Question

A particle of mass ${10}^{-2}$is moving along the positive x - axis under the influence of a force $F\left(x\right)=-\frac{1}{2x^2}.$. At time, t = 0, it is at x = 1.0 m and it is at rest. Find the velocity when it reaches x = 0.5m.

• A. $\frac{10m}{s}$
• B. $\sqrt[10]{102}\frac{m}{s}$
• C. $\sqrt[10]{103}\frac{m}{s}$
• D. $\frac{10}{\sqrt{3}}\frac{m}{s}$

Answer 1

The correct option is A $\frac{10m}{s}$

$F\left(x\right)=m\frac{dv}{dt}=mv\frac{dv}{dx}=-\frac{1}{2x^2}$
thus, $M{v}^2\frac{\left(x\right)}{2}={\int }_1^x(-\frac{1}{2x^2})=\frac{1}{2}[\frac{1}{x}-1]$
Thus, $v(v=\frac{1}{2}m)=\frac{10m}{s}$