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 = - Ax, where a is acceleration, x is displacement and k is a proportionality constant. $\frac{vdx}{dx}=-kx$ $\Rightarrow vdv=-kx dx$ Let for any displacement from 0 to x, the velocity changes from $v_0 \,to\, v$ $\Rightarrow \int ^2_{v_0}v dv =-\int ^x_0 kx dx$ $\Rightarrow \frac{v^2-v^2_0}{2}=-\frac{kx^2}{2}$ $\Rightarrow m\left(\frac{v^2-v^2_0}{2}\right)=-\frac{mk x^2}{2}$ $\Rightarrow \Delta K \propto x^2$ $(\Delta K \,is \,loss\,in\,KE)$