A particle of mass m moves on the x-axis under the influence of a force of attraction towards the origin O given by `F=-(k)/(x^(2))hat(i)`. If the particle starts from rest at x = a. The speed of it will attain to reach at distance x from the origin O will be

Options

A. `sqrt((2k)/(m))[(a-x)/(ax)]^(1//2)`
B. `sqrt((2k)/(m))[(a+x)/(ax)]^(1//2)`
C. `sqrt(k/m)[(ax)/(a-x)]`
D. `sqrt((m)/(2k))[(a-x)/(ax)]^(1//2)`

Answer

A

Explanation

`F= -(k)/(x^(2))`
`m (vdv)/(dx)=-(k)/(x^(2))`
`vdv=-k/m (dx)/(x^(2))`
`rArr int_(0)^(v)vdv=-k/m int_(a)^(x)(dx)/(x^(2))`
`rArr (v^(2))/(2)=[+ k/m 1/x]_(a)^(x)=k/m (1/x-1/a)`
`rArr v= sqrt((2k)/(m)((a-x))/(ax))`.
Hence (A) is correct.