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A force acts on a 2 kg object, so that its position is given as a function of time as $x = 3 t^{2} + 5$ . What is the work done by this force in first 5 seconds?

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850 J

950 J

875 J

900 J

answer is B.

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Detailed Solution

$x = 3 t^{2} + 5$

$v = \frac{d x}{d t} = 6 t$

Work done is the change in kinetic energy

$W = \frac{1}{2} 2 (30^{2} - 0^{2}) = 900 J$

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