A force `F=-(K)/(x^(2))(xne0)` acts on a particle along the X-axis . Find the work done by the force in displacing the particale from x = +a to x = + 2a. Take K as a positive constant .

Explanation

Force acting on the particle , `F=-(K)/(x^(2))`
Total amount of work done to displace the particle from x = +a to x = +2a is ,
`W=intdW=int_(x=a)^(x=2a)Fbx`
`rArrW=int_(x=a)^(2a)(-K)/(x^(2))dx=-Kint_(x=a)^(x=2a)(1)/(x^(2))dx`
`=-Kint_(x=a)^(x=2a)x^(-2)dx=K[x^(-1)]_(x=a)^(x=2a)`
`=K[(1)/(2a)-(1)/(a)]=K[(-1)/(2a)]=(-K)/(2a)`