# B Why isn't Kinetic Energy always equal to Potential Energy?

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1. May 16, 2017

### phasacs

K= ∫mvdv = ∫m dx/dt dv = ∫m dx/dv dv/dt dv = ∫m dv/dt dx = ∫Fdx = U
=> K=U, why isn't this true? If it is, wouldn't that mean that Kinetic Energy is always equal to Potential Energy?

2. May 16, 2017

### Orodruin

Staff Emeritus
The force is -dU/dx, not dU/dx. Furthermore, the potential is only defined up to an arbitrary constant. If you account for these things, what you get is just the work-energy theorem.

3. May 16, 2017

### nasu

What you found is that the change in KE is equal to the negative of the change in PE. Or the total change is zero. This is "conservation of "mechanical energy".
The fact that changes are equal in magnitude do not grand equality of the actual quantities.

Every time I pay taxes, the amount i give is equal to the amount they receive. But my bank balance is not even close to the one of the internal revenue agency.:)

4. May 17, 2017

### STAR GIRL

Kinetic energy is not integral sum of momentum.. Then how u did this?

5. May 17, 2017

### STAR GIRL

How he found kinetic energy in terms of integral sum of momentum?? Can you explain sir?

6. May 17, 2017

### cnh1995

Isn't it the "change" in KE as nasu said earlier?
kinetic energy E=(1/2)*mv2
You can see dE/dv=mv.
So, dE=mv*dv.

7. May 17, 2017

### STAR GIRL

Oh I see

8. May 17, 2017

### robphy

Note that you have made an assumption that the net force (m dv/dt= F) is a conservative force ( ∫Fdx = [-] U ising @Orodruin 's correction in obtaining a potential energy), which isn't necessarily true... for example, if there is friction that is doing nonzero work.
And as @nasu mentioned, you have really calculated the change-in-K (on the left) and the change-in-[minus]U (on the right), assuming the net force is conservative.

From a more abstract viewpoint [which generalizes to special relativity], it is better to start off with $\Delta K\equiv\int v\ dp$.
For non-relativistic physics, you will then see that (since v and p are linearly related) $\int v\ dp=\int p\ dv$.