A conservative force $\overrightarrow{\mathbf{F}}\mathbf{=}\left(6.0x-12\right)\hat{\mathbf{i}}\mathbf{N}$where xis in meters, acts on a particle moving along an xaxis. The potential energy Uassociated with this force is assigned a value of 27J at $\mathit{x}\mathbf{=}\mathbf{0}$. (a) Write an expression for Uas a function of x, with Uin joules and xin meters. (b) what is the maximum positive potential energy? At what (c) negative value (d) positive value of xis the potential energy equal to zero?

1. The conservative force acting on a particle is$\overrightarrow{F}=\left(6.0\mathrm{x}-12\right)\hat{\mathrm{i}}\mathrm{N}$
2. The potential energy of the particle at $x=0$is $U\left(0\right)=27\mathrm{J}$

The problem deals with the concept of gravitational potential energy. It is the energy possessed or acquired by an object due to a change in its position when it is present in a gravitational field.Use the concept of gravitational potential energy to find maximum potential energy. Find the first derivative of it and when it is equal to zero. To find the negative and positive value of xat which potential energy is equal to zero, solve quadratic equations.

Formulae are as follow:

1. $∆U=-{\int }_{xi}^{xf}F\left(x\right)dx$
2. $\frac{dU}{dx}=0$
3. $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

The change in potential energy is,

$∆U=U\left(x\right)-U\left(0\right)$ (i)

The expression for the gravitational potential energy is,

$$\begin{gathered} ∆U=-{\int }_{xi}^{xf}F\left(x\right)dx \\ ∆U=-{\int }_{xi}^{xf}\left(6.0x-12\right)dx \\ ∆U=12x-3.0x^2 \end{gathered}$$

From equation (i),

role="math" localid="1661398896247" $$\begin{gathered} U\left(x\right)-U\left(0\right)=12x-3.0x^2 \\ U\left(x\right)-27\mathrm{J}=12x-3.0x^2 \\ U\left(x\right)=27+12x-3.0x^2 \end{gathered}$$

Hence, an expression for U as a function of x is $U\left(x\right)=27+12x-3.0x^2$.

To find the maximum potential energy, find the first derivative of the potential energy and is equal to zero. Then,

$$\begin{gathered} \frac{dU}{dx}=0 \\ \frac{d\left(27+12x-3.0x^2\right)}{dx}=0 \\ 12-6.0x=0 \\ 12=6.0x \\ x=2.0\mathrm{m} \end{gathered}$$

The maximum potential energy is,

$$\begin{gathered} U=27+12\times \left(2.0\mathrm{m}\right)-3.0\times {\left(2.0\mathrm{m}\right)}^2 \\ U=39\mathrm{J} \end{gathered}$$

Hence, the maximum potential energy is $U=39\mathrm{J}$.

To find the negative and positive value of x , solve the quadratic equation.

$$\begin{gathered} 27+12x-3.0x^2=0 \\ 3.0x^2-12x-27=0 \end{gathered}$$

where, $a=3.0,b=-12$and $c=-27$

The quadratic equation is,

$$\begin{gathered} x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ x=\frac{-\left(-12\right)\pm \sqrt{{\left(-12\right)}^2-4\times 3.0\times \left(-27\right)}}{2\times 3.0} \\ x=-1.6\mathrm{m} \end{gathered}$$

or

$x=5.6\mathrm{m}$

Hence, the negative value of x at which the potential energy equal to zero is $x=-1.6\mathrm{m}$

The positive value of x at which the potential energy equal to zero is $x=5.6\mathrm{m}$.

Therefore, the gravitational potential energy can be found by using an expression of the gravitational potential energy.