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

Solution:

$x = 3t^2 + 5$
$v = \frac{dx}{dt} $
$ v = 6t +0 $
at t = 0 v = 0
t = 5 sec v = 30 m/s
W.D. = $\Delta$KE
W.D. $ = \frac{1}{2 } mv^{2} - 0 = \frac{1}{2}\left(2\right) \left(30\right)^{2} = 900 J $

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