Forces from Potential Energy Functions using Calculus: Videos & Practice Problems

In Forces from Potential Energy Functions using Calculus, the key idea is that a force can be found directly from a potential energy function by taking the negative derivative with respect to the changing position variable: \(F(x)=-\frac{dU}{dx}\) . If the motion is along a different coordinate, use that variable instead, such as \(F_y=-\frac{dU}{dy}\) .

This relationship reproduces familiar results like Hooke’s law from elastic potential energy and the weight force from gravitational potential energy. The sign of the derivative is especially important: a positive force points in the positive coordinate direction, while a negative force points in the negative direction. For a general U(x), careful differentiation, including power rules and the chain rule when needed, gives the full force function and helps determine whether an interaction is attractive or repulsive from the sign of F(x).

Calculating Forces from Potential Energy Functions

When analyzing forces derived from potential energy functions, the fundamental relationship is that the force is the negative derivative of the potential energy with respect to position. Mathematically, this is expressed as \(F(x) = -\frac{dU}{dx}\), where \(U(x)\) is the potential energy function. This principle applies universally, whether dealing with elastic potential energy in springs or gravitational potential energy.

For example, the spring force follows Hooke's law, which can be derived from the elastic potential energy \(U(x) = \frac{1}{2}kx^2\). Taking the derivative and applying the negative sign yields \(F(x) = -\frac{d}{dx}\left(\frac{1}{2}kx^2\right) = -kx\). Similarly, the gravitational force can be found by differentiating the gravitational potential energy \(U(y) = mgy\) with respect to the vertical position \(y\), resulting in \(F(y) = -\frac{d}{dy}(mgy) = -mg\), which corresponds to the weight force acting downward.

In cases involving more complex potential energy functions, the same rule applies. Consider a potential energy function defined as \(U(x) = -2x^2 + 0.3x^3\), where constants \(A=2\) and \(B=0.3\) are given. To find the force at a specific position, such as \(x=4\(, first compute the derivative:

\[\frac{dU}{dx} = \frac{d}{dx}(-2x^2 + 0.3x^3) = -4x + 0.9x^2\]

Then, apply the negative sign to find the force function:

\[F(x) = -\frac{dU}{dx} = 4x - 0.9x^2\]

Evaluating this at \)x=4\( gives:

\[F(4) = 4(4) - 0.9(4)^2 = 16 - 14.4 = 1.6 \text{ N}\]

The positive value indicates the force points in the positive \)x\)-direction with a magnitude of 1.6 newtons. This approach highlights the importance of carefully applying the negative derivative rule and correctly handling constants and exponents when differentiating potential energy functions to determine forces.

A particle moves along the $x$-axis. the potential energy function is given by $U\left(x\right)=e^{-x^2}$. What is the force acting on this particle?

• A. $-2x{e}^{-x^2}$
• B. $x^2{e}^{-x^2}$
• C. $2x{e}^{-x^2}$
• D. $-x^2{e}^{-x^2}$

Answer: (C)

Calculating Forces from Potential Energy Functions Example 1

When analyzing the interaction between two atoms separated by a distance x, the potential energy function can be expressed as U(x) = -A / x⁶, where A is a positive constant. To determine the force that one atom exerts on the other, we use the fundamental relationship between force and potential energy: the force is the negative derivative of the potential energy with respect to position. Mathematically, this is represented as F(x) = -dU/dx.

To differentiate U(x) = -A x⁻⁶, apply the power rule for derivatives. Taking the derivative yields:

\[\frac{dU}{dx} = -A \cdot (-6) x^{-7} = 6A x^{-7}\]

Applying the negative sign from the force equation, the force becomes:

\[F(x) = -\frac{dU}{dx} = -6A x^{-7} = -\frac{6A}{x^{7}}\]

This expression shows that the force magnitude depends inversely on the seventh power of the distance between the atoms, scaled by the positive constant A. The negative sign indicates the direction of the force.

Interpreting the sign of the force is crucial to understanding the nature of the interaction. Since A is positive and x⁷ is always positive for positive distances, the force F(x) is negative. This negative force implies that the force vector points toward decreasing x, meaning one atom pulls the other closer rather than pushing it away. Therefore, the force is attractive.

In summary, the force derived from the potential energy function U(x) = -A / x⁶ is given by F(x) = -6A / x⁷, and this force is inherently attractive, drawing the atoms together as they interact.

Here's what students ask on this topic:

How do you find the force from a potential energy function using calculus?

To find the force from a potential energy function, you use the relationship $F=-\frac{\mathrm{dU}}{\mathrm{dx}}$, where $U$ is the potential energy and $x$ is the position variable. This means you take the negative derivative of the potential energy function with respect to the coordinate that changes. For example, if $U$ depends on $x$, then $F$ is the negative derivative of $U$ with respect to $x$. This method applies to any potential energy function and allows you to calculate the force acting on a particle at any position.

What is the force derived from the elastic potential energy of a spring?

The force derived from the elastic potential energy of a spring is given by Hooke's law. The elastic potential energy is $U=\frac{1}{2}\mathrm{kx}{x}^2$, where $k$ is the spring constant and $x$ is the displacement. Taking the negative derivative, the force is $F=-\frac{d}{\mathrm{dx}}\frac{1}{2}\mathrm{kx}{x}^2=-\mathrm{kx}$. This shows the force is proportional to the displacement and acts in the opposite direction, restoring the spring to equilibrium.

How do you calculate the force from gravitational potential energy?

Gravitational potential energy near Earth's surface is $U=\mathrm{mgy}$, where $m$ is mass, $g$ is gravitational acceleration, and $y$ is height. To find the force, take the negative derivative with respect to $y$: $F=-\frac{d}{\mathrm{dy}}\mathrm{mgy}=-\mathrm{mg}$. This force points downward, consistent with the weight of the object.

What is the significance of the negative sign in the force and potential energy relationship?

The negative sign in the relationship $F=-\frac{\mathrm{dU}}{\mathrm{dx}}$ indicates that the force acts in the direction of decreasing potential energy. This means the force tends to move the particle toward positions where the potential energy is lower, which is a stable equilibrium. Without the negative sign, the force would point toward increasing potential energy, which is physically incorrect. Thus, the negative sign ensures the force is a restoring force that opposes displacement from equilibrium.

How do you find the magnitude and direction of force from a general potential energy function?

Given a general potential energy function $U(x)$, the force is found by taking the negative derivative: $F(x)\; =-\frac{\mathrm{dU}}{\mathrm{dx}}$. To find the magnitude and direction at a specific position, first compute the derivative, then substitute the position value. The magnitude is the absolute value of $F$, and the sign indicates direction: positive means force points in the positive $x$ direction, negative means it points in the negative direction. For example, if $U(x)\; =\; -2x^2\; +\; 0.3x^3$, then $F(x)\; =\; 4x\; -\; 0.9x^2$. Evaluating at $x=4$ gives $F(4)\; =\; 1.6$, so the force magnitude is 1.6 N pointing in the positive $x$ direction.