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Under the action of a force, a 2 kg body moves such that its position ‘x’ in meters as a function of time ‘t’ in seconds given by: $x = t^{2} / 2$ . The work done by the force in the first 5 seconds is

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25 J

250 J

0.25 J

2.5 J

answer is C.

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Detailed Solution

$W = \frac{1}{2} {mv}^{2} where v = \frac{dx}{dt}$

$x = \frac{t^{2}}{2}; \frac{dx}{dt} = t; when t = 0 u = 0, when t = 5 v = 5$

$W = \frac{1}{2} m (v^{2} - u^{2}) = 25 J$

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