Question

A potential energy function is given by U(x)=(3.00J)x+(1.00J/m2)x3. What is the force function F(x) (in newtons) that is associated with this potential energy function?

Answer 1

The function of the force associated with the given energy function is $\boxed{ - 3 - 3{x^2}}$ .

Further Explanation:

Given:

The potential energy function for the system is  $U=\left( {3.00\,{\text{J}}} \right)x + \left( {1.00\,{{\text{J}} \mathord{\left/{\vphantom {{\text{J}} {{{\text{m}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace} {{{\text{m}}^{\text{2}}}}}} \right){x^3}$.

Concept:

From the work-energy theorem, the force function for a given potential function can be expressed as the first derivative of the potential energy function with respect to the position of the body.

The force function is represented mathematically as:

$\boxed{F= -\dfrac{{dU}}{{dx}}}$              ......(1)

The potential energy function for the system is given as:

$U = 3x + 1{x^3}$

Substitute $3x + 1{x^3}$ for $U$ in equation (1).

$\begin{aligned}F&=-\frac{{d\left({3x + 1{x^3}}\right)}}{{dx}}\\&=-3- 3{x^2}\\\end{aligned}$

Thus, the function of the force associated with the given energy function is  $\boxed{ - 3 - 3{x^2}}$.

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Answer Details:

Grade: College

Subject: Physics

Chapter: Work and Force

Keywords:  Potential energy function, force function, associated with, $u(x)=(3.00j)x+(1.00j/m^2)x^3, F=-dU/dx$, derivative of potential function.

Answer 2

Answer:

$F(x)=-3 N - (3N)x^2$

Explanation:

The force is defined as the negative of the derivative of the potential energy:

$F=-\frac{dU}{dx}$

If we use the potential energy function given in this problem:

$U(x)=3.00 x + 1.00 x^3$

and we calculate the force, we get:

$F(x)=-\frac{d}{dx}(3x+x^3)=-3-3x^2$

So, the force is

$F(x)=-3 N - (3N)x^2$