Question

Part complete a potential energy function is given by u(x)=(3.00n)x−(1.00n/m2)x3. at what position or positions is the force equal to zero?

Answer 1

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.

Answer 2

The force function associated with the given potential energy function is zero at $\boxed{x=+ 1\,\&\,-1}$ .

Further Explanation:

Given:

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

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 $x$  covered by the body.

The function for the force is represented mathematically as:

$\boxed{F=-\frac{{dU}}{{dx}}}$                                          …… (1)

The potential function for the system is given as:

$U=3x+1{x^3}$  

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

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

Since the force of the body is zero, substitute $0$  for $F$  in above expression.

$\begin{aligned}0&=-3\left({1+{x^2}}\right)\\x&=\sqrt 1\\&=\pm 1\\\end{aligned}$

Thus, the force function associated with the given potential energy function is zero at $\boxed{x=+1\,\&\,-1}$ .

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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.