Why is change in potential energy equal to negative work?

[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.

 

[Naty1]

Check out:

http://en.wikipedia.org/wiki/Potential_energy#General_formula

it's not so much 'proof' as 'explanation'...

 

[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.

 

[CrazyNeutrino]

Thanks that helps.!

 

[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?

 

[rudolfstr]

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?

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

 

[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