# Deriving Acceleration from Potential Energy?

1. Aug 3, 2004

### cj

I have a known potential energy, V, expression:

V(x,y,z) = α·x + β·y2 + γ·z3

I'm given: @(0,0,0), v = v0 and then asked to find v at (1,1,1).

I can determine v from Conservation of Energy:

v2 = v02 - (2/m)·(α + β + γ)2

In general, what is the expression for the accelerations ax, ay, az?

Do I find F from -∇V?

If so, what's next (as far as finding the acceleration's x, y and z-components)?

Thanks!

2. Aug 3, 2004

### arildno

1.Why have you squared the potential energy term??

2. Yes, and divide F by m to find the accelerations.

3. Aug 3, 2004

### ZapperZ

Staff Emeritus
arildno sorta told you how to start it off. You should have F (and a) from the gradient of V. However, you will notice that "a" has a dependence on x, y, and z. If a is a function of t, then it is trivial to find v. But you don't have that here.

So what you need to do to find v is to use some calculus gymnastics by invoking the chain rule, i.e.

a = dv/dt = (dv_x/dx * dx/dt)i^ + (dv_y/dy * dy/dt)j^ + (dv_z/dz * dz/dt)k^

It is easier to solve this component by component, so for the x-component, you have

a_x = dv_x/dx * v_x (since dx/dt = v_x)

Thus, a_x dx = v_x dv_x

I think you should be able to handle the baby integral here using the initial conditions given. Do the same thing for the other 2 components.

Zz.

4. Aug 3, 2004

### cj

Thanks a lot Zz.

When solving, for example, az, I arrive at:

az = -(3γ/m)·z2

In general, is this a sufficient expression for az,
or should it be reduced or otherwise expressed differently?

5. Aug 3, 2004

### ZapperZ

Staff Emeritus
ASSUMING you did the gradient correctly, that should be a sufficient expression for the a_z to play with.

Zz.