Question

A body of mass $2\text{kg}$ is projected from origin with a velocity $3\text{m/s}$ along positive $x-$ axis on which a force $-x^2$ is acting. Find the maximum distance from origin upto which the body can go. ($x$ is the position of particle)

• A. $6\text{m}$
• B. $5\text{m}$
• C. $3\text{m}$
• D. $4\text{m}$

Answer 1

The correct option is C $3\text{m}$

Given:
Mass of the body, $m=2\text{kg}$
Initial velocity of the body, $v=3\text{m/s}$

The body will reach maximum distance when its velocity becomes zero for an instant.
The force acting on the body, $F=-x^2=ma$
$\Rightarrow 2a=-x^2$
$\Rightarrow v\frac{dv}{dx}=\frac{-x^2}{2}$
$\Rightarrow \underset{3}{\overset{0}{\int }}vdv=\underset{0}{\overset{x}{\int }}\frac{-x^2}{2}dx$
$\Rightarrow {\left[\frac{v^2}{2}\right]}_3^0=\frac{-1}{2}{\left[\frac{x^3}{3}\right]}_0^x$
$\Rightarrow [0-\frac{9}{2}]=\frac{-1}{6}[x^3-0]$
$\Rightarrow x^3=\frac{9\times 6}{2}\Rightarrow x=3\text{m}$

Hence, option (c) is correct.

$\begin{array}{c}\text{Why this question ?}\\ \text{Concept - At maximum distance from origin,}\\ \text{the body will come at instant rest.}\\ \text{Tip - Integrate the equation taking limit from}\\ v_0\text{to}v=0\text{and}x=0\text{to}{x}_0\end{array}$