The displacement of a particle moving in one dimension under a constant force is given by $t = \sqrt{x} + 3$ where x is in meters and t is seconds. The work done by the force in first 6 seconds is.

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36 J

6 J

12 J

0 J

answer is D.

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Detailed Solution

$t = \sqrt{x} + 3 \Rightarrow \sqrt{x} = t - 3 \Rightarrow x = t^{2} + 9 - 6 t v = \frac{d x}{d t} = 2 t - 6 a n d a = 2 t = 0, V_{1} = - 6, t = 6, V_{2} = 12 - 6 = 6$
$W = \frac{1}{2} m (V_{2}^{2} - V_{1}^{2}) = \frac{1}{2} \times m (36 - 36) = 0$

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