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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 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 .

Text Solution

Verified by Experts

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