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A particle of mass 0.1 kg is subjected to a force which varies with distance as shown in figure. It starts its journey from rest at x = 0 , then its velocity at x =12m is

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$20 \sqrt{3} m / s$

40 m/s

$20 \sqrt{2} m / s$

answer is D.

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Detailed Solution

W=area of curve $= \frac{1}{2} \times 4 \times 10 + 4 \times 10 + \frac{1}{2} \times 4 \times 10 = 80 \Rightarrow \frac{1}{2} {mv}^{2} = area = 80$
$\Rightarrow v^{2} = \frac{80 \times 2}{0 .1} = 1600 \Rightarrow v = 40 m / s$

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