Question

The velocity (v) of a particle (under a force F) depends on its distance (x) from the origin (with x > 0) v `oo 1/sqrt(x)`. Find how the magnitude of the force (F) on the particle depends on x.

Options

• `F  oo  1/x^(3/2)`

• `F  oo  1/x`

• `F  oo  1/x^2`

• `F  oo  x`

Answer 1

`F  oo  1/x^2`

Explanation:

Given: `v  oo  1/sqrt(x)`

∴ `(dv)/(dx) = 1/(2x^(3/2))`

Dividing throughout by dt, we get,

`((dv)/(dt))/((dx)/(dt)) = - 1/(2x^(3/2))`

∴ `(dv)/(dt) = 1/(2x^(3/2)) (dx)/(dt)`

But v = `(dx)/(dt) = k 1/(x^(1/2))`

∴ `(dv)/(dt)  oo  1/(2x^(3/2)) xx 1/(x^(1/2))`  .....[Considering constant of proportionality as (–1)]

∴ `(dv)/(dt)  oo  1/x^2`

∴ `F  oo  1/x^2`  ......`[F = ma = m  (dv)/(dt)]`