Question

A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement $ x $ is proportional to

• A. $ {{x}^{2}} $
• B. $ {{e}^{x}} $
• C. $ x $
• D. $ {{\log }_{e}}x $

Answer 1

Answer: (A)

From given information $ a=-\,kx, $
where a is acceleration, $ x $ is displacement and $ k $ is a proportionality constant.
$ \frac{v\,dv}{dx}=-\,k\,x $
$ \Rightarrow $ $ v\,dv=-\,k\,x\,dx $
Let for any displacement from 0 to $ x, $
the velocity changes from
$ {{v}_{0}} $ to v.
$ \Rightarrow $ $ \int_{{{v}_{0}}}^{v}{v\,dv=-\int_{0}^{x}{k\,x\,dx}} $
$ \Rightarrow $ $ \frac{{{v}^{2}}-v_{0}^{2}}{2}=-\frac{k\,{{x}^{2}}}{2} $
$ \Rightarrow $ $ m\left( \frac{{{v}^{2}}-v_{0}^{2}}{2} \right)=-\frac{mk\,{{x}^{2}}}{2} $
$ \Rightarrow $ $ \Delta K\propto {{x}^{2}} $ [ $ \Delta K $ is loss in KE]