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Force acts on a 20g particle in such a way that the position of the particle as a function of time is given by $x = 2 t - 3 t^{2} + 2 t^{3}$ where x is in metre and t in second. The work done in first three seconds is

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14.4 J

5.28 J

7.2 J

28.8 J

answer is C.

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Detailed Solution

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

$V = 2 - 6 t + 6 t^{2}$

Initial velocity $V_{1} = 2 m / s$

velocity at t = 3 sec, $V_{2} = 38 m / s$

Using work energy theorem,

$W = Δk = \frac{1}{2} (\frac{20}{1000}) (38^{2} - 2^{2}) = 14.4 J$

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