A force acts on a 2 kg object so that its position is given as a function of time as x=3t2+5 , where x is in meter and t is in second, then the work done by this force in first 5 seconds is N×100 J then N = _______

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A force acts on a 2 kg object so that its position is given as a function of time as $x = 3 t^{2} + 5$ , where x is in meter and t is in second, then the work done by this force in first 5 seconds is $N \times 100 J$ then N = ________.

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answer is 9.

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Detailed Solution

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

$V (t = 0) = 0$

$V (t = 5) = 30 m / s$

$Δ k E = \frac{1}{2} \times 2 \times 30^{2} = 900 J$

$N \times 100 = 900 \Rightarrow N = 9$

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