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)

${{x}^{2}}$

Answer 2

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]