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Under the action of a force, a 2 kg body moves such that its position x as a function of time t is given by x = $\frac{t^{2}}{3}$ , where x is in meter and t in seconds. The work done by the force in the first two seconds is:

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160 J

1600 J

16 J

1.6 J

answer is B.

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Detailed Solution

v = $\frac{d x}{d t}$ = $\frac{d}{d t} (\frac{t^{3}}{3})$ = $t^{2}$

When t = 0, then v = 0, When t = 2, then v = 4 m/s

Work done in first two seconds = change in K.E.

W = $\frac{1}{2} m$ [ ${(4)}^{2} - {(0)}^{2}$ ] = $\frac{1}{2}$ 2 $\times$ 16 = 16 J

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