Respuesta :
Given,
Potential energy, U(x)=3x-x^{3}
Force is given by, F=dU/dx
F= d(3x-3x^{3})/dx
=3-3x^{2}
For F=0 N,
3-3x^{2} =0
3x^{2} =3
x=√1
Therefore, at x=+1 and x= -1, the force for the given potential energy is zero.
The force function associated with the given potential energy function is zero at [tex]\boxed{x=+ 1\,\&\,-1}[/tex] .
Further Explanation:
Given:
The function of potential energy for the body is [tex]U=\left({3.00\,{\text{N}}}\right)x+\left({1.00\,{{\text{N}}\mathord{\left/{\vphantom{{\text{N}}{{{\text{m}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{m}}^{\text{2}}}}}}\right){x^3}[/tex] .
Concept:
The function for force of the given potential function can be given as the first derivative of the energy function with respect to the distance [tex]x[/tex] covered by the body.
The function for the force 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]
Since the force of the body is zero, substitute [tex]0[/tex] for [tex]F[/tex] in above expression.
[tex]\begin{aligned}0&=-3\left({1+{x^2}}\right)\\x&=\sqrt 1\\&=\pm 1\\\end{aligned}[/tex]
Thus, the force function associated with the given potential energy function is zero at [tex]\boxed{x=+1\,\&\,-1}[/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, force is zero, position.
