Question 194: 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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