Question

What is the relationship between force and potential energy?

Answer 1

Loosely speaking, work done by a force is stored as potential energy.

I lift a box up, the box gains potential energy.

I work to compress the spring, the spring gains potential energy.

Answer 2

Is there any equation to determine their relationship?

Answer 3

\[dW=Fdx\]

Then take the integral.

Answer 4

But , I found this equation online , can you explain to me,
F = -du/ds

Answer 5

Why is their a -sign in front?

Answer 6

The minus sign is because of how potential is defined.

Generally, we have a good way to describe the force. For hooke springs it's -kx, for electric charges it's -kqQ/r^2.

a way to look at this is: as you compress the spring, it gains potential energy,
\[du=Fdx = -(-kxdx)\] The minus sign makes it positive.

another way to look at this is that we have defined the force of how the system will react, but we don't have a good way to define how we are going to do work on the system. so it's better we let the force do the talking.

take the electric charge for example, between 2 positive charge. 2 positive charge will repel.
so as i bring the charge from infinity to a point, i do positive work: force and movement is in the same direction. But the electric force does negative work. it pushes the charge away but it still ends up moving closer. thus:
\[dW=F_{me}dx = -(\frac{-kQq}{x^{2}})dx\]

Answer 7

Kind of understand, can you help me with this question,
A body moving along x-axis experience a resistive force of magnitude kx. Deduce an equation for the potential energy of the body. Assume the P.E is zero when x=0.

Answer 8

to put it in a blunt way, the minus sign it to make the physics correct.

since it is a resisting force this force must be negative.
\[F=-kx\]

the resistive force does negative work on the object. how do i know this? well, it's resisting the motion, pushing it back to x=0 (described above) but the object is still going forward, so work done by this force is negative.

Thus the object loses kinetic energy and gains potential energy. at each small interval the relation is:
\[du=-(-kx)dx\]
and so the potential energy is:
\[\int_{0}^{U}du=\int_{0}^{x}kxdx\]\[U-0=\frac{1}{2}kx^{2}-\frac{1}{2}k0^{2}\]\[U=\frac{1}{2}kx^{2}\]

Answer 9

Thank

Answer 10

Force is considered the change in potential energy, U, over a change in position, x. In other words it is the derivative of potential energy with respect to time.

F =- dU/dx