Question

A force $F=\frac{-k}{x^2}(x\ne 0,(k\left)\text{is a positive constant}\right),$ acts on a particle in $x$ direction. The work done by this force in displacing the particle from $x=+a$ to $x=+2a$ is

• A. $+\frac{k}{2a}$
• B. $\frac{-k}{a}$
• C. $\frac{k}{a}$
• D. $\frac{-k}{2a}$

Answer 1

The correct option is D $\frac{-k}{2a}$

Given variable force $\left(F\right)=\frac{-k}{x^2}(k>0)$
Workdone by variable force on the particle to displace it from the position $x=a$ to $x=2a$.
$W={\int }_{x_1}^{x_2}{F}_{x}dx$
$W={\int }_{+a}^{+2a}\frac{-k}{x^2}dx=\frac{k}{x}{|}_{+a}^{+2a}=\frac{-k}{2a}$