Question

A conservative force $\vec F = (6.0x - 12)\hat{i}\,N$ where, $x$ is in metres, acts on a particle moving along an X axis. The potential energy $U$ associated with this force is assigned a value of $27J$ at $x = 0$ .
(a) Write an expression for U as a function of $x$ ,with U in Joules and $x$ in metres.
(b) What is the maximum positive potential energy? At what (c) negative value and (d) positive value of $x$ is the potential energy equal to zero?

Answer 1

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 $F = - \vec \nabla U$