Question

The force acting on a particle of mass $m$ moving along the x-axis is given by $F\left(x\right)=A{x}^2-Bx$. Which one of the following is the potential energy of the particle?

• A. $2Ax-B$
• B. $-\frac{X^2}{6}(2Ax-3B)$
• C. $A{x}^3-B{x}^2$
• D. Zero

Answer 1

The correct option is B $-\frac{X^2}{6}(2Ax-3B)$

The relation between the Force $F\left(x\right)$ acting on a particle of mass $m$ moving in a linear direction and the Potentional Energy $U\left(x\right)$ is given by ;

$F\left(x\right)$ = $-\frac{dU}{dx}$............(1)

Here, $F\left(x\right)$ = $A{x}^2-Bx$............(2),

From equation (1) and (2) and by integrating the equation (2) between the limits 0 and $x$. we get;

$U\left(x\right)$ = $-\frac{X^2}{6}(2Ax-3B)$