# Qns on potential energy and forces

#### coffeebeans

Hello, i hope i posted this in the correct section..

I have a question:
the potential energy of a body when it is at point P a distance x from a reference point O is given by V = kx^2, where k is a constant. what is the force acting on the body when it is at P?

the correct ans is 2kx in the direction of PO

the ans i got is kx in the direction OP. as work = force x distance, so force required to bring the body to point p is kx^2 / x .

could some please explain why is the correct ans so? thnk u so much!!

Last edited:
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#### HallsofIvy

Homework Helper
Work is "force times distance" as long as the force is constant- if the force is a variable then the work is given by $\int f(x)dx$. Notice that this involves a "constant of integration". It is standard to choose that constant so that potential energy is 0 at some specific point; potential energy is always relative to some given point.

Going the other way, if work is a constant times distance, W= Cx then force is that constant: F= Cx/x= C. But if a more general function then $F= \frac{dW}{dx}$.

#### borisleprof

The potential energy is positive. So there is an attraction toward the origin.

The force = - (dW/dx) = -2kx.
So the force is directed from the point P to the point O, the origin.
bye

Thnx alot!!!!!

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