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Under the action of a force, a 2 kg body...

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=(t^(3))/(3)`, where x is in metre and t in second. The work done by the force in the first two seconds is

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From work-energy theorem, `W = Delta KE , x = t^(3)//3 therefore "velocity v" = (dx)/(dt) = t^(2)`
At t = 0, `v_(i) = 0^(2) = 0` , At t = 2, `v_(f) = 2^(2)` = 4 m/s
work done `W = (1)/(2) m(v_(f)^(2)-v_(i)^(2)) = (1)/(2) xx 2(4^(2) - 0) = 16 J`
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