Question

A particle moves in a potential region given by $U = 8{x^2} - 4x + 400$. Its state of equilibrium will be
A. $x = 25{\text{ }}m$
B. $x = 0.25{\text{ }}m$ a
C. $x = 0.025{\text{ }}m$
D. $x = 2.5{\text{ }}m$

Answer 1

Hint:Equilibrium of a body is defined as the state of the body in which there is no any change in the motion of the body and neither any change of the internal energy with time. In this question, we have to differentiate potential with respect to $x$ and hence find force $F$. Then by putting force as zero as it is in equilibrium, we will find the answer.

Complete step by step answer:
In the given question the potential of the particle is given as $U = 8{x^2} - 4 x+400$.The first derivative of potential corresponds to Force. The state of equilibrium states that the condition in which the body does not move or it is not in any kind of motion and there isn’t any change of internal energy with time.Let the force be $F$.Thus, we get,
$\dfrac{{dU}}{{dx}} = F$
Differentiating the equation we get,
$U = 8{x^2} - 4x + 400 \\
\Rightarrow \dfrac{{dU}}{{dx}} = 16x - 4 $
Therefore, we get,
$F = 16x - 4$
In equilibrium condition $F = 0$,
$16x - 4 = 0$
$ \therefore x = 0.25\,m$

Therefore, the correct option is B.

Note:It must be noted that at the equilibrium condition of a body the force acting on it is zero as the body has no motion, which in turn means that there is no velocity, hence there is no acceleration. So, if acceleration is zero, the force also becomes zero. The first derivative of potential is force.