# Why does delta PE = -W?

1. Oct 10, 2012

### CrazyNeutrino

Can someone prove that the change in potential energy is negative work.
I have a very basic understanding of the concept. I do not understand where it is derived from.

2. Oct 10, 2012

### Naty1

3. Oct 10, 2012

### rudolfstr

Unfortunately you can't prove that in the general case. For gravity it's easy. What you can prove is that work done=1/2*m*v^2= kinetic energy, and from conservation of energy (dE/dt=0) you can derive the remaining stuff.

4. Oct 10, 2012

### CrazyNeutrino

Thanks that helps.!

5. Oct 10, 2012

### CrazyNeutrino

So the proof would be...

mgy(b)+KE(b)=mgy(a)+KE(a)

That is: U(b)+K(b)=U(a)+K(a)
So U(b)-U(a)=K(a)-K(b)=-(K(b)-K(a)
So U(b)-U(a)=-W ( By work energy theorem )
Therefore:

ΔU=-W

Is this proof correct?

6. Oct 11, 2012

### rudolfstr

Yeah it's ok. But another important question is wheather you know where work energy theorem comes from?

7. Oct 11, 2012

### CrazyNeutrino

Yeah.
W= ∫from a to b of Fdx
=∫from a to b of madx
=∫from va to vb of mdv/dt vdt. (dx/dt =v so dx =vdt)
=∫from va to vb of mvdv
=1/2mv^2 evaluated at va and vb
= 1/2mvb^2-1/2mva^2
=KEb-KEa
Therefore
W=KEb-KEa

Last edited: Oct 11, 2012