# Elastic potential

1. Nov 15, 2013

### James Starligh

Collegues,

if I have elastic potential U=( K(x-X0)^2 )/2 for the flexible spring moved on X distance from the initial X0 position So I have I have elastic force F=K(x0+vt-x) wich is the derivative of the U. Its not clear for me why vt term which should be equal to x in this equation is present ? From the general assumption (x-X0)^2 should be x^2-2*X*X0+X0^2 where is the vt term here?

James

2. Nov 15, 2013

### mikeph

what's vt?

looks like you differentiated incorrectly.

3. Nov 15, 2013

### James Starligh

x = V (velosity) * t (time)
both of that formulas are from textbooks.

Reasonable U= (Kx^2)/2 so force = kx. wehat fould be force if U=( K(x-X0)^2 )/2 in case when we moved body from the initial position (Xo) on X. The delta X should give us velosity should it?

4. Nov 15, 2013

### Philip Wood

x = vt is for a body moving at constant velocity for a time t. I wouldn't have thought this was the case here. So we have simply

$U = \frac{1}{2}k(x - x_0)^2$

$F = -\frac{dU}{dx} = -k(x - x_0)$ (having used chain rule)

The minus sign in front of the k indicates that the force is in the -x direction.

5. Nov 15, 2013

### James Starligh

Thanks!

Some calculus:
In the F=K(x0+vt-x) dealing with the constant velosity vt should be equal to 0, shouldnt it?

In the chain rule we have f(x)= x-x0 and g(y)=y^2 so solving it we obtain that (x-xo)' should be 1. I dont understand why its not 0.

James

6. Nov 15, 2013

### Philip Wood

Don't understand why you're bringing in vt, especially since you seem to have both x and vt in this equation. Stick with just x, unless you're sure that x really does vary with time as x = vt.

$\frac{d}{dx}\frac{1}{2}k(x - x_0)^2 = \frac{1}{2}\frac{d}{du}u^2 \frac {du}{dx}$

in which $u = (x - x_0)$ so $\frac{du}{dx} = 1$

and $\frac{d}{du}u^2 = 2u$

So $\frac{d}{dx}\frac{1}{2}k(x - x_0)^2 =k(x - x_0)$.