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PhysicsMan
PhysicsMan
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The force on a particle can also be described in terms of a scalar potential function $U(x)$, it has nothing to do with coordinate frames.

$\begin{align} F(x)=-\frac{dU}{dx};\hspace{5mm} \vec{F}(\vec{r})=-\vec{\nabla}U(\vec{r}) \end{align}$

The potential energy $U$ means the amount of work needed to move an object from some point to another, the force applied needs to be equal but opposite, thats why there is a negative sign.

The force exerted by the force field always tends toward lower energy and will act to reduce the potential energy.

The negative sign on the derivative shows that if the potential U increases with increasing x, the force will tend to move it toward smaller x to decrease the potential energy.

The minus sign is there because we know forces want to minimize the potential energy.

The force on a particle can also be described in terms of a scalar potential function $U(x)$, it has nothing to do with coordinate frames.

$\begin{align} F(x)=-\frac{dU}{dx};\hspace{5mm} \vec{F}(\vec{r})=-\vec{\nabla}U(\vec{r}) \end{align}$

The force exerted by the force field always tends toward lower energy and will act to reduce the potential energy.

The negative sign on the derivative shows that if the potential U increases with increasing x, the force will tend to move it toward smaller x to decrease the potential energy.

The minus sign is there because we know forces want to minimize the potential energy.

The force on a particle can also be described in terms of a scalar potential function $U(x)$, it has nothing to do with coordinate frames.

$\begin{align} F(x)=-\frac{dU}{dx};\hspace{5mm} \vec{F}(\vec{r})=-\vec{\nabla}U(\vec{r}) \end{align}$

The potential energy $U$ means the amount of work needed to move an object from some point to another, the force applied needs to be equal but opposite, thats why there is a negative sign.

The force exerted by the force field always tends toward lower energy and will act to reduce the potential energy.

The negative sign on the derivative shows that if the potential U increases with increasing x, the force will tend to move it toward smaller x to decrease the potential energy.

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PhysicsMan
PhysicsMan
  • 178
  • 11

The force on a particle can also be described in terms of a scalar potential function $U(x)$, it has nothing to do with coordinate frames.

$\begin{align} F(x)=-\frac{dU}{dx} \end{align}$$\begin{align} F(x)=-\frac{dU}{dx};\hspace{5mm} \vec{F}(\vec{r})=-\vec{\nabla}U(\vec{r}) \end{align}$

The force exerted by the force field always tends toward lower energy and will act to reduce the potential energy.

The negative sign on the derivative shows that if the potential U increases with increasing x, the force will tend to move it toward smaller x to decrease the potential energy.

The minus sign is there because we know forces want to minimize the potential energy.

The force on a particle can also be described in terms of a potential function $U(x)$.

$\begin{align} F(x)=-\frac{dU}{dx} \end{align}$

The force exerted by the force field always tends toward lower energy and will act to reduce the potential energy.

The negative sign on the derivative shows that if the potential U increases with increasing x, the force will tend to move it toward smaller x to decrease the potential energy.

The force on a particle can also be described in terms of a scalar potential function $U(x)$, it has nothing to do with coordinate frames.

$\begin{align} F(x)=-\frac{dU}{dx};\hspace{5mm} \vec{F}(\vec{r})=-\vec{\nabla}U(\vec{r}) \end{align}$

The force exerted by the force field always tends toward lower energy and will act to reduce the potential energy.

The negative sign on the derivative shows that if the potential U increases with increasing x, the force will tend to move it toward smaller x to decrease the potential energy.

The minus sign is there because we know forces want to minimize the potential energy.

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PhysicsMan
PhysicsMan
  • 178
  • 11

The force on a particle can also be described in terms of a potential function $U(x)$.

$\begin{align} F(x)=-\frac{dU}{dx} \end{align}$

The force exerted by the force field always tends toward lower energy and will act to reduce the potential energy.

The negative sign on the derivative shows that if the potential U increases with increasing x, the force will tend to move it toward smaller x to decrease the potential energy.