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**Hint:**In physics, when the force acts on a body and it causes a displacement in the body , it’s called body has done some work and when this work done by a force is independent of path taken, then it’s called conservative force and this conservative force is related with potential energy as $F = - \dfrac{{\partial U}}{{\partial x}}$.

**Complete step by step answer:**

(a) It’s given us that $U(0) = 27J$ and we can write $F = - \dfrac{{\partial U}}{{\partial x}}$ as

\[ - U = \int\limits_0^x {(6x - 12)dx} \]

$\Rightarrow - U = (3{x^2} - 12x)$

$\therefore U = 27 + 12x - 3{x^2}$

Hence, the function of U is $U(x) = 27 + 12x - 3{x^2}$

(b) To find maximum potential energy its derivative must be zero which is the given force, hence

\[(6.0x - 12) = 0\]

$x = 2$

Hence finding $U(x) = 27 + 12x - 3{x^2}$ at $x = 2$

$U(2) = 27 + 24 - 12$

${U_{\max }}(2) = 39Joule$

Hence, maximum potential energy is ${U_{\max }}(2) = 39\,Joule$

(c) Equating this equation $U(x) = 27 + 12x - 3{x^2}$ to zero we get,

$27 + 12x - 3{x^2} = 0$

Or Taking common factor we can write

$(x + 1.6)(x - 5.6) = 0$

Equate both factors to zero we get,

$\therefore x = - 1.6\,m$

(d) From part (c) we get, $(x + 1.6)(x - 5.6) = 0$

$x = 5.6$

Hence, $x = 5.6$ is the positive value of $x$ at which potential energy is zero.

**Note:**It should be remembered that, partial derivative of potential energy is taken because force is a vector quantity and its derivative has to be taken in every component’s direction. The relation between force and potential energy can also be written in form of gradient at $F = - \vec \nabla U$

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