Q.

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 = t33, 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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a

1.6 J

b

16 J

c

160 J

d

1600 J

answer is B.

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Detailed Solution

v = dxdt = ddt(t33) = t2When t = 0, then v = o, When t = 2, then v = 4 m/sWork done in first two seconds = change in KEW = 12m[(4)2-(0)2] = 122×16 = 16
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