Question

A particle experiences a position-dependent force given by
$F\left(x\right)=-6x^2+4x+3/x^2$
where $x$ is in meters and $F$ is in Newtons (units have been abbreviated).
At $x=1m$, what is the potential energy of the particle relative to the potential energy at the origin?

• A. $+5J$
• B. $+3J$
• C. $-3J$
• D. $-5J$
• E. Cannot be determined

Answer 1

The correct option is C $-3J$

The force acting on a particle $F\left(x\right)=-6x^2+4x+\frac{3}{x^2}$ N

Potential energy of the particle $U\left(x\right)=\int F\left(x\right).dx$

$\therefore$ $U\left(x\right)=\int (-6x^2+4x+\frac{3}{x^2})dx=-2x^3+2x^2-\frac{3}{x}$ J

$⟹$ $U{|}_{x=1}=-2(1{)}^3+2(1{)}^2-3=-3$ J