Question

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

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

Answer 1

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

$W=\int Fdx={\int }_{+a}^{+2a}\left(\frac{-k}{x^2}\right)dx={\left[\frac{k}{x}\right]}_{+a}^{+2a}$
$=-\frac{k}{2a}$
Note: It is important to note that work comes out to be negative which is quite obvious as the force acting on the particle is in negative x-direction $(F=-\frac{k}{x^2})$ while displacement is along positive x-direction. (from x=a to x=2a)