Question

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

Answer 1

Correct Answer - A
`becauseF= -(k)/(x^(2))`
`therefore" Acceleration,"f=-(k)/(mx^(2))`
When x decreases, v increases.
`therefore " "f=-v(dv)/(dx)`
`therefore " "-v(dv)/(dx)=-(k)/(mx^(2))`
`or " "int_(0)^(v)vdv=(k)/(m)int_(a)^(x)(1)/(x^(2))dx`
`therefore " "v=sqrt((2k)/(m))((x-a)/(ax))^(1//2)`