Interpreting this dU/dx as a force in the x direction makes no sense. mgy
is the potential energy associated with the VERTICAL gravitational force mg.
What you did would be like considering a mass in free fall, and treating the
parameter t in the same way as you are here treating the parameter x ==>
U = mgy = mg(.5gt^2)
F_t =dU/dt = mg^2t is nonsense! (Not even the dimension of force.)
From: LaMontagne, Bob
Sent: Wednesday, April 04, 2012 3:11 PM
To: Forum for Physics Educators
Subject: [Phys-l] Force and potential
Deep in the sugar shock of eating too many Easter "Peeps", I came up with
the following:
Consider a bowl formed from the lower half of a sphere. In 2D I can write
the equation for the bowl as x^2 + (y-1)^2 = 1.
This gives y = 1 - sqrt(1-x^2)
If I place a marble at the inside rim (x=1, y=1) of the bowl and release it,
the marble will oscillate back and forth, repeatedly coming back to the
release point.
The potential energy is given by U=mgh=mgy = mg[1 - sqrt(1-x^2)]
Using F_x = - dU/dx , then we have F_x = -x/sqrt(1-x^2) which implies that
F_x is infinitely strong when x = 1 (the release point).