energy

by moderate
Tags: potential energy
 P: 21 I have a problem with my train of thought. I have narrowed it down to several premises, and I think that one of them must be wrong. Premise 1 Δ total energy (E) = Δ kinetic energy (K) + Δ potential energy (U) (assuming that there is no change in any of the other forms of energy in the system) Premise 2 Δ E = q + w Consider the system of a block. The block is initially traveling upwards (+y direction), away from the earth's surface. At t=0 it has a kinetic energy of 1/2mv2 (v is non zero). It initially also has gravitational potential energy of 0. The velocity eventually decreases to 0 (at t1). The whole time (from t=0 to t1), the earth's gravitational attraction is exerting a force of mg on the block. No other force is acting on the block, since it is already in motion at t=0. It appears that the work done by the gravitational force on the block is -mgh. But, ΔE = 0, because the kinetic energy has simply been transformed into potential energy. So, how can work be non zero (-mgh)? It seems that (looking at the block as the system) ΔE = 0 = W= -mgh (no heat transfer, so q=0, no friction, no other energy transfer) Since this is impossible, I know that I did something wrong. Thanks for your help!
 Sci Advisor HW Helper Thanks PF Gold P: 26,107 Hi moderate! Welcome to PF! You've been double-counting … Potential energy is just another name for (minus) work done (by a conservative field such as gravity). From the PF Library … Is potential energy energy? There is confusion over whether "energy" includes "potential energy". On the one hand, in the work-energy equation, potential energy is part of the work done. On the other hand, in the conservation-of-energy equation (and conservation of course only applies to conservative forces), potential energy is part of the energy.
 P: 21 Hey - thanks for the quick response and the welcome! Let me make sure I understand what you are saying. For the 1st law of thermodynamics ("conservation-of-energy equation"), ΔE=q+W, potential energy is already accounted for (on the left side), so I do not need to add another work term? Essentially: ΔE=q+W and ΔE=ΔK therefore: ΔK=q+W ΔK=0 + (-mgh) ΔK + mgh = 0 And, mgh = ΔU (this is what it is called by convention, "gravitational potential energy") Correct? This makes me slightly uncomfortable. Let me think about why. Ok, so when I was taught that ΔE=ΔK+ΔU, this can not be equated to the 1st law of thermodynamics? The two terms with the same name (ΔE) in different equations include different components of what is called "mechanical energy"? One includes gravitational potential energy and kinetic energy, and the other doesn't. In other words, "mechanical energy" is not some inherent form of energy, but it is made up of kinetic energy (which is "inherent" to the object), and the work due to gravity that was brought over from the other side of the ΔE=q+W equation? This seems awkward.
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P: 26,107

## Conservation of mechanical energy

Now you're just confusing me.

The simple rule to remember is that you can't count anything twice.

So, since potential energy and work done are the same thing, you can't use both.
 P: 21 Yeah, sorry, I posted before I had my thoughts in order. My main point can be summed up like this: ΔE is a term that appears in two equations: ΔE=q+W (1st law of thermodynamics) and ΔE=ΔK+ΔU (total mechanical energy) However, ΔE is not the same in both cases. In one case (left equation) it includes only kinetic energy. In the right equation, it includes both kinetic and potential energy. So, the two can not be equated. I hope that this was more clear. (what you are saying makes sense)