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.
$\frac{vdv}{dx}=-kx$
$\Rightarrow$ $vdv=-kxdx$
Let for any displacement from 0 to $x,$
the velocity changes from
$v_0$ to v.
$\Rightarrow$ ${\int }_{v_0}^{v}vdv=-{\int }_0^{x}kxdx$
$\Rightarrow$ $\frac{v^2-v_0^2}{2}=-\frac{k{x}^2}{2}$
$\Rightarrow$ $m\left(\frac{v^2-v_0^2}{2}\right)=-\frac{mk{x}^2}{2}$
$\Rightarrow$ $\Delta K\propto x^2$ [ $\Delta K$ is loss in KE]