Question 204: Work, Energy and Power & Circular Motion

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

Accepted Answer

Given that, retardation $ \propto $ displacement
$ \Rightarrow $ $a=-kx$
But we know $a = v{{dv} \over {dx}}\,\,\,\,\,\,\,\,\,$
$	herefore$ ${{vdv} \over {dx}} =  - kx $
$\Rightarrow \int\limits_{{v_1}}^{{v_2}} v \,dv =  - k\int\limits_0^x {xdx} $
$\left( {v_2^2 - v_1^2} ight) =  - k{{{x^2}} \over 2}$
$ \Rightarrow {1 \over 2}m\left( {v_2^2 - v_1^2} ight) = {1 \over 2}mk\left( {{{ - x^2} \over 2}} ight)$
$	herefore$ Loss in kinetic energy is proportional to ${x^2}$.
$	herefore$ $\Delta K \propto {x^2}$

Answer

$x$

Answer

${e^x}$

Answer

${x^2}$

Answer

${\log _e}x$

Work, Energy and Power & Circular Motion

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

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