The position (x) of a particle of mass 1 kg moving along x-axis at time t is given by `(x=(1)/(2)t^(2))` metre. Find the work done by force acting on it in time interval from t=0 to t=3 s.

Explanation

`x = 1/2 t^(2) Rightarrow v = (dx)/(dt) = 1/2(2t) = t`
`therefore At t = 0, v_(i) = 0 `
At ` t = 3s, v_(f) = 3ms^(-1)`
According to W-E theorem, `W = Delta K = K_(f) - K_(i) = 1/2mv_(f)^(2) - 1/2(mv_(i)^(2)`
` = 1/2 xx 1 xx 3^(2) = 4.5 J`