JEE MAIN - Physics (2004 - No. 41)
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
$$x$$
$${e^x}$$
$${x^2}$$
$${\log _e}x$$
Explanation
Given that, retardation $$ \propto $$ displacement
$$ \Rightarrow $$ $$a=-kx$$
But we know $$a = v{{dv} \over {dx}}\,\,\,\,\,\,\,\,\,$$
$$\therefore$$ $${{vdv} \over {dx}} = - kx $$
$$\Rightarrow \int\limits_{{v_1}}^{{v_2}} v \,dv = - k\int\limits_0^x {xdx} $$
$$\left( {v_2^2 - v_1^2} \right) = - k{{{x^2}} \over 2}$$
$$ \Rightarrow {1 \over 2}m\left( {v_2^2 - v_1^2} \right) = {1 \over 2}mk\left( {{{ - x^2} \over 2}} \right)$$
$$\therefore$$ Loss in kinetic energy is proportional to $${x^2}$$.
$$\therefore$$ $$\Delta K \propto {x^2}$$
$$ \Rightarrow $$ $$a=-kx$$
But we know $$a = v{{dv} \over {dx}}\,\,\,\,\,\,\,\,\,$$
$$\therefore$$ $${{vdv} \over {dx}} = - kx $$
$$\Rightarrow \int\limits_{{v_1}}^{{v_2}} v \,dv = - k\int\limits_0^x {xdx} $$
$$\left( {v_2^2 - v_1^2} \right) = - k{{{x^2}} \over 2}$$
$$ \Rightarrow {1 \over 2}m\left( {v_2^2 - v_1^2} \right) = {1 \over 2}mk\left( {{{ - x^2} \over 2}} \right)$$
$$\therefore$$ Loss in kinetic energy is proportional to $${x^2}$$.
$$\therefore$$ $$\Delta K \propto {x^2}$$
Comments (0)
