Respuesta :
Given potential energy as a function of time is
[tex]U = (3.00J)x + (1.00J/m^2)x^3[/tex]
now for all conservative forces we know that
[tex]F = - \frac{dU}{dx}[/tex]
now we will differentiate the potential energy with respect to x
so now we will have
[tex]F = -\frac{d(3x + 1x^3)}{dx}[/tex]
now we will have
[tex]F = - 3 - 3x^2[/tex]
so the force associated with the given potential energy is given as
[tex]F = -3(1 + x^2)[/tex]
The force function associated with the given potential energy function is [tex]\boxed{-3\left({1+{x^2}}\right)}[/tex] .
Further Explanation:
Given:
The potential energy function is [tex]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}[/tex] .
Concept:
The force function for a given potential function is expressed as the derivative of the potential energy function with respect to the position of the body.
The force function is represented mathematically as:
[tex]\boxed{F=-\frac{{dU}}{{dx}}}[/tex] …… (1)
The potential function for the system is given as:
[tex]U=3x+1{x^3}[/tex]
Substitute [tex]3x+1{x^3}[/tex] for [tex]U[/tex] in equation (1).
[tex]\begin{aligned}F&=-\frac{{d\left({3x+1{x^3}}\right)}}{{dx}}\\&=-3-3{x^2}\\&=-3\left({1+{x^2}}\right)\\\end{aligned}[/tex]
Thus, the force function associated with the given potential energy function is [tex]\boxed{-3\left({1+{x^2}}\right)}[/tex] .
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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.
