Question

A particle is moving on x-axis has potential energy $U=x^2-4x+16$. If particle is released from $x=0$, locate the position where acceleration of the particle is zero.

• A. $x=0$
• B. $x=2$
• C. $x=4$
• D. $x=-2$

Answer 1

The correct option is B $x=2$

$U=x^2-4x+16$
$U=(x-2{)}^2+12$
$U=12+(x-2{)}^2$
We know that at $U_{min}$, the force will be zero.
$U_{min}=12$ at $x=2$.

So, at $x=2$; the acceleration of the particle will be zero.