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A force acts on a 30 gm particle in such a way that the position of the particle as a function of time is given by x = (3t – 4t² + t³), where x is in metres and t is in seconds. The work done during the first 4 seconds is

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$5.28 J$

$4.5 J$

$4.9 J$

$5.76 J$

answer is A.

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Detailed Solution

$x = 3 t - 4 t^{2} + t^{3}$ , $\frac{dx}{dt} = 3 - 8 t + 3 t^{2}$

$Velocity = 3 - 8 t + 3 t^{2}$
$At t = 0, u = 3 m / s; At t = 4, v = 19 m / s$
$Work done,$ $W = \frac{1}{2} m (v_{}^{2} - u_{}^{2})$ $= 528 x 10^{- 2} J$

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